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  • CONVERT(int, (datepart(month, @search)), (datepart(day, @search)), DateAdd(year, Years.Year - (datepart(year, @search)))

    - by MyHeadHurts
    In the query the top part is getting all the years that will run in the stored procedure. Works fine But at first i just wanted to run the queries for yesterdays date for all the years, but now i realized i want the user to select a date that will be in a parameter @search Booked <= CONVERT(int,DateAdd(year, Years.Year - Year(getdate()), DateAdd(day, DateDiff(day, 2, getdate()), 1))) this should be easy because normally it would just be Booked <= CONVERT(int,@search) but the problem is i want to do something like a Booked <= CONVERT(int, (datepart(month, @search)), (datepart(day, @search)), DateAdd(year, Years.Year - (datepart(year, @search))) would something like that work i dont need to worry about subtracting days but i still need to worry about the years WITH Years AS ( SELECT DATEPART(year, GETDATE()) [Year] UNION ALL SELECT [Year]-1 FROM Years WHERE [Year]>@YearToGet ), q_00 as ( select DIVISION , DYYYY , sum(PARTY) as asofPAX , sum(APRICE) as asofSales from dbo.B101BookingsDetails INNER JOIN Years ON B101BookingsDetails.DYYYY = Years.Year where Booked <= CONVERT(int,DateAdd(year, Years.Year - Year(getdate()), DateAdd(day, DateDiff(day, 2, getdate()), 1))) and DYYYY = Years.Year group by DIVISION, DYYYY, years.year having DYYYY = years.year ),

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  • SQL SERVER – Various Leap Year Logics

    - by pinaldave
    Earlier I wrote one article on Leap Year and created one video about Leap Year. My point of view was to demonstrate how we can use SQL Server 2012 features to identify Leap year. How ever during the conversation I had some really good conversation. Here are updates for those who have missed reading the excellent comments on the blog. Incorrect Logic There are so many people still think Leap Year is the event which is consistently happening at every four year and the way to find it is divide the year with 4 and if the remainder is 0. That year is leap year. Well, it is not correct. Comment by David Bridge Check out this excerpt from wikipedia page http://en.wikipedia.org/wiki/Leap_year “most years that are evenly divisible by 4 are leap years…” “…Some exceptions to this rule are required since the duration of a solar year is slightly less than 365.25 days. Years that are evenly divisible by 100 are not leap years, unless they are also evenly divisible by 400, in which case they are leap years. For example, 1600 and 2000 were leap years, but 1700, 1800 and 1900 were not. Similarly, 2100, 2200, 2300, 2500, 2600, 2700, 2900 and 3000 will not be leap years, but 2400 and 2800 will be.” If you use logic of divide by 4 and remainder is 0 to find leap year, you will may end up with inaccurate result. The correct way to identify the year is to figure out the days of February and if the count is 29, the year is for sure leap year. Valid Alternate Solutions Comment by sainswor99insworth IIF((@Year%4=0 AND @Year%100 != 0) OR @Year%400=0, 1,0) Comment by Madhivanan Madhivanan has written a blog post about an year ago where he listed multiple ways to find leap year. Comment by Jayan DECLARE @year INT SET @year = 2012 IF (((@year % 4 = 0) AND (@year % 100 != 0)) OR (@year % 400 = 0)) PRINT ’1' ELSE print ’0' Comment by David DECLARE @Year INT = 2012 SELECT ISDATE('2/29/' + CAST(@Year AS CHAR(4))) Comment by David Bridge Incidentally – Another approach would be to take one day off March 1st and see if it is 29. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL DateTime, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Correct expression for checking leap years

    - by Debanjan
    What will be the exact definition of leap year? AFAIK "A year which is divisible by 4 is a leap year. But for century years' the years which are divisible by 400 is a leap year." But that definition makes 100, 200, 300, 400.... upto 1700 NOT LEAP years! But in Gregorian calendar all of them are all leap year, check this out. You can also try "call 1700" in Linux to verify. So the correct algorithm for leap years would be: if ( (year % 4 == 0) && ( year <= 1700 || year % 100 != 0 || year % 400 == 0 )) printf("%d is a leap year.\n", year); else printf("%d is not a leap year.\n", year); But is this specific to Gregorian calendar? Even if that is the case why is it not mentioned here? Regards, PS:The history of Gregorian callender seems interesting check out the September month of 1752.

