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  • Most efficient way to check if a date falls between two dates?

    - by Dave Jarvis
    Given: Start Month & Start Day End Month & End Day Any Year What SQL statement results in TRUE if a date lands between the Start and End days? 1st example: Start Date = 11-22 End Date = 01-17 Year = 2009 Specific Date = 2010-01-14 TRUE 2nd example: Start Date = 11-22 End Date = 11-16 Year = 2009 Specific Date = 2010-11-20 FALSE 3rd example: Start Date = 02-25 End Date = 03-19 Year = 2004 Specific Date = 2004-02-29 TRUE I was thinking of using the MySQL functions datediff and sign plus a CASE condition to determine whether the year wraps, but it seems rather expensive. Am looking for a simple, efficient calculation. Update The problem is the end date cannot simply use the year. The year must be increased if the end month/day combination happens before the start date. The start date is easy: Start Date = date( concat_ws( '-', year, Start Month, Start Day ) The end date is not so simple. Thank you!

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  • Range partition skip check

    - by user289429
    We have large amount of data partitioned on year value using range partition in oracle. We have used range partition but each partition contains data only for one year. When we write a query targeting a specific year, oracle fetches the information from that partition but still checks if the year is what we have specified. Since this year column is not part of the index it fetches the year from table and compares it. We have seen that any time the query goes to fetch table data it is getting too slow. Can we somehow avoid oracle comparing the year values since we for sure know that the partition contains information for only one year.

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  • linq multiple order DESCENDING

    - by ile
    .OrderBy(y => y.Year).ThenBy(m => m.Month); How to set descending order? EDIT: I tried this: var result = (from dn in db.DealNotes where dn.DealID == dealID group dn by new { month = dn.Date.Month, year = dn.Date.Year } into date orderby date.Key.year descending orderby date.Key.month descending select new DealNoteDateListView { DisplayDate = date.Key.month + "-" + date.Key.year, Month = date.Key.month, Year = date.Key.year, Count = date.Count() }) //.OrderBy(y => y.Year).ThenBy(m => m.Month) ; And it seems working. Is it wrong to use orderby twice like I used it here?

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  • Stupid newbie c++ two-dimensional array problem.

    - by paulson scott
    I've no idea if this is too newbie or generic for stackoverlflow. Apologies if that's the case, I don't intend to waste time. I've just started working through C++ Primer Plus and I've hit a little stump. This is probably super laughable but: for (int year = 0; year < YEARS; year++) { cout << "Year " << year + 1 << ":"; for (int month = 0; month < MONTHS; month++) { absoluteTotal = (yearlyTotal[year][year] += sales[year][month]); } cout << yearlyTotal[year][year] << endl; } cout << "The total number of books sold over a period of " << YEARS << " years is: " << absoluteTotal << endl; I wish to display the total of all 3 years. The rest of the code works fine: input is fine, individual yearly output is fine but I just can't get 3 years added together for one final total. I did have the total working at one point but I didn't have the individual totals working. I messed with it and reversed the situation. I've been messing with it for God knows how long. Any idea guys? Sorry if this isn't the place!

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  • bubble sort on array of c structures not sorting properly

    - by xmpirate
    I have the following program for books record and I want to sort the records on name of book. the code isn't showing any error but it's not sorting all the records. #include "stdio.h" #include "string.h" #define SIZE 5 struct books{ //define struct char name[100],author[100]; int year,copies; }; struct books book1[SIZE],book2[SIZE],*pointer; //define struct vars void sort(struct books *,int); //define sort func main() { int i; char c; for(i=0;i<SIZE;i++) //scanning values { gets(book1[i].name); gets(book1[i].author); scanf("%d%d",&book1[i].year,&book1[i].copies); while((c = getchar()) != '\n' && c != EOF); } pointer=book1; sort(pointer,SIZE); //sort call i=0; //printing values while(i<SIZE) { printf("##########################################################################\n"); printf("Book: %s\nAuthor: %s\nYear of Publication: %d\nNo of Copies: %d\n",book1[i].name,book1[i].author,book1[i].year,book1[i].copies); printf("##########################################################################\n"); i++; } } void sort(struct books *pointer,int n) { int i,j,sorted=0; struct books temp; for(i=0;(i<n-1)&&(sorted==0);i++) //bubble sort on the book name { sorted=1; for(j=0;j<n-i-1;j++) { if(strcmp((*pointer).name,(*(pointer+1)).name)>0) { //copy to temp val strcpy(temp.name,(*pointer).name); strcpy(temp.author,(*pointer).author); temp.year=(*pointer).year; temp.copies=(*pointer).copies; //copy next val strcpy((*pointer).name,(*(pointer+1)).name); strcpy((*pointer).author,(*(pointer+1)).author); (*pointer).year=(*(pointer+1)).year; (*pointer).copies=(*(pointer+1)).copies; //copy back temp val strcpy((*(pointer+1)).name,temp.name); strcpy((*(pointer+1)).author,temp.author); (*(pointer+1)).year=temp.year; (*(pointer+1)).copies=temp.copies; sorted=0; } *pointer++; } } } My Imput The C Programming Language X Y Z 1934 56 Inferno Dan Brown 1993 453 harry Potter and the soccers stone J K Rowling 2012 150 Ruby On Rails jim aurther nil 2004 130 Learn Python Easy Way gmaps4rails 1967 100 And the output ########################################################################## Book: Inferno Author: Dan Brown Year of Publication: 1993 No of Copies: 453 ########################################################################## ########################################################################## Book: The C Programming Language Author: X Y Z Year of Publication: 1934 No of Copies: 56 ########################################################################## ########################################################################## Book: Ruby On Rails Author: jim aurther nil Year of Publication: 2004 No of Copies: 130 ########################################################################## ########################################################################## Book: Learn Python Easy Way Author: gmaps4rails Year of Publication: 1967 No of Copies: 100 ########################################################################## ########################################################################## Book: Author: Year of Publication: 0 No of Copies: 0 ########################################################################## We can see the above sorting is wrong? What I'm I doing wrong?

