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  • How to "redefine search" or correct "misspelling" from the database

    - by From.ME.to.YOU
    Hello i want to add new feature to the search in my website. i'm using PHP and MYSQL. mysql database containing a table to the items that the user will search for, for each item there is a "keyword" column that's comma separated keywords "EXAMPLE: cat,dog,horse". after the user search in my website i want to get the words that are let me say "85%" similar to his search keyword, this is for redefine search. and for misspelling i want a service or something that provide if the keyword is correct or misspelled so i get some corrections and check if those exists in the database and then give those corrections to user to change his search keyword. i'm not asking for a solution here ... but if you can direct me in a one way or another that will be great Thanks guys Cheers

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  • Won't connect to the database

    - by user1657958
    I'm confused...I'm using the same code in a different document and in there it's not a problem to get a connection to the database. But in the new document it's just not working...(password, username, database name...all is checked and correct) :-/ <?php define ("DB_HOST", "db1234567.db.hello.com"); // set database host define ("DB_USER", "db1234567"); // set database user define ("DB_PASS","password123"); // set database password define ("DB_NAME","db1234567"); // set database name $link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die("Couldn't make connection."); $db = mysql_select_db(DB_NAME, $link) or die("Couldn't select database"); ?> In the browser I get this: "Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'db1234567'@'123.123.12.12 (using password: YES) in /homepages/12/1234567/test/test.php on line 8 Couldn't make connection." Would be cool if someone could help me :) I'm not seeing any error... Thx!

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  • Synchronizing non-DB SQL Server objects

    - by DigDoug
    There are a number of tools available for synchronizing Tables, Indexes, Views, Stored Procedures and objects within a database. (We love RedGate here, and throw a lot of money their way). However, I'm having a very difficult time finding tools that will help with Jobs, Logins and Linked Servers. Do these things exist? Am I missing something obvious?

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  • DB Strategy for inserting into a high read table (Sql Server)

    - by Tom
    Looking for strategies for a very large table with data maintained for reporting and historical purposes, a very small subset of that data is used in daily operations. Background: We have Visitor and Visits tables which are continuously updated by our consumer facing site. These tables contain information on every visit and visitor, including bots and crawlers, direct traffic that does not result in a conversion, etc. Our back end site allows management of the visitor's (leads) from the front end site. Most of the management occurs on a small subset of our visitors (visitors that become leads). The vast majority of the data in our visitor and visit tables is maintained only for a much smaller subset of user activity (basically reporting type functionality). This is NOT an indexing problem, we have done all we can with indexing and keeping our indexes clean, small, and not fragmented. ps: We do not currently have the budget or expertise for a data warehouse. The problem: We would like the system to be more responsive to our end users when they are querying, for instance, the list of their assigned leads. Currently the query is against a huge data set of mostly irrelevant data. I am pondering a few ideas. One involves new tables and a fairly major re-architecture, I'm not asking for help on that. The other involves creating redundant data, (for instance a Visitor_Archive and a Visitor_Small table) where the larger visitor and visit tables exist for inserts and history/reporting, the smaller visitor1 table would exist for managing leads, sending lead an email, need leads phone number, need my list of leads, etc.. The reason I am reaching out is that I would love opinions on the best way to keep the Visitor_Archive and the Visitor_Small tables in sync... Replication? Can I use replication to replicate only data with a certain column value (FooID = x) Any other strategies?

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  • Codeigniter Active record help

    - by sea_1987
    Hello, I am trying to increment a INT column by 1 if a certain field is not null on an update request, currently I have this update too columns, public function updateCronDetails($transaction_reference, $flag, $log) { $data = array ( 'flag' => $flag, 'log' => "$log" ); $this->db->where('transaction_reference', $transaction_reference); $this->db->update('sy_cron', $data); } What I need to know is how I can check if the value being sent to the log field is NULL and if it is how could I increment a column called count by 1?

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  • Correcting an UPDATE statement (and making it more secure!)