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  • Calculating for leap year [migrated]

    - by Bradley Bauer
    I've written this program using Java in Eclipse. I was able to utilize a formula I found that I explained in the commented out section. Using the for loop I can iterate through each month of the year, which I feel good about in that code, it seems clean and smooth to me. Maybe I could give the variables full names to make everything more readable but I'm just using the formula in its basic essence :) Well my problem is it doesn't calculate correctly for years like 2008... Leap Years. I know that if (year % 400 == 0 || (year % 4 == 0 && year % 100 != 0)) then we have a leap year. Maybe if the year is a leap year I need to subtract a certain amount of days from a certain month. Any solutions, or some direction would be great thanks :) package exercises; public class E28 { /* * Display the first days of each month * Enter the year * Enter first day of the year * * h = (q + (26 * (m + 1)) / 10 + k + k/4 + j/4 + 5j) % 7 * * h is the day of the week (0: Saturday, 1: Sunday ......) * q is the day of the month * m is the month (3: March 4: April.... January and Feburary are 13 and 14) * j is the century (year / 100) * k is the year of the century (year %100) * */ public static void main(String[] args) { java.util.Scanner input = new java.util.Scanner(System.in); System.out.print("Enter the year: "); int year = input.nextInt(); int j = year / 100; // Find century for formula int k = year % 100; // Find year of century for formula // Loop iterates 12 times. Guess why. for (int i = 1, m = i; i <= 12; i++) { // Make m = i. So loop processes formula once for each month if (m == 1 || m == 2) m += 12; // Formula requires that Jan and Feb are represented as 13 and 14 else m = i; // if not jan or feb, then set m to i int h = (1 + (26 * (m + 1)) / 10 + k + k/4 + j/4 + 5 * j) % 7; // Formula created by a really smart man somewhere // I let the control variable i steer the direction of the formual's m value String day; if (h == 0) day = "Saturday"; else if (h == 1) day = "Sunday"; else if (h == 2) day = "Monday"; else if (h == 3) day = "Tuesday"; else if (h == 4) day = "Wednesday"; else if (h == 5) day = "Thursday"; else day = "Friday"; switch (m) { case 13: System.out.println("January 1, " + year + " is " + day); break; case 14: System.out.println("Feburary 1, " + year + " is " + day); break; case 3: System.out.println("March 1, " + year + " is " + day); break; case 4: System.out.println("April 1, " + year + " is " + day); break; case 5: System.out.println("May 1, " + year + " is " + day); break; case 6: System.out.println("June 1, " + year + " is " + day); break; case 7: System.out.println("July 1, " + year + " is " + day); break; case 8: System.out.println("August 1, " + year + " is " + day); break; case 9: System.out.println("September 1, " + year + " is " + day); break; case 10: System.out.println("October 1, " + year + " is " + day); break; case 11: System.out.println("November 1, " + year + " is " + day); break; case 12: System.out.println("December 1, " + year + " is " + day); break; } } } }

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  • Java Code for calculating Leap Year, is this code correct ?

    - by Ibn Saeed
    Hello I am following "The Art and Science of Java" book and it shows how to calculate a leap year. The book uses ACM Java Task Force's library. Here is the code the books uses: import acm.program.*; public class LeapYear extends ConsoleProgram { public void run() { println("This program calculates leap year."); int year = readInt("Enter the year: "); boolean isLeapYear = ((year % 4 == 0) && (year % 100 != 0) || (year % 400 == 0)); if (isLeapYear) { println(year + " is a leap year."); } else println(year + " is not a leap year."); } } Now, this is how I calculated the leap year. import acm.program.*; public class LeapYear extends ConsoleProgram { public void run() { println("This program calculates leap year."); int year = readInt("Enter the year: "); if ((year % 4 == 0) && year % 100 != 0) { println(year + " is a leap year."); } else if ((year % 4 == 0) && (year % 100 == 0) && (year % 400 == 0)) { println(year + " is a leap year."); } else { println(year + " is not a leap year."); } } } Is there anything wrong with my code or should i use the one provided by the book ? EDIT :: Both of the above code works fine, What i want to ask is which code is the best way to calculate the leap year.

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  • The Year 2010, The Year of Change