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  • Announcing Fusion Middleware Innovation Awards

    - by Michelle Kimihira
    Author: Neela Chaudhari Every year at OpenWorld, Oracle announces the winners to its most prestigious awards in Middleware, the Fusion Middleware Innovation Awards. This year, we’ll be announcing the winners and highlighting a few of their original implementations during this key session in the Middleware stream: 11:45 AM on Tuesday, October 2nd, CON9162 Oracle Fusion Middleware: Meet This Year's Most Impressive Customer Projects in Moscone West, 3001. In addition, we’ll give a sneak peak of a few winners during GEN9394: Fusion Middleware General Session with Hasan Rizvi at 10:15 AM on Tuesday, October 2nd in Moscone West, Hall D! What kinds of customers win the Fusion Middleware Innovation Awards? Winners are selected based on the uniqueness of their business case, business benefits, level of impact relative to the size of the organization, complexity and magnitude of implementation, and the originality of architecture. The winners are selected by a panel of judges that score each entry across multiple different scoring categories. This year, the following categories included: Oracle Exalogic Cloud Application Foundation Service Integration (SOA) and BPM WebCenter Identity Management Data Integration Application Development Framework and Fusion Development Business Analytics (BI, EPM and Exalytics) Last year at OpenWorld 2011 we had standing room only in our session, so come early!  We had over 30 innovative customers that won the award, including companies like BT, Choice Hotels, Electronic Arts, Clorox Company, ING, Dunkin Brands, Telenor, Haier, AT&T, Manpower, Herbal Life and many others. Did you miss your chance this year to nominate your company? Come join with us in the awards session to get an edge in your next year’s submission and watch for the next opportunity for 2013 on this blog. There’s other awards as part of Oracle’s Excellence awards program or subscribe to our regular Fusion Middleware newsletter to get the latest reminders.

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  • Top Partners 2010 Specialization Awards

    - by Paulo Folgado
    Portugal Top Partners 2010 Specialization Awards Be Recognized Caro Parceiro, Vão ter lugar mais uma vez os prémios Top Partners, que anualmente visam reconhecer as realizações dos parceiros Oracle em termos de negócio. Este ano, sob o signo de Partner Specialization Awards, pretendemos, a par com os resultados de negócio, premiar igualmente o esforço dos nossos parceiros em termos de especialização e de desenvolvimento da sua auto-suficiência. Outras das inovações deste ano é o processo ser baseado numa entrega de candidaturas dos parceiros, e de estas serem avaliadas, com base em critérios definidos, por um júri incluindo entidades externas. Categorias dos Prémios: ·       Database Partner of the Year ·       Middleware Partner of the Year ·       Applications Partner of the Year ·       ISV Partner of the Year ·       Midsize Partner of the Year ·       Industry Partner of the Year ·       Accelerate Partner of the Year   Um dos objectivos dos Top Partners 2010 Specialization Awards é destacar e incentivar a cooperação entre a Oracle e os seus parceiros. Por esta razão, para além dos vários critérios específicos, os parceiros têm de cumprir um conjunto de critérios gerais: ·       Cooperação com a Oracle ·       Investimento no desenvolvimento dos conhecimentos em Oracle ·       Aproximação conjunta ao mercado ·       Utilização activa dos recursos e ferramentas do Oracle PartnerNetwork através do Portal OPN   Os Top Partners 2010 Specialization Awards estão abertos a todos os parceiros em Portugal que estejam registados no OPN Specialized em uma ou mais especializações. As candidaturas podem ser enviadas até 31 de Maio de 2010. Be recognized! Para mais informação clique aqui

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  • 25 Favorite JCP Award Memories

    - by heathervc
    As we celebrated the 10th Annual JCP Awards and Party at JavaOne last week, we asked attendees to share their favorite memories.  Add yours to the retrospective list below... The 10th Award party will be the best :-) I won a DSLR camera at the 2011 JCP party and have taken many awesome photos of my family with it ever since!  Thanks JCP! Remembering the password to get in! It was very fascinating talking to all those JUG Members of last years' (2011) party and hearing about their hopes & expectations.  Especially from members of SouJava and LJC. Hanging out with my friends Best food and one of my colleagues won the raffle prize. My friend Brian won a jacket 3 years ago and my friend Craig won a camera last year. 2010 when I took home 2 awards on behalf of JSRs I'm on. When Patrick & Scott sang 'Light My Fire'! Catch up with friends! Being able to attend my first JCP party and and joining JCP community. Of course it's when some people won the award (SouJava and LJC)!   Meeting Crazy Bob! This is my first. Mike  to be JCP Member of the Year in 2011. When SouJava and London Java Community won Member of the Year award! JBoss making CDI Everything! When SouJava won the JCP Member of the Year award. I love feeling like it is the Oscars! First Party! Winning JCP Member of the Year last year. The year I was running for it (JCP Award). 2009 music and hostess. Obscured on legal advice.

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  • The Start of a Blog

    - by dbradley
    So, here's my new blog up and running, who am I and what am I planning to write here?First off - here's a little about me:I'm a recent graduate from university (coming up to a year ago since I finished) studying Software Engineering on a four year course where the third year was an industrial placement. During the industrial placement I went to work for a company called Adfero in a "Technical Consultant" role as well as a junior "Information Systems Developer". Once I completed my placement I went back to complete my final year but also continued in my developer role 2/3 days a week with the company.Working part time while at uni always seems like a great idea until you get half way through the year. For me the problem was not so much having a lack of time, but rather a lack of interest in the course content having got a chance at working on real projects in a live environment. Most people who have been graduated a little while also find this - when looking back at uni work, it seem to be much more trivial from a problem solving point of view which I found to be true and I found key to uni work to actually be your ability to prove though how you talk about something that you comprehensively understand the basics.After completing uni I then returned full time to Adfero purely in the developer role which is where I've now been for almost a year and have now also taken on the title of "Information Systems Architect" where I'm working on some of the more high level design problems within the products.What I'm wanting to share on this blog is some of the interesting things I've learnt myself over the last year, the things they don't teach you in uni and pretty much anything else I find interesting! My personal favorite areas are text indexing, search and particularly good software engineering design - good design combined with good code makes the first step towards a well-written, maintainable piece of software.Hopefully I'll also be able to share a few of the products I've worked on, the mistake I've made and the software problems I've inherited from previous developers and had to heavily re-factor.