    - by Jess
    I'm trying to a single value in my DB...When I run it through the console, it works correctly (as I'm replacing the variables with numbers and text).. However, My query is not running correctly. It's just prompting a syntax error Here is what I have: "UPDATE books SET readstatus='".$readstatus."' WHERE book_id=".$book_id; This won't work, I also tried doing something like this as I'm told this makes it a bit more secure? : "UPDATE books SET readstatus='{$readstatus}', WHERE read_id='{read_id}'"; This does not prompt any errors, but no change is happeneing to the value in the DB, I'm guessing the syntax is incorrect.

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  • MySql - JSON data not showing in html

    - by Ramzie
    I'm trying to create a drop down list from a MySql. The php is successfully fetching the data from the MySql. But my problem is the data is not showing on the drop down list in my HTML page? json_mysql_data2.php header("Content-Type: application/json"); require_once("con.php"); $i=0; $jsonData = array(); foreach ($conn_db->query("SELECT customerID FROM customers WHERE furniture='33' ") as $result){ $i++; $jsonData["article".$i]=$result['customerID']; } echo json_encode($jsonData); myJS.js $(document).ready(function(){ var ddlist = document.getElementById("ddlist"); var hr = new XMLHttpRequest(); hr.open("GET", "json_mysql_data2.php", true); hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); hr.onreadystatechange = function() { if(hr.readyState == 4 && hr.status == 200) { var d = JSON.parse(hr.responseText); for(var o in d){ if(d[o].title){ ddlist.innerHTML += '</option><option value='+d[o].title+'</option>'; } } } } hr.send("null"); ddlist.innerHTML = "Loading Customer ID...."; }); html <script src="myJS.js" type="text/javascript"></script> </head> <body> <div class="dlist"> Customer ID: <select id='EmpLst' name="dwlist" onchange='document.getElementById("val1").value = this.value;'><option value="">SELECT STUDENT ID</option> <div id="ddlist"></div> </select> </div>

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  • Is this a ridiculous way to structure a DB schema, or am I completely missing something?

    - by Jim
    I have done a fair bit of work with relational databases, and think I understand the basic concepts of good schema design pretty well. I recently was tasked with taking over a project where the DB was designed by a highly-paid consultant. Please let me know if my gut intinct - "WTF??!?" - is warranted, or is this guy such a genius that he's operating out of my realm? DB in question is an in-house app used to enter requests from employees. Just looking at a small section of it, you have information on the users, and information on the request being made. I would design this like so: User table: UserID (primary Key, indexed, no dupes) FirstName LastName Department Request table RequestID (primary Key, indexed, no dupes) <...> various data fields containing request details UserID -- foreign key associated with User table Simple, right? Consultant designed it like this (with sample data): UsersTable UserID FirstName LastName 234 John Doe 516 Jane Doe 123 Foo Bar DepartmentsTable DepartmentID Name 1 Sales 2 HR 3 IT UserDepartmentTable UserDepartmentID UserID Department 1 234 2 2 516 2 3 123 1 RequestTable RequestID UserID <...> 1 516 blah 2 516 blah 3 234 blah The entire database is constructed like this, with every piece of data encapsulated in its own table, with numeric IDs linking everything together. Apparently the consultant had read about OLAP and wanted the 'speed of integer lookups' He also has a large number of stored procedures to cross reference all of these tables. Is this valid design for a small to mid-sized SQL DB? Thanks for comments/answers...

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  • Implementing set operations in TSQL