    As I look back on the year of 2010, I could have never predicted the wonderful changes that have occurred for my wife and me. The beginning of this year started out as the 9th year that we lived in South Florida, and my fourth year working for DentalPlans.com as a software engineer/network admin. About 3 months in to the year I was given an excellent opportunity to work for MovieTickets.com in the software engineering department. This opportunity allowed me to gain experience with jQuery due to one of my projects was to reengineering MovieTickets.com existing Marketing Panel System. About 3 months after working at MovieTickets.com, my wife and I were offered an opportunity of a life time. I was offered a Job in a large background\information security company located in Nashville, TN as software engineer II.  I must note that after living in South Florida for 9 years, my wife and I really had a strong distaste for the South Florida life style and the general attitude/culture of the area. Even though we shared a strong dislike for the area in which we lived I must admit that it was a tough decision to leave MovieTickets.com because I was really doing well and I made some great new friends like Chris Catto, and Tyson Nero.  In fact, they introduced me to Local Microsoft User Groups, and software development podcast like DotNetRocks.com and Hanselminutes.com.  In addition, we also went to my first Microsoft launch down in Miami for Visual Studios 2010. I must admit it was a cool experience.  I truly hope to keep in touch with them to see how their careers grow, and I know they will. I must admit I was nervous and excited to start the next chapter in our live as I started up the 26 foot U-Haul truck and got on the road for Nashville from Boca Raton. I knew that the change was going to lead to new adventures and new opportunities that I could never imagine.  As we pulled in to the long driveway of our rental house, we knew that this was the right place for my wife and I. Natalie, my wife had actually come up to Nashville and within one week of my job offer had set up a nice rental home for us to restart our lives in TN.  I must admit that the wonderful southern hospitality took a bit to get use to due to the type of people we were used to dealing with on a regular basis. Our first 2 months seemed like we were living a dream because of our new area and the wonderful people we live around. So far my new job is going really well and I really like the people on my team and department. In fact after 6 months I am now in charge of all application builds for our new deployment process. I am also leading up a push for setting up of continuous integration within our new build process.  In addition to starting my new job, I was also offered a position as an adjust instructor at ITT Tech teaching course like VB.net, Java Script, Ajax, and database development. So far I have really like teaching at the college level.  Information technology has really been great for my life so I am really glad to be able to give back. That is actually why I started DotNetBlocks. This site allows me to document things I have learned as I work with technology, and allows others to borrow from my experiences.  I hope that this site can help others as others have helped me get where I am. Finally, I am glade to report that I only have 4 classes left for my master’s degree at Capella University. I am proud to announce that I am still on track to graduate with 3.91 GPA.  This last class was really a test because I had a crazy idea that I could work full time as a software engineer, teach two college courses as a first time teacher and also take an advanced masters class in application architecture. I have no idea how I actually survived, but I am really surprised how well I actually did. I was invited back to reach again at ITT Tech, and I passed my masters class with an “A”.  I have decided to take this next term off from my master’s program so that I do not get burned out.  Also, so that my new current employer will pay for more of my education, tuition reimbursement is an awesome benefit. This was my year 2010, how was yours?

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  • ASP.NET – Function to Fill Month, Date and Year into Dropdown lists

    - by SAMIR BHOGAYTA
    public void fillMonthList(DropDownList ddlList) { ddlList.Items.Add(new ListItem("Month", "Month")); ddlList.SelectedIndex = 0; DateTime month = Convert.ToDateTime("1/1/2000"); for (int intLoop = 0; intLoop { DateTime NextMont = month.AddMonths(intLoop); //ddlList.Items.Add(new ListItem(NextMont.ToString("MMMM"), NextMont.Month.ToString())); ddlList.Items.Add(new ListItem(NextMont.ToString("MMMM"), NextMont.ToString("MMMM"))); } } public void fillDayList(DropDownList ddlList) { ddlList.Items.Add(new ListItem("Day", "Day")); ddlList.SelectedIndex = 0; int totalDays = DateTime.DaysInMonth(DateTime.Now.Year, DateTime.Now.Month); for (int intLoop = 1; intLoop { ddlList.Items.Add(new ListItem(intLoop.ToString(), intLoop.ToString())); } } public void fillYearList(DropDownList ddlList) { ddlList.Items.Add(new ListItem("Year", "Year")); ddlList.SelectedIndex = 0; int intYearName = 1900; for (int intLoop = intYearName; intLoop { ddlList.Items.Add(new ListItem(intLoop.ToString(), intLoop.ToString())); } }

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  • 2 year degree plus experience vs 4 year degree

    - by CenterOrbit
    Alright, I have searched around a bit on this site and found two somewhat similar questions: Computer Science Programming Certificate vs. Computer Science Degree? Is it possible/likely to be paid fairly without a college degree? But these do not provide an answer specifically to what I am seeking. I have my 2 year A.A.S. Degree in computer programming, along with a networking certificate from a technical college. I also have been working at a small educational game development company for 3 years now in various positions, but steadily moving up and now as a lead programmer on a few projects. Some of the higher programmers I work with claim that no matter how much experience I develop it still will not mean as much as someone with a 4 year degree. Their argument is that most employers will look over my resume because of the common '4 yr' minimum requirement. I have also heard people state (not as many though) that experience is everything and that an employer would rather have someone that has worked in the field instead of a rookie fresh out of college. I have heard both sides of this argument, but am looking for a general consensus, or more arguments from both sides from the people who have been there, or are there.