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  • Working with Reporting Services Filters–Part 1

    - by smisner
    There are two ways that you can filter data in Reporting Services. The first way, which usually provides a faster performance, is to use query parameters to apply a filter using the WHERE clause in a SQL statement. In that case, the structure of the filter depends upon the syntax recognized by the source database. Another way to filter data in Reporting Services is to apply a filter to a dataset, data region, or a group. Using this latter method, you can even apply multiple filters. However, the use of filter operators or the setup of multiple filters is not always obvious, so in this series of posts, I'll provide some more information about the configuration of filters. First, why not use query parameters exclusively for filtering? Here are a few reasons: You might want to apply a filter to part of the report, but not all of the report. Your dataset might retrieve data from a stored procedure, and doesn't allow you to pass a query parameter for filtering purposes. Your report might be set up as a snapshot on the report server and, in that case, cannot be dynamically filtered based on a query parameter. Next, let's look at how to set up a report filter in general. The process is the same whether you are applying the filter to a dataset, data region, or a group. When you go to the Filters page in the Properties dialog box for whichever of these items you selected (dataset, data region, group), you click the Add button to create a new filter. The interface looks like this: The Expression field is usually a field in the dataset, so to make it easier for you to make a selection,the drop-down list displays all of the current dataset fields. But notice the expression button to the right, which means that you can set up any type of expression-not just a dataset field. To the right of the expression button, you'll find a data type drop-down list. It's important to specify the correct data type for the field or expression you're using. Now for the operators. Here's a list of the options that you have: This Operator Performs This Action =, <>, >, >=, <, <=, Like Compares expression to value Top N, Bottom N Compares expression to Top (Bottom) set of N values (N = integer) Top %, Bottom % Compares expression to Top (Bottom) N percent of values (N = integer or float) Between Determines whether expression is between two values, inclusive In Determines whether expression is found in list of values Last, the Value is what you're comparing to the expression using the operator. The construction of a filter using some operators (=, <>, >, etc.) is fairly simple. If my dataset (for AdventureWorks data) has a Category field, and I have a parameter that prompts the user for a single category, I can set up a filter like this: Expression Data Type Operator Value [Category] Text = [@Category] But if I set the parameter to accept multiple values, I need to change the operator from = to In, just as I would have to do if I were using a query parameter. The parameter expression, [@Category], which translates to =Parameters!Category.Value, doesn’t need to change because it represents an array as soon as I change the parameter to allow multiple values. The “In” operator requires an array. With that in mind, let’s consider a variation on Value. Let’s say that I have a parameter that prompts the user for a particular year – and for simplicity’s sake, this parameter only allows a single value, and I have an expression that evaluates the previous year based on the user’s selection. Then I want to use these two values in two separate filters with an OR condition. That is, I want to filter either by the year selected OR by the year that was computed. If I create two filters, one for each year (as shown below), then the report will only display results if BOTH filter conditions are met – which would never be true. Expression Data Type Operator Value [CalendarYear] Integer = [@Year] [CalendarYear] Integer = =Parameters!Year.Value-1 To handle this scenario, we need to create a single filter that uses the “In” operator, and then set up the Value expression as an array. To create an array, we use the Split function after creating a string that concatenates the two values (highlighted in yellow) as shown below. Expression Data Type Operator Value =Cstr(Fields!CalendarYear.Value) Text In =Split( CStr(Parameters!Year.Value) + ”,” + CStr(Parameters!Year.Value-1) , “,”) Note that in this case, I had to apply a string conversion on the year integer so that I could concatenate the parameter selection with the calculated year. Pay attention to the second argument of the Split function—you must use a comma delimiter for the result to work correctly with the In operator. I also had to change the Expression value from [CalendarYear] (or =Fields!CalendarYear.Value) so that the expression would return a string that I could compare with the values in the string array. More fun with filter expressions in future posts!

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  • HTG Explains: Should You Buy Extended Warranties?

    - by Chris Hoffman
    Buy something at an electronics store and you’ll be confronted by a pushy salesperson who insists you need an extended warranty. You’ll also see extended warranties pushed hard when shopping online. But are they worth it? There’s a reason stores push extended warranties so hard. They’re almost always pure profit for the store involved. An electronics store may live on razor-thin product margins and make big profits on extended warranties and overpriced HDMI cables. You’re Already Getting Multiple Warranties First, back up. The product you’re buying already includes a warranty. In fact, you’re probably getting several different types of warranties. Store Return and Exchange: Most electronics stores allow you to return a malfunctioning product within the first 15 or 30 days and they’ll provide you with a new one. The exact period of time will vary from store to store. If you walk out of the store with a defective product and have to swap it for a new one within the first few weeks, this should be easy. Manufacturer Warranty: A device’s manufacturer — whether the device is a laptop, a television, or a graphics card — offers their own warranty period. The manufacturer warranty covers you after the store refuses to take the product back and exchange it. The length of this warranty depends on the type of product. For example, a cheap laptop may only offer a one-year manufacturer warranty, while a more expensive laptop may offer a two-year warranty. Credit Card Warranty Extension: Many credit cards offer free extended warranties on products you buy with that credit card. Credit card companies will often give you an additional year of warranty. For example, if you buy a laptop with a two year warranty and it fails in the third year, you could then contact your credit card company and they’d cover the cost of fixing or replacing it. Check your credit card’s benefits and fine print for more information. Why Extended Warranties Are Bad You’re already getting a fairly long warranty period, especially if you have a credit card that offers you a free extended warranty — these are fairly common. If the product you get is a “lemon” and has a manufacturing error, it will likely fail pretty soon — well within your warranty period. The extended warranty matters after all your other warranties are exhausted. In the case of a laptop with a two-year warranty that you purchase with a credit card giving you a one-year warranty extension, your extended warranty will kick in three years after you purchase the laptop. In that many years, your current laptop will likely feel pretty old and laptops that are as good — or better — will likely be pretty cheap. If it’s a television, better television displays will be available at a lower price point. You’ll either want to upgrade to a newer model or you’ll be able to buy a new, just-as-good product for very cheap. You’ll only have to pay out-of-pocket if your device fails after the normal warranty period — in over two or three years for typical laptops purchased with a decent credit card. Save the money you would have spent on the warranty and put it towards a future upgrade. How Much Do Extended Warranties Cost? Let’s look at an example from a typical pushy retail outlet, Best Buy. We went to Best Buy’s website and found a pretty standard $600 Samsung laptop. This laptop comes with a one-year warranty period. If purchased with a fairly common credit card, you can easily get a two-year warranty period on this laptop without spending an additional penny. (Yes, such credit cards are available with no yearly fees.) During the check-out process, Best Buy tries to sell you a Geek Squad “Accidental Protection Plan.” To get an additional year of Best Buy’s extended warranty, you’d have to pay $324.98 for a “3-Year Accidental Protection Plan”. You’d basically be paying more than half the price of your laptop for an additional year of warranty — remember, the standard warranties would cover you anyway for the first two years. If this laptop did break sometime between two and three years from now, we wouldn’t be surprised if you could purchase a comparable laptop for about $325 anyway. And, if you don’t need to replace it, you’ve saved that money. Best Buy would object that this isn’t a standard extended warranty. It’s a supercharged warranty plan that will also provide coverage if you spill something on your laptop or drop it and break it. You just have to ask yourself a question. What are the odds that you’ll drop your laptop or spill something on it? They’re probably pretty low if you’re a typical human being. Is it worth spending more than half the price of the laptop just in case you’ll make an uncommon mistake? Probably not. There may be occasional exceptions to this — some Apple users swear by Apple’s AppleCare, for example — but you should generally avoid buying these things. There’s a reason stores are so pushy about extended warranties, and it’s not because they want to help protect you. It’s because they’re making lots of profit from these plans, and they’re making so much profit because they’re not a good deal for customers. Image Credit: Philip Taylor on Flickr     