    - by dotneteer
    SQL excels at operating on dataset. In this post, I will discuss how to implement basic set operations in transact SQL (TSQL). The operations that I am going to discuss are union, intersection and complement (subtraction).   Union Intersection Complement (subtraction) Implementing set operations using union, intersect and except We can use TSQL keywords union, intersect and except to implement set operations. Since we are in an election year, I will use voter records of propositions as an example. We create the following table and insert 6 records into the table. declare @votes table (VoterId int, PropId int) insert into @votes values (1, 30) insert into @votes values (2, 30) insert into @votes values (3, 30) insert into @votes values (4, 30) insert into @votes values (4, 31) insert into @votes values (5, 31) Voters 1, 2, 3 and 4 voted for proposition 30 and voters 4 and 5 voted for proposition 31. The following TSQL statement implements union using the union keyword. The union returns voters who voted for either proposition 30 or 31. select VoterId from @votes where PropId = 30 union select VoterId from @votes where PropId = 31 The following TSQL statement implements intersection using the intersect keyword. The intersection will return voters who voted only for both proposition 30 and 31. select VoterId from @votes where PropId = 30 intersect select VoterId from @votes where PropId = 31 The following TSQL statement implements complement using the except keyword. The complement will return voters who voted for proposition 30 but not 31. select VoterId from @votes where PropId = 30 except select VoterId from @votes where PropId = 31 Implementing set operations using join An alternative way to implement set operation in TSQL is to use full outer join, inner join and left outer join. The following TSQL statement implements union using full outer join. select Coalesce(A.VoterId, B.VoterId) from (select VoterId from @votes where PropId = 30) A full outer join (select VoterId from @votes where PropId = 31) B on A.VoterId = B.VoterId The following TSQL statement implements intersection using inner join. select Coalesce(A.VoterId, B.VoterId) from (select VoterId from @votes where PropId = 30) A inner join (select VoterId from @votes where PropId = 31) B on A.VoterId = B.VoterId The following TSQL statement implements complement using left outer join. select Coalesce(A.VoterId, B.VoterId) from (select VoterId from @votes where PropId = 30) A left outer join (select VoterId from @votes where PropId = 31) B on A.VoterId = B.VoterId where B.VoterId is null Which one to choose? To choose which technique to use, just keep two things in mind: The union, intersect and except technique treats an entire record as a member. The join technique allows the member to be specified in the “on” clause. However, it is necessary to use Coalesce function to project sets on the two sides of the join into a single set.

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  • Join 2 children tables with a parent tables without duplicated