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  • HCM: North America: Year End Knowledge Content References

    - by CaroleB
    As we all know, the next couple of months will be busy ones for the Payroll and IT department in relation to preparing for Year End,as a means of assisting you to find documented knowledge in reference to North American (NA) Year End, the following reference guide has been put together: General Knowledge: Doc ID 404478.1 Americas (US, CA, MX) HCM High Priority Alert Doc ID 1577601.1 North American Year End 2013 / 2014 Year Begin Patch Information and Useful Links. Monitor this note as it will be updated as new information becomes available NA Year End Processing: Document 255466.1 - End of Year Processing Using Oracle HRMS (US)  Document 260344.1 - End Of Year Processing Using Oracle HRMS (Canada) Document 395622.1 - End Of Year Processing Using Oracle HRMS (Mexican) Patching : Document 216109.1 - Oracle Human Resources (HRMS) Payroll North America Annual Patching Schedule Document 1160507.1 - Oracle E-Business Suite - Consolidated HRMS Mandatory Patch List Document 1144633.1 - US Year End Patch Flow Advisor: E-Business Suite (EBS) Human Capital Management (HCM) for US Legislation patching 2013 YE Phase I Readme's US Document 1584795.1 Release 11i   - 2013 US Payroll Year End Phase 1 Readme Document 1584796.1 Release 12.0 - 2013 US Payroll Year End Phase 1 Readme Document 1584797.1 Release 12.1 - 2013 US Payroll Year End Phase 1 Readme CA Document 1585365.1 2013 Canadian Payroll Year End Phase 1 Readme Release 11i Document 1585366.1 2013 Canadian Payroll Year End Phase 1 Readme Release 12.0 Document 1585367.1 2013 Canadian Payroll Year End Phase 1 Readme Release 12.1 Known Issues / How To: Document 1527958.2 - Information Center: Oracle HRMS (US) (All Application Versions) Look specifically at the US- Year End Tab for information on: Year End Pre-Processor 1099R Federal, State, and Local Magnetic Media W-2 Paper Reports W-2 PDF W-2 Register Additional Resources: Webcast: Document 1455851.1 - Advisor Webcasts for Oracle E-Business Suite- Human Capital Management (HCM) Document 1592483.1 - Webcast: EBS North American Payroll Year End Process Flow November 20, 2013 at 3:30 pm ET, 2:30 pm CT, 1:30 pm MT, 12:30 pm PT Communities: Payroll – EBS HCM - EBS Community E-Business Patching Community

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  • PHP strtotime without leap-years

    - by Christian Sciberras
    With regards to this thread, I've developed a partial solution: function strtosecs($time,$now=null){ static $LEAPDIFF=86400; $time=strtotime($time,$now); return $time-((date('Y',$time)-1968)/4*$LEAPDIFF); } The function is supposed to get the number of seconds given a string without checking leap-years. It does this calculating the number of leap-years 1970 [(year-1986)/4], multiplying it by the difference in seconds between a leap-year and a normal year (which in the end, it's just the number of seconds in a day). Finally, I simply remove all those excess leap-year seconds from the calculated time. Here's some examples of the inputs/outputs: // test code echo strtosecs('+20 years',0).'=>'.(strtosecs('+20 years',0)/31536000); echo strtosecs('+1 years',0).'=>'.(strtosecs('+1 years',0)/31536000); // test output 630676800 => 19.998630136986 31471200 => 0.99794520547945 You will probably ask why am I doing a division on the output? It's to test it out; 31536000 is the number of seconds in a year, so that 19.99... should be 20 and 0.99... should be a 1. Sure, I could round it all and get "correct" answer, but I'm worried about the inaccuracies.

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  • Excel chart with year-to-year comparison

    - by Craig
    Given this data: Date Year Month Usage (Kw-h) Cost/Month 02/19/08 2008 2 501 59.13 03/18/08 2008 3 404 48.49 04/16/08 2008 4 387 45.67 05/22/08 2008 5 319 37.85 06/23/08 2008 6 363 43.81 07/23/08 2008 7 372 48.86 08/21/08 2008 8 435 59.74 09/23/08 2008 9 358 49.9 10/16/08 2008 10 313 42.01 11/20/08 2008 11 328 39.99 12/16/08 2008 12 374 44.7 01/20/09 2009 1 474 55.35 02/19/09 2009 2 444 52.85 03/19/09 2009 3 398 49.25 04/17/09 2009 4 403 51.05 05/19/09 2009 5 405 49.61 06/18/09 2009 6 373 45.18 07/20/09 2009 7 337 44.67 08/18/09 2009 8 369 50.73 09/17/09 2009 9 377 52.36 10/16/09 2009 10 309 43.4 11/17/09 2009 11 249 34.14 12/16/09 2009 12 327 41.79 01/20/10 2010 1 356 45.66 I would like to produce a report that displays a Usage (Kw-h) line for each year. Features: Y axis: Usage (Kw-h) X axis: Month Line 0..n: lines representing each year's monthly Usage (Kw-h) Bonus points: instead of a line for each year, each month would have a high-low-close (HLC) bar; 'close' would be replaced by the average second Y axis and HLC bar that represents cost/month Questions: Can this be done without a Pivot table? Do I need to have the Year and Month column or can Excel automatically determine this? Current chart:

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  • Sort Order With End Year and Start Year

    - by Maletor
    I'm looking to write a comparator to sort my items in a list. For items without an end year they should be at the top. For items with an end year they should be next. For items with the same end year the one with the lowest start year should be next. Something I have so far [item.get('end_year'), item.get('start_year')] Test sceanrios first is end year second is start year ("" is present) "", "" "", 2012 "", 2011 2012, 2005 2012, 2008 2011, 2011 2010, 2005

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  • How could the year 2010 be mistaken as a leap year? [closed]

    - by Templar
    According to news reports such as this one, older Playstation 3 game consoles incorrectly considered the year 2010 to be a leap year. The problem persisted through most of the Monday, March 1, which older PlayStations thought was Feb. 29, 2010, a date that doesn't exist. The standard leap year formula states that a leap year is any year that is divisible by 4, but not by 100, except if divisible by 400. The year 2010 doesn't fit any of those criteria. Any idea what programming error would have caused this?

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  • mysql birthday reminder, leap year

    - by moodh
    Hi, I'm trying to sort out a result set that gives the 5 closest users sorted by upcoming birthday. This works perfectly until leap years comes into play. For example: May 15th - 96 days left May 15th - 97 days left The top result is a birth at 1987 and the lower is from 1988. u_birth is stored as yyyy-mm-dd. Is there a simple way to sort this problem without having to rewrite the entire query? SELECT u_birth, IF( DAYOFYEAR( u_birth ) >= DAYOFYEAR( NOW() ), DAYOFYEAR( u_birth ) - DAYOFYEAR( NOW() ), DAYOFYEAR( u_birth ) - DAYOFYEAR( NOW() ) + DAYOFYEAR( CONCAT( YEAR( NOW() ), '-12-31' ) ) ) AS distance FROM (blog_users) WHERE `s_agehide` = 0 ORDER BY distance ASC LIMIT 5 This query is taken and modified from the mysql manual: http://dev.mysql.com/doc/refman/5.1/en/date-and-time-functions.html#c7489

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  • How to create database records for a financial year

    - by David A Gibson
    Hello, I need to create entities (specifically contracts) in a database table that are associated with a financial year. These contracts will be applied to projects. Any contract variation will be recorded by creating a new contract record for the same financial year but the original will remain associated with the project as a history. The projects can last several years and so at any one time a project will have a live contract record for each year as well as any number of historic contracts for that year. All of which is incidental but I'm trying to provide some context. If the contracts where for the year - it would be easy and I'd just store the Year either as a date field with the 1st of January or just an Integer. However the contracts run for financial years and I don't know how to approach this. I don't want a separate table containing the financial years as I don't want the users to have to maintain this. I don't want to store the financial year as a string "2009/2010" as this is not ideal for sorting/extracting the data. Any ideas will be helpful, my best so far is to have starting and ending year in 2 columns and just "KNOW" that starting is April of the year etc Thanks

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  • how to identify an argument as a "Year" in PERL

    - by dexter
    i have created a file argument.pl which takes several arguments first of which should be in form of a year example of argument : 2010 23 type here 2010 is a year my code does something like: use strict; use warning use Date::Calc qw(:all); my ($startyear, $startmonth, $startday) = Today(); my $weekofyear = (Week_of_Year ($startyear,$startmonth,$startday))[0]; my $Year = $startyear; ... ... if ($ARGV[0]) { $Year = $ARGV[0]; } here this code fill $Year with "current year" if if $ARGV[0] is null or doesn't exists now here instead of if ($ARGV[0]) is it possible to check that the value in $ARGV[0] is a valid year (like 2010, 1976,1999 etc.)

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  • Converting a year from 4 digit to 2 digit and back again in C#

    - by Mike Wills
    My credit card processor requires I send a two-digit year from the credit card expiration date. Here is how I a currently processing: I put a DropDownList of the 4-digit year on the page. I validate the expiration date in a DateTime field to be sure that the expiration date being passed to the CC processor isn't expired. I send a two-digit year to the CC processor (as required). I do this via a substring of the value from the year DDL. Is there a method out there to convert a four-digit year to a two-digit year. I am not seeing anything on the DateTime object. Or should I just keep processing it as I am?