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  • How to get the age from a birthdate using PHP & MySQL?

    - by TaG
    I ask my users for their birthdate and store it in my database in the following way $month $day $year output May 6 1901 but I was wondering how can I get the age from the stored birthdate using PHP & MySQL? Here is the PHP code. if (isset($_POST['submitted'])) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT users.* FROM users WHERE user_id=3"); $month_options = array("Month", "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"); $day_options = array("Day", "1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17", "18", "19", "20", "21", "22", "23", "24", "25", "26", "27", "28", "29", "30", "31"); $month = mysqli_real_escape_string($mysqli, htmlentities(strip_tags($_POST['month']))); $day = mysqli_real_escape_string($mysqli, htmlentities(strip_tags($_POST['day']))); $year = mysqli_real_escape_string($mysqli, htmlentities(strip_tags($_POST['year']))); if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"INSERT INTO users (user_id, month, day, year) VALUES ('$user_id', '$month', '$day', '$year')"); } if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE users SET month = '$month', day = '$day', year = '$year' WHERE user_id = '$user_id'"); echo '<p class="changes-saved">Your changes have been saved!</p>'; } if (!$dbc) { print mysqli_error($mysqli); return; } } Here is the html. <form method="post" action="index.php"> <fieldset> <ul> <li><label>Date of Birth: </label> <label for="month" class="hide">Month: </label> <?php // month options echo '<select name="month" id="month">' . "\n"; foreach($month_options as $option) { if ($option == $month) { echo '<option value="' . stripslashes(htmlentities(strip_tags($option))) . '" selected="selected">' . stripslashes(htmlentities(strip_tags($option))) . '</option>' . "\n"; } else { echo '<option value="'. stripslashes(htmlentities(strip_tags($option))) . '">' . stripslashes(htmlentities(strip_tags($option))) . '</option>'."\n"; } } echo '</select>'; ?> <label for="day" class="hide">Day: </label> <?php // day options echo '<select id="day" name="day">' . "\n"; foreach($day_options as $option) { if ($option == $day) { echo '<option value="' . stripslashes(htmlentities(strip_tags($option))) . '" selected="selected">' . stripslashes(htmlentities(strip_tags($option))) . '</option>' . "\n"; } else { echo '<option value="'. stripslashes(htmlentities(strip_tags($option))) . '">' . stripslashes(htmlentities(strip_tags($option))) . '</option>'."\n"; } } echo '</select>'; ?> <label for="year" class="hide">Year: </label><input type="text" name="year" id="year" size="4" maxlength="4" value="<?php if (isset($_POST['year'])) { echo stripslashes(htmlentities(strip_tags($_POST['year']))); } else if(!empty($year)) { echo stripslashes(htmlentities(strip_tags($year))); } ?>" /></li> <li><input type="submit" name="submit" value="Save Changes" class="save-button" /> <input type="hidden" name="submitted" value="true" /> <input type="submit" name="submit" value="Preview Changes" class="preview-changes-button" /></li> </ul> </fieldset> </form>

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  • using java Calendar

    - by owca
    I have a simple task. There are two classes : Ticket and Date. Ticket contains event, event place and event date which is a Date object. I also need to provide a move() method for Date object, so I used Calendar and Calendar's add(). Everything looks fine apart of the output. I constantly get 5,2,1 as the date's day,month,year. Lines with asterix return proper date. The code : Ticket class : public class Ticket { private String what; private String where; private Date when; public Ticket(String s1, String s2, Data d){ this.what = s1; this.where = s2; this.when = d; } *public Date giveDate(){ System.out.println("when in giveDate() "+this.when); return this.when; } public String toString(){ return "what: "+this.what+"\n"+"where: "+this.where+"\n"+"when: "+this.when; } } Date class: import java.util.Calendar; import java.util.GregorianCalendar; public class Date { public int day; public int month; public int year; public Date(int x, int y, int z){ *System.out.println("x: "+x); *System.out.println("y: "+y); *System.out.println("z: "+z); this.day = x; this.month = y; this.year = z; *System.out.println("this.day: "+this.day); *System.out.println("this.month: "+this.month); *System.out.println("this.year: "+this.year); } public Date move(int p){ *System.out.println("before_change: "+this.day+","+this.month+","+this.year); Calendar gc = new GregorianCalendar(this.year, this.month, this.day); System.out.println("before_adding: "+gc.DAY_OF_MONTH+","+gc.MONTH+","+gc.YEAR); gc.add(Calendar.DAY_OF_YEAR, p); System.out.println("after_adding: "+gc.DAY_OF_MONTH+","+gc.MONTH+","+gc.YEAR); this.year = gc.YEAR; this.day = gc.DAY_OF_MONTH; this.month = gc.MONTH; return this; } @Override public String toString(){ return this.day+","+this.month+","+this.year; } } Main for testing : public class Main { public static void main(String[] args) { Date date1=new Date(30,4,2002); Ticket event1=new Ticket("Peter Gabriel's gig", "London",date1 ); Ticket event2=new Ticket("Diana Kroll's concert", "Glasgow",date1 ); Date date2=event2.giveDate(); date2.move(30); Ticket event3=new Ticket("X's B-day", "some place",date2 ); System.out.println(date1); System.out.println(event1); System.out.println(event2); System.out.println(event3); } } And here's my output. I just can't get it where 5,2,1 come from :/ x: 30 y: 4 z: 2002 this.day: 30 this.month: 4 this.year: 2002 when in giveDate() 6,12,2004 before_change: 6,12,2004 before_adding: 5,2,1 after_adding: 5,2,1 5,2,1 what: Peter Gabriel's gig where: London when: 5,2,1 (...)