    - by user1847866
    Problem I have 3 tables: People, Phones and Emails. Each person has an UNIQUE ID, and each person can have multiple numbers or multiple emails. Simplified it looks like this: +---------+----------+ | ID | Name | +---------+----------+ | 5000003 | Amy | | 5000004 | George | | 5000005 | John | | 5000008 | Steven | | 8000009 | Ashley | +---------+----------+ +---------+-----------------+ | ID | Number | +---------+-----------------+ | 5000005 | 5551234 | | 5000005 | 5154324 | | 5000008 | 2487312 | | 8000009 | 7134584 | | 5000008 | 8451384 | +---------+-----------------+ +---------+------------------------------+ | ID | Email | +---------+------------------------------+ | 5000005 | [email protected] | | 5000005 | [email protected] | | 5000008 | [email protected] | | 5000008 | [email protected] | | 5000008 | [email protected] | | 8000009 | [email protected] | | 5000004 | [email protected] | +---------+------------------------------+ I am trying to joining them together without duplicates. It works great, when I try to join only Emails with People or only Phones with People. SELECT People.Name, People.ID, Phones.Number FROM People LEFT OUTER JOIN Phones ON People.ID=Phones.ID ORDER BY Name, ID, Number; +----------+---------+-----------------+ | Name | ID | Number | +----------+---------+-----------------+ | Steven | 5000008 | 8451384 | | Steven | 5000008 | 24887312 | | John | 5000005 | 5551234 | | John | 5000005 | 5154324 | | George | 5000004 | NULL | | Ashley | 8000009 | 7134584 | | Amy | 5000003 | NULL | +----------+---------+-----------------+ SELECT People.Name, People.ID, Emails.Email FROM People LEFT OUTER JOIN Emails ON People.ID=Emails.ID ORDER BY Name, ID, Email; +----------+---------+------------------------------+ | Name | ID | Email | +----------+---------+------------------------------+ | Steven | 5000008 | [email protected] | | Steven | 5000008 | [email protected] | | Steven | 5000008 | [email protected] | | John | 5000005 | [email protected] | | John | 5000005 | [email protected] | | George | 5000004 | [email protected] | | Ashley | 8000009 | [email protected] | | Amy | 5000003 | NULL | +----------+---------+------------------------------+ However, when I try to join Emails and Phones on People - I get this: SELECT People.Name, People.ID, Phones.Number, Emails.Email FROM People LEFT OUTER JOIN Phones ON People.ID = Phones.ID LEFT OUTER JOIN Emails ON People.ID = Emails.ID ORDER BY Name, ID, Number, Email; +----------+---------+-----------------+------------------------------+ | Name | ID | Number | Email | +----------+---------+-----------------+------------------------------+ | Steven | 5000008 | 8451384 | [email protected] | | Steven | 5000008 | 8451384 | [email protected] | | Steven | 5000008 | 8451384 | [email protected] | | Steven | 5000008 | 24887312 | [email protected] | | Steven | 5000008 | 24887312 | [email protected] | | Steven | 5000008 | 24887312 | [email protected] | | John | 5000005 | 5551234 | [email protected] | | John | 5000005 | 5551234 | [email protected] | | John | 5000005 | 5154324 | [email protected] | | John | 5000005 | 5154324 | [email protected] | | George | 5000004 | NULL | [email protected] | | Ashley | 8000009 | 7134584 | [email protected] | | Amy | 5000003 | NULL | NULL | +----------+---------+-----------------+------------------------------+ What happens is - if a Person has 2 numbers, all his emails are shown twice (They can not be sorted! which means they can not be removed by @last) What I want: Bottom line, playing with the @last, I want to end up with somethig like this, but @last won't work if I don't arrange ORDER columns in the righ way - and this seems like a big problem..Orderin the email column. Because seen from the example above: Steven has 2 phone number and 3 emails. The JOIN Emails with Numbers happens with each email - thus duplicated values that can not be sorted (SORT BY does not work on them). **THIS IS WHAT I WANT** +----------+---------+-----------------+------------------------------+ | Name | ID | Number | Email | +----------+---------+-----------------+------------------------------+ | Steven | 5000008 | 8451384 | [email protected] | | | | 24887312 | [email protected] | | | | | [email protected] | | John | 5000005 | 5551234 | [email protected] | | | | 5154324 | [email protected] | | George | 5000004 | NULL | [email protected] | | Ashley | 8000009 | 7134584 | [email protected] | | Amy | 5000003 | NULL | NULL | +----------+---------+-----------------+------------------------------+ Now I'm told that it's best to keep emails and number in separated tables because one can have many emails. So if it's such a common thing to do, what isn't there a simple solution? I'd be happy with a PHP Solution aswell. What I know how to do by now that satisfies it, but is not as pretty. If I do it with GROUP_CONTACT I geat a satisfactory result, but it doesn't look as pretty: I can't put a "Email type = work" next to it. SELECT People.Ime, GROUP_CONCAT(DISTINCT Phones.Number), GROUP_CONCAT(DISTINCT Emails.Email) FROM People LEFT OUTER JOIN Phones ON People.ID=Phones.ID LEFT OUTER JOIN Emails ON People.ID=Emails.ID GROUP BY Name; +----------+----------------------------------------------+---------------------------------------------------------------------+ | Name | GROUP_CONCAT(DISTINCT Phones.Number) | GROUP_CONCAT(DISTINCT Emails.Email) | +----------+----------------------------------------------+---------------------------------------------------------------------+ | Steven | 8451384,24887312 | [email protected],[email protected],[email protected] | | John | 5551234,5154324 | [email protected],[email protected] | | George | NULL | [email protected] | | Ashley | 7134584 | [email protected] | | Amy | NULL | NULL | +----------+----------------------------------------------+---------------------------------------------------------------------+

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  • Having a problem inserting a foreign key data into a table using a PHP form