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  • Correct algorith for checking leap year

    - by Debanjan
    What will be the exact definition of leap year ? AFAIK "A year which is divisible by 4 is a leap year.But for century years' the years which are divisible by 400 is a leap year." But that definition makes 100, 200, 300, 400.... upto 1700 NOT LEAP years! But in Gregorian calender all of them are all leap year,check this out. You can also try "call 1700" in Linux to verify. So the correct algorithm foe leap year would be if ( (year % 4 == 0) && ( year <= 1700 || year % 100 != 0 || year % 400 == 0 )) printf("%d is a leap year.\n", year); else printf("%d is not a leap year.\n", year); But is this specific to Gregorian calender ? Even if that is the case why is it not mentioned here ? Regards, PS:The history of Gregorian callender seems interesting check out the September month of 1752.

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  • Xpath question Xml Xpath

    - by Ibrar Afzal
    I need an xpath expression that would return the value of I need to get the value of this node. the value to extract is my xpath expression is //rates/rate[loantype='30-Year Fixed Rate'] The issue hre is that there are three value each node has a subtype element. Beside fileter for loantype I also need to filter for subtype. I am not sure how to do it in xpath. I have the following xml 40-Year Fixed Rate A 3 5.375 1.000 5.491 0 1 40-Year Fixed Rate B 5.500 0.500 5.579 0 1 40-Year Fixed Rate C 5.625 0.000 5.667 0 1 30-Year Fixed Rate A 3 5.000 1.000 5.134 0 1 30-Year Fixed Rate B 5.125 0.500 5.215 0 1 30-Year Fixed Rate C 5.250 0.000 5.297 0 1 20-Year Fixed Rate A 3 4.875 1.000 5.055 0 1 20-Year Fixed Rate B 5.000 0.500 5.121 0 1 20-Year Fixed Rate C 5.125 0.000 5.187 0 1 15-Year Fixed Rate A 3 4.250 1.000 4.467 0 1 15-Year Fixed Rate B 4.375 0.500 4.512 0 1 15-Year Fixed Rate C 4.500 0.000 4.570 0 1 10-Year Fixed Rate A 3 4.125 1.000 4.435 0 1 10-Year Fixed Rate B 4.250 0.500 4.454 0 1 10-Year Fixed Rate C 4.375 0.000 4.473 0 1 High-Balance 15-Year Fixed Rate D 3 4.250 1.000 4.461 0 1 High-Balance 15-Year Fixed Rate B 4.375 0.500 4.512 0 1 High-Balance 15-Year Fixed Rate C 4.500 0.000 4.563 0 1 High-Balance 30-Year Fixed Rate D 3 5.000 1.000 5.130 0 1 High-Balance 30-Year Fixed Rate B 5.125 0.500 5.211 0 1 High-Balance 30-Year Fixed Rate C 5.250 0.000 5.293 0 1 30-Year Fixed Rate Jumbo A 2 5.125 1.000 5.254 1 1 30-Year Fixed Rate Jumbo B 5.250 0.500 5.336 1 1 30-Year Fixed Rate Jumbo C 5.375 0.000 5.417 1 1 -- 15-Year Fixed Rate Jumbo A 2 5.000 1.000 5.220 1 1 15-Year Fixed Rate Jumbo B 5.125 0.500 5.270 1 1 15-Year Fixed Rate Jumbo C 5.250 0.000 5.320 1 1 -- 3/1 30-Year Adjustable Rate A 3 3.625 1.000 3.431 0 0 3/1 30-Year Adjustable Rate B 3.875 0.500 3.448 0 0 3/1 30-Year Adjustable Rate C 4.125 0.000 3.465 0 0 3/1 40-Year Adjustable Rate A 3 3.875 1.000 3.438 0 0 3/1 40-Year Adjustable Rate B 4.125 0.500 3.453 0 0 3/1 40-Year Adjustable Rate C 4.375 0.000 3.467 0 0 5/1 30-Year Adjustable Rate A 3 3.375 1.000 3.401 0 0 5/1 30-Year Adjustable Rate B 3.625 0.500 3.457 0 0 5/1 30-Year Adjustable