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  • TOTD #166: Using NoSQL database in your Java EE 6 Applications on GlassFish - MongoDB for now!

    - by arungupta
    The Java EE 6 platform includes Java Persistence API to work with RDBMS. The JPA specification defines a comprehensive API that includes, but not restricted to, how a database table can be mapped to a POJO and vice versa, provides mechanisms how a PersistenceContext can be injected in a @Stateless bean and then be used for performing different operations on the database table and write typesafe queries. There are several well known advantages of RDBMS but the NoSQL movement has gained traction over past couple of years. The NoSQL databases are not intended to be a replacement for the mainstream RDBMS. As Philosophy of NoSQL explains, NoSQL database was designed for casual use where all the features typically provided by an RDBMS are not required. The name "NoSQL" is more of a category of databases that is more known for what it is not rather than what it is. The basic principles of NoSQL database are: No need to have a pre-defined schema and that makes them a schema-less database. Addition of new properties to existing objects is easy and does not require ALTER TABLE. The unstructured data gives flexibility to change the format of data any time without downtime or reduced service levels. Also there are no joins happening on the server because there is no structure and thus no relation between them. Scalability and performance is more important than the entire set of functionality typically provided by an RDBMS. This set of databases provide eventual consistency and/or transactions restricted to single items but more focus on CRUD. Not be restricted to SQL to access the information stored in the backing database. Designed to scale-out (horizontal) instead of scale-up (vertical). This is important knowing that databases, and everything else as well, is moving into the cloud. RBDMS can scale-out using sharding but requires complex management and not for the faint of heart. Unlike RBDMS which require a separate caching tier, most of the NoSQL databases comes with integrated caching. Designed for less management and simpler data models lead to lower administration as well. There are primarily three types of NoSQL databases: Key-Value stores (e.g. Cassandra and Riak) Document databases (MongoDB or CouchDB) Graph databases (Neo4J) You may think NoSQL is panacea but as I mentioned above they are not meant to replace the mainstream databases and here is why: RDBMS have been around for many years, very stable, and functionally rich. This is something CIOs and CTOs can bet their money on without much worry. There is a reason 98% of Fortune 100 companies run Oracle :-) NoSQL is cutting edge, brings excitement to developers, but enterprises are cautious about them. Commercial databases like Oracle are well supported by the backing enterprises in terms of providing support resources on a global scale. There is a full ecosystem built around these commercial databases providing training, performance tuning, architecture guidance, and everything else. NoSQL is fairly new and typically backed by a single company not able to meet the scale of these big enterprises. NoSQL databases are good for CRUDing operations but business intelligence is extremely important for enterprises to stay competitive. RDBMS provide extensive tooling to generate this data but that was not the original intention of NoSQL databases and is lacking in that area. Generating any meaningful information other than CRUDing require extensive programming. Not suited for complex transactions such as banking systems or other highly transactional applications requiring 2-phase commit. SQL cannot be used with NoSQL databases and writing simple queries can be involving. Enough talking, lets take a look at some code. This blog has published multiple blogs on how to access a RDBMS using JPA in a Java EE 6 application. This Tip Of The Day (TOTD) will show you can use MongoDB (a document-oriented database) with a typical 3-tier Java EE 6 application. Lets get started! The complete source code of this project can be downloaded here. Download MongoDB for your platform from here (1.8.2 as of this writing) and start the server as: arun@ArunUbuntu:~/tools/mongodb-linux-x86_64-1.8.2/bin$./mongod./mongod --help for help and startup optionsSun Jun 26 20:41:11 [initandlisten] MongoDB starting : pid=11210port=27017 dbpath=/data/db/ 64-bit Sun Jun 26 20:41:11 [initandlisten] db version v1.8.2, pdfile version4.5Sun Jun 26 20:41:11 [initandlisten] git version:433bbaa14aaba6860da15bd4de8edf600f56501bSun Jun 26 20:41:11 [initandlisten] build sys info: Linuxbs-linux64.10gen.cc 2.6.21.7-2.ec2.v1.2.fc8xen #1 SMP Fri Nov 2017:48:28 EST 2009 x86_64 BOOST_LIB_VERSION=1_41Sun Jun 26 20:41:11 [initandlisten] waiting for connections on port 27017Sun Jun 26 20:41:11 [websvr] web admin interface listening on port 28017 The default directory for the database is /data/db and needs to be created as: sudo mkdir -p /data/db/sudo chown `id -u` /data/db You can specify a different directory using "--dbpath" option. Refer to Quickstart for your specific platform. Using NetBeans, create a Java EE 6 project and make sure to enable CDI and add JavaServer Faces framework. Download MongoDB Java Driver (2.6.3 of this writing) and add it to the project library by selecting "Properties", "LIbraries", "Add Library...", creating a new library by specifying the location of the JAR file, and adding the library to the created project. Edit the generated "index.xhtml" such that it looks like: <h1>Add a new movie</h1><h:form> Name: <h:inputText value="#{movie.name}" size="20"/><br/> Year: <h:inputText value="#{movie.year}" size="6"/><br/> Language: <h:inputText value="#{movie.language}" size="20"/><br/> <h:commandButton actionListener="#{movieSessionBean.createMovie}" action="show" title="Add" value="submit"/></h:form> This page has a simple HTML form with three text boxes and a submit button. The text boxes take name, year, and language of a movie and the submit button invokes the "createMovie" method of "movieSessionBean" and then render "show.xhtml". Create "show.xhtml" ("New" -> "Other..." -> "Other" -> "XHTML File") such that it looks like: <head> <title><h1>List of movies</h1></title> </head> <body> <h:form> <h:dataTable value="#{movieSessionBean.movies}" var="m" > <h:column><f:facet name="header">Name</f:facet>#{m.name}</h:column> <h:column><f:facet name="header">Year</f:facet>#{m.year}</h:column> <h:column><f:facet name="header">Language</f:facet>#{m.language}</h:column> </h:dataTable> </h:form> This page shows the name, year, and language of all movies stored in the database so far. The list of movies is returned by "movieSessionBean.movies" property. Now create the "Movie" class such that it looks like: import com.mongodb.BasicDBObject;import com.mongodb.BasicDBObject;import com.mongodb.DBObject;import javax.enterprise.inject.Model;import javax.validation.constraints.Size;/** * @author arun */@Modelpublic class Movie { @Size(min=1, max=20) private String name; @Size(min=1, max=20) private String language; private int year; // getters and setters for "name", "year", "language" public BasicDBObject toDBObject() { BasicDBObject doc = new BasicDBObject(); doc.put("name", name); doc.put("year", year); doc.put("language", language); return doc; } public static Movie fromDBObject(DBObject doc) { Movie m = new Movie(); m.name = (String)doc.get("name"); m.year = (int)doc.get("year"); m.language = (String)doc.get("language"); return m; } @Override public String toString() { return name + ", " + year + ", " + language; }} Other than the usual boilerplate code, the key methods here are "toDBObject" and "fromDBObject". These methods provide a conversion from "Movie" -> "DBObject" and vice versa. The "DBObject" is a MongoDB class that comes as part of the mongo-2.6.3.jar file and which we added to our project earlier.  The complete javadoc for 2.6.3 can be seen here. Notice, this class also uses Bean Validation constraints and will be honored by the JSF layer. Finally, create "MovieSessionBean" stateless EJB with all the business logic such that it looks like: package org.glassfish.samples;import com.mongodb.BasicDBObject;import com.mongodb.DB;import com.mongodb.DBCollection;import com.mongodb.DBCursor;import com.mongodb.DBObject;import com.mongodb.Mongo;import java.net.UnknownHostException;import java.util.ArrayList;import java.util.List;import javax.annotation.PostConstruct;import javax.ejb.Stateless;import javax.inject.Inject;import javax.inject.Named;/** * @author arun */@Stateless@Namedpublic class MovieSessionBean { @Inject Movie movie; DBCollection movieColl; @PostConstruct private void initDB() throws UnknownHostException { Mongo m = new Mongo(); DB db = m.getDB("movieDB"); movieColl = db.getCollection("movies"); if (movieColl == null) { movieColl = db.createCollection("movies", null); } } public void createMovie() { BasicDBObject doc = movie.toDBObject(); movieColl.insert(doc); } public List<Movie> getMovies() { List<Movie> movies = new ArrayList(); DBCursor cur = movieColl.find(); System.out.println("getMovies: Found " + cur.size() + " movie(s)"); for (DBObject dbo : cur.toArray()) { movies.add(Movie.fromDBObject(dbo)); } return movies; }} The database is initialized in @PostConstruct. Instead of a working with a database table, NoSQL databases work with a schema-less document. The "Movie" class is the document in our case and stored in the collection "movies". The collection allows us to perform query functions on all movies. The "getMovies" method invokes "find" method on the collection which is equivalent to the SQL query "select * from movies" and then returns a List<Movie>. Also notice that there is no "persistence.xml" in the project. Right-click and run the project to see the output as: Enter some values in the text box and click on enter to see the result as: If you reached here then you've successfully used MongoDB in your Java EE 6 application, congratulations! Some food for thought and further play ... SQL to MongoDB mapping shows mapping between traditional SQL -> Mongo query language. Tutorial shows fun things you can do with MongoDB. Try the interactive online shell  The cookbook provides common ways of using MongoDB In terms of this project, here are some tasks that can be tried: Encapsulate database management in a JPA persistence provider. Is it even worth it because the capabilities are going to be very different ? MongoDB uses "BSonObject" class for JSON representation, add @XmlRootElement on a POJO and how a compatible JSON representation can be generated. This will make the fromXXX and toXXX methods redundant.