    - by Gideon
    I am newbee with PHP and MySQL and need help... I have two tables (Staff and Position) in the database. The Staff table (StaffId, Name, PositionID (fk)). The Position table is populated with different positions (Manager, Supervisor, and so on). The two tables are linked with a PositionID foreign key in the Staff table. I have a staff registration form with textfields asking for the relevant attributes and a dynamically populated drop down list to choose the position. I need to insert the user's entry into the staff table along with the selected position. However, when inserting the data, I get the following error (Cannot add or update a child row: a foreign key constraint fails). How do I insert the position selected by the user into the staff table? Here is some of my code... ... echo "<tr>"; echo "<td>"; echo "*Position:"; echo "</td>"; echo "<td>"; //dynamically populate the staff position drop down list from the position table $position="SELECT PositionId, PositionName FROM Position ORDER BY PositionId"; $exeposition = mysql_query ($position) or die (mysql_error()); echo "<select name=position value=''>Select Position</option>"; while($positionarray=mysql_fetch_array($exeposition)) { echo "<option value=$positionarray[PositionId]>$positionarray[PositionName]</option>"; } echo "</select>"; echo "</td>"; echo "</tr>" //the form is processed with the code below $FirstName = $_POST['firstname']; $LastName = $_POST['lastname']; $Address = $_POST['address']; $City = $_POST['city']; $PostCode = $_POST['postcode']; $Country = $_POST['country']; $Email = $_POST['email']; $Password = $_POST['password']; $ConfirmPass = $_POST['confirmpass']; $Mobile = $_POST['mobile']; $NI = $_POST['nationalinsurance']; $PositionId = $_POST[$positionarray['PositionId']]; //format the dob for the database $dobpart = explode("/", $_POST['dob']); $formatdob = $dobpart[2]."-".$dobpart[1]."-".$dobpart[0]; $DOB = date("Y-m-d", strtotime($formatdob)); $newReg = "INSERT INTO Staff (FirstName, LastName, Address, City, PostCode, Country, Email, Password, Mobile, DOB, NI, PositionId) VALUES ('".$FirstName."', '".$LastName."', '".$Address."', '".$City."', '".$PostCode."', '".$Country."', '".$Email."', '".$Password."', ".$Mobile.", '".$DOB."', '".$NI."', '".$PostionId."')"; Your time and help is surely appreciated.

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  • PHP-How to Pass Multiple Value In Form Field

    - by Tall boY
    hi i have a php based sorting method with drop down menu to sort no of rows, it is working fine. i have another sorting links to sort id & title, it is also working fine. but together they are not working fine. what happens is that when i sort(say by title) using links, result gets sorted by title, then if i sort rows using drop down menu rows get sorted but result gets back to default of id sort. sorting codes for id & tite is if ($orderby == 'title' && $sortby == 'asc') {echo " <li id='scurrent'><a href='?rpp=$rowsperpage&order=title&sort=asc'>title-asc:</a></li> ";} else {echo " <li><a href='?rpp=$rowsperpage&order=title&sort=asc'>title-asc:</a></li> ";} if ($orderby == 'title' && $sortby == 'desc') {echo " <li id='scurrent'><a href='?rpp=$rowsperpage&order=title&sort=desc'>title-desc:</a></li> ";} else {echo " <li><a href='?rpp=$rowsperpage&order=title&sort=desc'>title-desc:</a></li> ";} if ($orderby == 'id' && $sortby == 'asc') {echo " <li id='scurrent'><a href='?rpp=$rowsperpage&order=id&sort=asc'>id-asc:</a></li> ";} else {echo " <li><a href='?rpp=$rowsperpage&order=id&sort=asc'>id-asc:</a></li> ";} if ($orderby == 'id' && $sortby == 'desc') {echo " <li id='scurrent'><a href='?rpp=$rowsperpage&order=id&sort=desc'>id-desc:</a></li> ";} else {echo " <li><a href='?rpp=$rowsperpage&order=id&sort=desc'>id-desc:</a></li> ";} ?> sorting codes for rows is <form action="is-test.php" method="get"> <select name="rpp" onchange="this.form.submit()"> <option value="10" <?php if ($rowsperpage == 10) echo 'selected="selected"' ?>>10</option> <option value="20" <?php if ($rowsperpage == 20) echo 'selected="selected"' ?>>20</option> <option value="30" <?php if ($rowsperpage == 30) echo 'selected="selected"' ?>>30</option> </select> </form> this method passes only rows per page(rpp) into url. i want it to pass order, sort& rpp. is there a way around to pass multiple values in form fields like this. <form action="is-test.php" method="get"> <select name="rpp, order, sort" onchange="this.form.submit()"> <option value="10, $orderby, $sortby" <?php if ($rowsperpage == 10) echo 'selected="selected"' ?>>10</option> <option value="20, $orderby, $sortby" <?php if ($rowsperpage == 20) echo 'selected="selected"' ?>>20</option> <option value="30, $orderby, $sortby" <?php if ($rowsperpage == 30) echo 'selected="selected"' ?>>30</option> </select> </form> this may seem silly but it just to give you an idea of what i am trying to implement,(i am very new to php) please suggest any way to make this work. thanks