Rate C 3.875 0.000 3.514 0 0 5/1 40-Year Adjustable Rate A 3 3.625 1.000 3.441 0 0 5/1 40-Year Adjustable Rate B 3.875 0.500 3.481 0 0 5/1 40-Year Adjustable Rate C 4.125 0.000 3.531 0 0 7/1 30-Year Adjustable Rate A 3 3.875 1.000 3.670 0 0 7/1 30-Year Adjustable Rate B 4.125 0.500 3.755 0 0 7/1 30-Year Adjustable Rate C 4.375 0.000 3.841 0 0 10/1 30-Year Adjustable Rate A 3 4.375 1.000 4.092 0 0 10/1 30-Year Adjustable Rate B 4.625 0.500 4.217 0 0 10/1 30-Year Adjustable Rate C 4.875 0.000 4.342 0 0 -- 2/2 ARM 30-Year (Purchase only) DH 5.250 0.000 3.709 0 0 -- High-Balance 5/1 30-Year Adjustable Rate D 3 3.375 1.000 3.366 0 0 High-Balance 5/1 30-Year Adjustable Rate B 3.625 0.500 3.404 0 0 High-Balance 5/1 30-Year Adjustable Rate C 3.875 0.000 3.454 0 0 High-Balance 7/1 30-Year Adjustable Rate D 3 3.875 1.000 3.670 0 0 High-Balance 7/1 30-Year Adjustable Rate B 4.125 0.500 3.755 0 0 High-Balance 7/1 30-Year Adjustable Rate C 4.375 0.000 3.841 0 0 3/1 30-Year Jumbo Adjustable Rate A 2 4.875 1.000 3.719 1 0 3/1 30-Year Jumbo Adjustable Rate B 5.000 0.500 3.708 1 0 3/1 30-Year Jumbo Adjustable Rate C 5.125 0.000 3.704 1 0 -- 3/1 40-Year Jumbo Adjustable Rate A 2 5.250 1.000 3.733 1 0 3/1 40-Year Jumbo Adjustable Rate B 5.375 0.500 3.727 1 0 3/1 40-Year Jumbo Adjustable Rate C 5.500 0.000 3.725 1 0 -- 5/1 30-Year Jumbo Adjustable Rate A 3 4.375 1.000 3.791 1 0 5/1 30-Year Jumbo Adjustable Rate B 4.500 0.500 3.803 1 0 5/1 30-Year Jumbo Adjustable Rate C 4.625 0.000 3.814 1 0 5/1 40-Year Jumbo Adjustable Rate A 2 5.000 1.000 3.922 1 0 5/1 40-Year Jumbo Adjustable Rate B 5.125 0.500 3.925 1 0 5/1 40-Year Jumbo Adjustable Rate C 5.250 0.000 3.936 1 0 -- 7/1 30-Year Jumbo Adjustable Rate A 3 4.950 1.000 4.261 1 0 7/1 30-Year Jumbo Adjustable Rate B 5.075 0.500 4.286 1 0 7/1 30-Year Jumbo Adjustable Rate C 5.200 0.000 4.311 1 0 2/2 ARM 30-Year Jumbo (Purchase only) DH 6.500 0.000 4.260 1 0 -- 30 Due in 7 Fixed Rate JUMBO Balloon A 6.375 1.000 6.613 1 0 30 Due in 7 Fixed Rate JUMBO Balloon B 6.500 0.500 6.625 1 0 40 due in 7 Fixed Rate offer1 5.250 0.000 5.374 0 0 1 40 Due in 7 Fixed Rate JUMBO Balloon offer2 6.500 0.000 6.625 1 0 1 Interest Only HELOC A To 80% LTV 3.250 0 1 Home Equity Loan - 7Yrs A Up to $100,000.00 Up to 75% LTV 6.000 6.000 0 2 Home Equity Loan - 7Yrs A $100,000.01 - $250,000.00 Up to 75% LTV 6.00 6.153 0 2 Home Equity Loan - 7Yrs A Up to $100,000.00 Up to 80% LTV 6.250 6.250 0 2 Home Equity Loan - 7Yrs A $100,000.01 - $250,000.00 Up to 80% LTV 6.25 6.403 0 2 Home Equity Loan - 7Yrs B $100,000.01 - $250,000.00 Up to 90% LTV 6.99 7.145 0 2 Home Equity Loan - 10,15Yrs C $5,000-$250,000.00 To 75% LTV 6.50 6.612 0 2 Home Equity Loan - 10,15Yrs C $5,000-$250,000.00 To 80% LTV 6.75 6.863 0 2 Home Equity Loan - 10,15Yrs D $5,000-$250,000.00 Up to 90% LTV 7.50 7.614 0 2 Home Equity Loan - 20Yrs E $5,000-$250,000.00 To 75% LTV 7.50 7.566 0 2 Home Equity Loan - 20Yrs E $5,000-$250,000.00 To 80% LTV 7.75 7.817 0 2 Home Equity Loan - 20Yrs F $5,000-$250,000.00 Up to 90% LTV 8.50 8.569 0 2 Equity Edge $5,000-$25,000.00 Up to 125% LTV 12.00 12.188 Current Index 0.350 Prime Index 3.250 03/26/2010