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  • TOTD #166: Using NoSQL database in your Java EE 6 Applications on GlassFish - MongoDB for now!

    - by arungupta
    The Java EE 6 platform includes Java Persistence API to work with RDBMS. The JPA specification defines a comprehensive API that includes, but not restricted to, how a database table can be mapped to a POJO and vice versa, provides mechanisms how a PersistenceContext can be injected in a @Stateless bean and then be used for performing different operations on the database table and write typesafe queries. There are several well known advantages of RDBMS but the NoSQL movement has gained traction over past couple of years. The NoSQL databases are not intended to be a replacement for the mainstream RDBMS. As Philosophy of NoSQL explains, NoSQL database was designed for casual use where all the features typically provided by an RDBMS are not required. The name "NoSQL" is more of a category of databases that is more known for what it is not rather than what it is. The basic principles of NoSQL database are: No need to have a pre-defined schema and that makes them a schema-less database. Addition of new properties to existing objects is easy and does not require ALTER TABLE. The unstructured data gives flexibility to change the format of data any time without downtime or reduced service levels. Also there are no joins happening on the server because there is no structure and thus no relation between them. Scalability and performance is more important than the entire set of functionality typically provided by an RDBMS. This set of databases provide eventual consistency and/or transactions restricted to single items but more focus on CRUD. Not be restricted to SQL to access the information stored in the backing database. Designed to scale-out (horizontal) instead of scale-up (vertical). This is important knowing that databases, and everything else as well, is moving into the cloud. RBDMS can scale-out using sharding but requires complex management and not for the faint of heart. Unlike RBDMS which require a separate caching tier, most of the NoSQL databases comes with integrated caching. Designed for less management and simpler data models lead to lower administration as well. There are primarily three types of NoSQL databases: Key-Value stores (e.g. Cassandra and Riak) Document databases (MongoDB or CouchDB) Graph databases (Neo4J) You may think NoSQL is panacea but as I mentioned above they are not meant to replace the mainstream databases and here is why: RDBMS have been around for many years, very stable, and functionally rich. This is something CIOs and CTOs can bet their money on without much worry. There is a reason 98% of Fortune 100 companies run Oracle :-) NoSQL is cutting edge, brings excitement to developers, but enterprises are cautious about them. Commercial databases like Oracle are well supported by the backing enterprises in terms of providing support resources on a global scale. There is a full ecosystem built around these commercial databases providing training, performance tuning, architecture guidance, and everything else. NoSQL is fairly new and typically backed by a single company not able to meet the scale of these big enterprises. NoSQL databases are good for CRUDing operations but business intelligence is extremely important for enterprises to stay competitive. RDBMS provide extensive tooling to generate this data but that was not the original intention of NoSQL databases and is lacking in that area. Generating any meaningful information other than CRUDing require extensive programming. Not suited for complex transactions such as banking systems or other highly transactional applications requiring 2-phase commit. SQL cannot be used with NoSQL databases and writing simple queries can be involving. Enough talking, lets take a look at some code. This blog has published multiple blogs on how to access a RDBMS using JPA in a Java EE 6 application. This Tip Of The Day (TOTD) will show you can use MongoDB (a document-oriented database) with a typical 3-tier Java EE 6 application. Lets get started! The complete source code of this project can be downloaded here. Download MongoDB for your platform from here (1.8.2 as of this writing) and start the server as: arun@ArunUbuntu:~/tools/mongodb-linux-x86_64-1.8.2/bin$./mongod./mongod --help for help and startup optionsSun Jun 26 20:41:11 [initandlisten] MongoDB starting : pid=11210port=27017 dbpath=/data/db/ 64-bit Sun Jun 26 20:41:11 [initandlisten] db version v1.8.2, pdfile version4.5Sun Jun 26 20:41:11 [initandlisten] git version:433bbaa14aaba6860da15bd4de8edf600f56501bSun Jun 26 20:41:11 [initandlisten] build sys info: Linuxbs-linux64.10gen.cc 2.6.21.7-2.ec2.v1.2.fc8xen #1 SMP Fri Nov 2017:48:28 EST 2009 x86_64 BOOST_LIB_VERSION=1_41Sun Jun 26 20:41:11 [initandlisten] waiting for connections on port 27017Sun Jun 26 20:41:11 [websvr] web admin interface listening on port 28017 The default directory for the database is /data/db and needs to be created as: sudo mkdir -p /data/db/sudo chown `id -u` /data/db You can specify a different directory using "--dbpath" option. Refer to Quickstart for your specific platform. Using NetBeans, create a Java EE 6 project and make sure to enable CDI and add JavaServer Faces framework. Download MongoDB Java Driver (2.6.3 of this writing) and