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  • Ruby on Rail using MYSQL database

    - by Joseph Misiti
    Hey guys, New to rails, trying to figure out something simple. Seems as though I cannot migrate a very simple mysql database using "rake db:migrate" command. Here is the issue: I know rails defaults to sqllite right now, but I need to use mysql for a series of reasons. Use the following commands rails -d mysql MyMoviesSQL cd MyMoviesSQL script/generate scaffold Movies title:string rating:integer rake db:migrate never get past here because i see the following error: in /Users/user/websites/MyMovieSQL) rake aborted! NoMethodError: undefined method `ord' for 0:Fixnum: SET NAMES 'utf8' (See full trace by running task with --trace) using trace XXXXX-macbook-pro:MyMovieSQL user$ rake db:migrate --trace (in /Users/user/websites/MyMovieSQL) ** Invoke db:migrate (first_time) ** Invoke environment (first_time) ** Execute environment ** Execute db:migrate rake aborted! NoMethodError: undefined method ord' for 0:Fixnum: SET NAMES 'utf8' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/abstract_adapter.rb:219:inlog' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/mysql_adapter.rb:323:in execute' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/mysql_adapter.rb:599:inconfigure_connection' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/mysql_adapter.rb:594:in connect' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/mysql_adapter.rb:203:ininitialize' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/mysql_adapter.rb:75:in new' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/mysql_adapter.rb:75:inmysql_connection' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/abstract/connection_pool.rb:223:in send' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/abstract/connection_pool.rb:223:innew_connection' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/abstract/connection_pool.rb:245:in checkout_new_connection' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/abstract/connection_pool.rb:188:incheckout' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/abstract/connection_pool.rb:184:in loop' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/abstract/connection_pool.rb:184:incheckout' /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/1.8/monitor.rb:242:in synchronize' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/abstract/connection_pool.rb:183:incheckout' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/abstract/connection_pool.rb:98:in connection' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/abstract/connection_pool.rb:326:inretrieve_connection' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/abstract/connection_specification.rb:123:in retrieve_connection' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/connection_adapters/abstract/connection_specification.rb:115:inconnection' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/migration.rb:435:in initialize' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/migration.rb:400:innew' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/migration.rb:400:in up' /Library/Ruby/Gems/1.8/gems/activerecord-2.3.5/lib/active_record/migration.rb:383:inmigrate' /Library/Ruby/Gems/1.8/gems/rails-2.3.5/lib/tasks/databases.rake:116 /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:636:in call' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:636:inexecute' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:631:in each' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:631:inexecute' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:597:in invoke_with_call_chain' /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/1.8/monitor.rb:242:insynchronize' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:590:in invoke_with_call_chain' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:583:ininvoke' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:2051:in invoke_task' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:2029:intop_level' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:2029:in each' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:2029:intop_level' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:2068:in standard_exception_handling' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:2023:intop_level' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:2001:in run' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:2068:instandard_exception_handling' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/lib/rake.rb:1998:in run' /Library/Ruby/Gems/1.8/gems/rake-0.8.7/bin/rake:31 /usr/bin/rake:19:inload' /usr/bin/rake:19 no clue what is going on, if they want me to add a patch because the methods does not exist, please tell me which file to add it to, and also, how in the future do i figure out which file I need to patch (I see it looks like its a method in FixNum class) here is a patch to a problem that looks similar, but its a different version of ruby http://www.mail-archive.com/[email protected]/msg00250.html versions rails 2.3.5 ruby 1.8.6 gem list yeilds: * LOCAL GEMS * actionmailer (2.3.5, 1.3.6) actionpack (2.3.5, 1.13.6) actionwebservice (1.2.6) activerecord (2.3.5, 1.15.6) activeresource (2.3.5) activesupport (2.3.5, 1.4.4) acts_as_ferret (0.4.1) capistrano (2.0.0) cgi_multipart_eof_fix (2.5.0) daemons (1.0.9) dbi (0.4.3) deprecated (2.0.1) dnssd (0.6.0) fastthread (1.0.1) fcgi (0.8.7) ferret (0.11.4) gem_plugin (0.2.3) highline (1.2.9) hpricot (0.6) libxml-ruby (0.9.5, 0.3.8.4) mongrel (1.1.4) needle (1.3.0) net-sftp (1.1.0) net-ssh (1.1.2) rack (1.0.1) rails (2.3.5) rake (0.8.7, 0.7.3) RedCloth (3.0.4) ruby-openid (1.1.4) ruby-yadis (0.3.4) rubygems-update (1.3.6) rubynode (0.1.3) sqlite3-ruby (1.2.1) termios (0.9.4) thanks in advanced