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  • MySQL date query only returns one year, when multiple exist

    - by Bowman
    I'm a part-time designer/developer with a part-time photography business. I've got a database of photos with various bits of metadata attached. I want to query the database and return a list of the years that photos were taken, and the quantity of photos that were taken in that year. In short, I want a list that looks like this: 2010 (35 photos) 2009 (67 photos) 2008 (48 photos) Here's the query I'm using: SELECT YEAR(date) AS year, COUNT(filename) as quantity FROM photos WHERE visible='1' GROUP BY 'year' ORDER BY 'year' DESC Instead of churning out all the possible years (the database includes photos from 2010-2008), this is the sole result: 2010 (35 photos) I've tried a lot of different syntax but at this point I'm giving in and asking for help!

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  • SQLAuthority News – Resolution for New Year 2011

    - by pinaldave
    Today is the first day of the year so I want to write something very light. Last Year: 2010 Last Year was a blast; really traveled a lot. My family and I went on vacation. There I enjoyed being father, rolling on the floor and playing with my daughter. Here is the list of the countries I visited throughout 2010: Singapore (twice) Malaysia (twice) Sri Lanka (thrice) Nepal (once) United States of America (twice) United Arab Emirates (UAE) (once) My daughter who just completed 1 year on September 1, 2010 has so far visited three countries: Singapore, Malaysia and Sri Lanka, where I have done lots of community activities. The list containing all my activities can be found at Pinal Dave’s Community Events. I have written nearly 380 blog posts last year. It would be difficult for me to pick a few. However, I keep a running list of all of my articles over here: All Articles on SQLAuthority.com. I have so far received more than 10,000 email questions during the year and consequently I have done my best to answer most of them. I strongly believe if one would Search SQLAuthority.com blog, they would have found the answer quickly. The best part of 2010 for me was working on SQL Server Health Check and SQL Server Performance Tuning. This Year: 2011 This year, I came up with two simple goals: 1. Personal Goal: Reduce Weight 2. Professional Goal: Stay busy for the entire year with SQL Server Performance Tuning Projects. (Currently January 2011 is booked with performance tuning projects and 40 other days are already booked throughout the year). Future The future is something one cannot exactly guess and one cannot see. I just want to wish all of you the very best for this coming New Year. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: About Me, Pinal Dave, PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, SQLAuthority News, T SQL, Technology

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  • Year dropdown range - when do we stop?

    - by Andrew Heath
    I attended a payroll software demo yesterday wherein the year dropdowns throughout the software ran from 2000 to 2200. Now, we've all been down this road before with 2 digit shortsight, but honestly - a 200 year service life for a Java & Oracle payroll system? Our Board of Directors would be thrilled if the company was even solvent for 1/4th that long. When forced to use a dropdown year select, where do you draw the line?

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  • Ensure Payroll Success with PeopleSoft Year-End Training for U.S. and Canada

    - by Breanne Cooley
    Year-end payroll processing and reporting is a requirement for your business. If you're responsible for completing these processes in either Canada or the United States using the PeopleSoft Payroll application, and if you're new to PeopleSoft Payroll or to performing these processes, consider enrolling in Oracle University's expert training. Our PeopleSoft Payroll specialists will guide you through the necessary steps to ensure you can smoothly and successfully perform your job. Training is specific to the country for which you are performing the processing and reporting. Training lasts one day and is delivered in our Live Virtual Class Format, which helps you avoid travel during this busy season. Here's the training we recommend: PeopleSoft Year-End Payroll - U.S. This course teaches you how to complete U.S. year-end processing and reporting using PeopleSoft Payroll for North America, step-by-step. Update tax reporting setup tables and update employees' income and tax records. Load each employee's year-end data into a single year-end record for processing and reporting.  Identify reports needed to reconcile the year-end data. Correct tax balances and other data as necessary. Generate final print and online W-2 forms and prepare the electronic file for the Social Security Administration.  Enter corrected W-2 information and print a W-2c form. Report periodic retirement distributions and related tax withholding amounts on form 1099-R.   Please Note: this course is intended for organizations using PeopleSoft release 8.81 or higher. PeopleSoft Year-End Payroll – Canada This course covers the steps necessary to perform Canadian year-end processing using Oracle's PeopleSoft Payroll for North America. Explore adjustments, balances, year-end slip processing, common pitfalls and errors and balancing reports.  Produce accurate year-end reporting results such as T4, T4A, RL-1 and RL-2.  Please Note: this course is intended for organizations using PeopleSoft release 8.81 or higher. See you in class! -Oracle University Marketing Team 

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