add it to the project library by selecting "Properties", "LIbraries", "Add Library...", creating a new library by specifying the location of the JAR file, and adding the library to the created project. Edit the generated "index.xhtml" such that it looks like: <h1>Add a new movie</h1><h:form> Name: <h:inputText value="#{movie.name}" size="20"/><br/> Year: <h:inputText value="#{movie.year}" size="6"/><br/> Language: <h:inputText value="#{movie.language}" size="20"/><br/> <h:commandButton actionListener="#{movieSessionBean.createMovie}" action="show" title="Add" value="submit"/></h:form> This page has a simple HTML form with three text boxes and a submit button. The text boxes take name, year, and language of a movie and the submit button invokes the "createMovie" method of "movieSessionBean" and then render "show.xhtml". Create "show.xhtml" ("New" -> "Other..." -> "Other" -> "XHTML File") such that it looks like: <head> <title><h1>List of movies</h1></title> </head> <body> <h:form> <h:dataTable value="#{movieSessionBean.movies}" var="m" > <h:column><f:facet name="header">Name</f:facet>#{m.name}</h:column> <h:column><f:facet name="header">Year</f:facet>#{m.year}</h:column> <h:column><f:facet name="header">Language</f:facet>#{m.language}</h:column> </h:dataTable> </h:form> This page shows the name, year, and language of all movies stored in the database so far. The list of movies is returned by "movieSessionBean.movies" property. Now create the "Movie" class such that it looks like: import com.mongodb.BasicDBObject;import com.mongodb.BasicDBObject;import com.mongodb.DBObject;import javax.enterprise.inject.Model;import javax.validation.constraints.Size;/** * @author arun */@Modelpublic class Movie { @Size(min=1, max=20) private String name; @Size(min=1, max=20) private String language; private int year; // getters and setters for "name", "year", "language" public BasicDBObject toDBObject() { BasicDBObject doc = new BasicDBObject(); doc.put("name", name); doc.put("year", year); doc.put("language", language); return doc; } public static Movie fromDBObject(DBObject doc) { Movie m = new Movie(); m.name = (String)doc.get("name"); m.year = (int)doc.get("year"); m.language = (String)doc.get("language"); return m; } @Override public String toString() { return name + ", " + year + ", " + language; }} Other than the usual boilerplate code, the key methods here are "toDBObject" and "fromDBObject". These methods provide a conversion from "Movie" -> "DBObject" and vice versa. The "DBObject" is a MongoDB class that comes as part of the mongo-2.6.3.jar file and which we added to our project earlier.  The complete javadoc for 2.6.3 can be seen here. Notice, this class also uses Bean Validation constraints and will be honored by the JSF layer. Finally, create "MovieSessionBean" stateless EJB with all the business logic such that it looks like: package org.glassfish.samples;import com.mongodb.BasicDBObject;import com.mongodb.DB;import com.mongodb.DBCollection;import com.mongodb.DBCursor;import com.mongodb.DBObject;import com.mongodb.Mongo;import java.net.UnknownHostException;import java.util.ArrayList;import java.util.List;import javax.annotation.PostConstruct;import javax.ejb.Stateless;import javax.inject.Inject;import javax.inject.Named;/** * @author arun */@Stateless@Namedpublic class MovieSessionBean { @Inject Movie movie; DBCollection movieColl; @PostConstruct private void initDB() throws UnknownHostException { Mongo m = new Mongo(); DB db = m.getDB("movieDB"); movieColl = db.getCollection("movies"); if (movieColl == null) { movieColl = db.createCollection("movies", null); } } public void createMovie() { BasicDBObject doc = movie.toDBObject(); movieColl.insert(doc); } public List<Movie> getMovies() { List<Movie> movies = new ArrayList(); DBCursor cur = movieColl.find(); System.out.println("getMovies: Found " + cur.size() + " movie(s)"); for (DBObject dbo : cur.toArray()) { movies.add(Movie.fromDBObject(dbo)); } return movies; }} The database is initialized in @PostConstruct. Instead of a working with a database table, NoSQL databases work with a schema-less document. The "Movie" class is the document in our case and stored in the collection "movies". The collection allows us to perform query functions on all movies. The "getMovies" method invokes "find" method on the collection which is equivalent to the SQL query "select * from movies" and then returns a List<Movie>. Also notice that there is no "persistence.xml" in the project. Right-click and run the project to see the output as: Enter some values in the text box and click on enter to see the result as: If you reached here then you've successfully used MongoDB in your Java EE 6 application, congratulations! Some food for thought and further play ... SQL to MongoDB mapping shows mapping between traditional SQL -> Mongo query language. Tutorial shows fun things you can do with MongoDB. Try the interactive online shell  The cookbook provides common ways of using MongoDB In terms of this project, here are some tasks that can be tried: Encapsulate database management in a JPA persistence provider. Is it even worth it because the capabilities are going to be very different ? MongoDB uses "BSonObject" class for JSON representation, add @XmlRootElement on a POJO and how a compatible JSON representation can be generated. This will make the fromXXX and toXXX methods redundant.