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  • T-SQL How To: Compare and List Duplicate Entries in a Table

    - by Dan7el
    SQL Server 2000. Single table has a list of users that includes a unique user ID and a non-unique user name. I want to search the table and list out any users that share the same non-unique user name. For example, my table looks like this: ID User Name Name == ========= ==== 0 parker Peter Parker 1 parker Mary Jane Parker 2 heroman Joseph (Joey) Carter Jones 3 thehulk Bruce Banner What I want to do is do a SELECT and have the result set be: ID User Name Name == ========= ==== 0 parker Peter Parker 1 parker Mary Jane Parker from my table. I'm not a T-SQL guru. I can do the basic joins and such, but I'm thinking there must be an elegant way of doing this. Barring elegance, there must be ANY way of doing this. I appreciate any methods that you can help me with on this topic. Thanks! ---Dan---

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  • Object Reference Error filling a datarow

    - by JPJedi
    This is the code: Dim dr() As DataRow = DataSet.Tables("TableName").Select("EVENTNAME = '" & name & "'") I get an "Object reference not set to an instance of an object." Error when this line is executed. It is looping through a list of selected items in a listbox. I think it has to do with how I have the datarow declared because I can look at the name and I see it ok and I also do a null check on the name before I use it. Visual Studio 2008, VB.NET. Any ideas? Yep it was a wrong table name. I guess after looking at the code for 8 hours that minor detail I just wasn't thinking to check. Thanks!

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  • create table from another table in different database in sql server 2005

    - by Greg
    Hi, I have a database "temp" with table "A". I created new database "temp2". I want to copy table "A" from "temp" to a new table in "temp2" . I tried this statement but it says I have incorrect syntax, here is the statement: CREATE TABLE B IN 'temp2' AS (SELECT * FROM A IN 'temp'); Here is the error: Msg 156, Level 15, State 1, Line 2 Incorrect syntax near the keyword 'IN'. Msg 156, Level 15, State 1, Line 3 Incorrect syntax near the keyword 'IN'. Anyone knows whats the problem? Thanks in advance, Greg

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  • Preserve onchange for a dropdown list when setting the value with Javascript.