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  • Which programming language is manageable by an 11 year old kid?

    - by tangens
    Possible Duplicates: What is the easiest language to start with? What are some recommended programming resources for pre-teens? My son is 11 years old and he would like to learn a programming language. Of course his primary goal is to develop some (simple) games. Do you know of a programming language that is suitable for this situation? Summary of languages recommended in the answers Snake Wrangling for Kids (answer) Scratch (answer) Small Basic (answer) (answer) Logo NXT-G for Lego Mindstorms (answer) Alice (answer) BlueJ (answer) Squeak Smalltalk (answer) (answer) (answer) Blender Game Engine (answer) PyGame (answer) (answer) (answer) Inform (answer) Phrogram (answer) Dr Scheme (answer) eToys (answer) runrev (answer) Karel Programming (answer) Hackety Hack (answer) Visual Basic (answer) (answer) Learn to Program (answer) QBasic (answer) (answer) Visual Basic Express (answer) Processing (answer) C# (answer) JavaScript (answer) (answer) Ruby (answer) ToonTalk (answer) Flash and ActionScript (answer) StarLogo (answer) Java (answer) Kodu (answer) XNA (answer) (answer) unity3D (answer) BlitzBasic (answer)(answer) Lua (answer)

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  • Is SQL DATEDIFF(year, ..., ...) an Expensive Computation?

    - by rlb.usa
    I'm trying to optimize up some horrendously complicated SQL queries because it takes too long to finish. In my queries, I have dynamically created SQL statements with lots of the same functions, so I created a temporary table where each function is only called once instead of many, many times - this cut my execution time by 3/4. So my question is, can I expect to see much of a difference if say, 1,000 datediff computations are narrowed to 100?

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  • How can change my code to can user insert numbers of year and days ?

    - by MANAL
    I make code to calculate the date of today and the date for two years ago before the day of today and also calculate day before 5 days. I want user insert the numbers of years and days to make compare between date of today . import java.util.Date; import java.util.Calendar; import java.text.SimpleDateFormat; import java.util.Scanner; public class Calendar1 { private static void doCalendarTime() { System.out.print("*************************************************"); Date now = Calendar.getInstance().getTime(); System.out.print(" \n Calendar.getInstance().getTime() : " + now); System.out.println(); } private static void doSimpleDateFormat() { System.out.print("*************************************************"); System.out.print("\n\nSIMPLE DATE FORMAT\n"); System.out.print("*************************************************"); // Get today's date Calendar now = Calendar.getInstance(); SimpleDateFormat formatter = new SimpleDateFormat("E yyyy.MM.dd 'at' hh:mm:ss a zzz"); System.out.print(" \n It is now : " + formatter.format(now.getTime())); System.out.println(); } private static void doAdd() { System.out.println("ADD / SUBTRACT CALENDAR / DATEs"); System.out.println("================================================================="); // Get today's date Calendar now = Calendar.getInstance(); Calendar working; SimpleDateFormat formatter = new SimpleDateFormat("E yyyy.MM.dd 'at' hh:mm:ss a zzz"); working = (Calendar) now.clone(); working.add(Calendar.DAY_OF_YEAR, - (365 * 2)); System.out.println (" Two years ago it was: " + formatter.format(working.getTime())); working = (Calendar) now.clone(); working.add(Calendar.DAY_OF_YEAR, + 5); System.out.println(" In five days it will be: " + formatter.format(working.getTime())); System.out.println(); } public static void main(String[] args) { System.out.println(); doCalendarTime(); doSimpleDateFormat(); doAdd(); } } help me .

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  • Create attribute in existing XML

    - by user560411
    Hello. I have the following php code that inserts data into XML and works correctly. However, I want to add an ID number like below <CD id="xxxx"> My question is how can i add an ID into CD for this to work ? I use form to parse the id. ** the main code ** <?php $CD = array( 'TITLE' => $_POST['title'], 'BAND' => $_POST['band'], 'YEAR' => $_POST['year'], ); $doc = new DOMDocument(); $doc->load( 'insert.xml' ); $doc->formatOutput = true; $r = $doc->getElementsByTagName("CATEGORIES")->item(0); $b = $doc->createElement("CD"); $TITLE = $doc->createElement("TITLE"); $TITLE->appendChild( $doc->createTextNode( $CD["TITLE"] ) ); $b->appendChild( $TITLE ); $BAND = $doc->createElement("BAND"); $BAND->appendChild( $doc->createTextNode( $CD["BAND"] ) ); $b->appendChild( $BAND ); $YEAR = $doc->createElement("YEAR"); $YEAR->appendChild( $doc->createTextNode( $CD["YEAR"] ) ); $b->appendChild( $YEAR ); $r->appendChild( $b ); $doc->save("insert.xml"); ?> the XML file <?xml version="1.0" encoding="utf-8"?> <MY_CD> <CATEGORIES> <CD> <TITLE>NEVER MIND THE BOLLOCKS</TITLE> <BAND>SEX PISTOLS</BAND> <YEAR>1977</YEAR> </CD> <CD> <TITLE>NEVERMIND</TITLE> <BAND>NIRVANA</BAND> <YEAR>1991</YEAR> </CD> </CATEGORIES> </MY_CD>

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