    - by Zac Altman
    I have a dropdown list with a piece of code that is run when the value is changed: <select name="SList" onchange="javascript:setTimeout('__doPostBack(\'SList\',\'\')', 0)" id="SList"> Everything works fine when manually done. As an option is selected, the onchange code is called. The problem begins when I try to change the selected value using a piece of Javscript. I want to be able to automatically change the selected option using JS, whilst still having the onchange code called, exactly as if done manually. I try calling this: form.SList.value = "33"; The right option is selected, but the onchange code does not get called. So then I try doing this: form.SList.value = "33"; javascript:setTimeout('__doPostBack(\'SList\',\'\')', 0); The right value is not selected and nothing happens. FYI, the code is done in ASP.NET and Javascript. What can I run to change the selected option whilst still calling the onchange code?

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  • SQL: many-to-many relationship, IN condition

    - by Maarten
    I have a table called transactions with a many-to-many relationship to items through the items_transactions table. I want to do something like this: SELECT "transactions".* FROM "transactions" INNER JOIN "items_transactions" ON "items_transactions".transaction_id = "transactions".id INNER JOIN "items" ON "items".id = "items_transactions".item_id WHERE (items.id IN (<list of items>)) But this gives me all transactions that have one or more of the items in the list associated with it and I only want it to give me the transactions that are associated with all of those items. Any help would be appreciated.

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  • Having trouble deselecting all jquery tabs

    - by Julian
    I set up some jQuery tabs to start off with no tabs selected like this: $('#tabs').tabs( { selected: -1 } ); Then I also have a separate link that when pressed needs to deselect all the tabs. $("#deselectButton").click(function(){ $('#tabs').tabs( 'select' , -1 ) }); or $("#deselectButton").click(function(){ $('#tabs').tabs( 'selected' , -1 ) }); The deselectButton click does deselect the tabs content, however the tabs title remains active with the class 'ui-tabs-selected ui-state-active'. What is the correct way to deselect all the tabs?

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  • postgres counting one record twice if it meets certain criteria

    - by Dashiell0415
    I thought that the query below would naturally do what I explain, but apparently not... My table looks like this: id | name | g | partner | g2 1 | John | M | Sam | M 2 | Devon | M | Mike | M 3 | Kurt | M | Susan | F 4 | Stacy | F | Bob | M 5 | Rosa | F | Rita | F I'm trying to get the id where either the g or g2 value equals 'M'... But, a record where both the g and g2 values are 'M' should return two lines, not 1. So, in the above sample data, I'm trying to return: $q = pg_query("SELECT id FROM mytable WHERE ( g = 'M' OR g2 = 'M' )"); 1 1 2 2 3 4 But, it always returns: 1 2 3 4

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  • How to make a php function contain mysql commands

    - by bob
    I want to create a simple menu function which can call it example get_menu() Here is my current code. <?php $select = 'SELECT * FROM pages'; $query = $db->rq($select); while ($page = $db->fetch($query)) { $id = $page['id']; $title = $page['title']; ?> <a href="page.php?id<?php echo $id; ?>" title="<?php echo $title; ?>"><?php echo $title; ?></a> <?php } ?> How to do that in? function get_menu() { } Let me know.

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  • SQL Server: query database user roles for all databases in server

    - by atricapilla
    I would like to make a query for database user roles for all databases in my sql server instance. I modified a query from sp_helpuser: select u.name ,case when (r.principal_id is null) then 'public' else r.name end ,l.default_database_name ,u.default_schema_name ,u.principal_id from sys.database_principals u left join (sys.database_role_members m join sys.database_principals r on m.role_principal_id = r.principal_id) on m.member_principal_id = u.principal_id left join sys.server_principals l on u.sid = l.sid where u.type <> 'R' How can I modify this to query from all databases? What is the link between sys.databases and sys.database_principals?

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  • How do I get a count of events each day with SQL?

    - by upl8
    I have a table that looks like this: Timestamp Event User ================ ===== ===== 1/1/2010 1:00 PM 100 John 1/1/2010 1:00 PM 103 Mark 1/2/2010 2:00 PM 100 John 1/2/2010 2:05 PM 100 Bill 1/2/2010 2:10 PM 103 Frank I want to write a query that shows the events for each day and a count for those events. Something like: Date Event EventCount ======== ===== ========== 1/1/2010 100 1 1/1/2010 103 1 1/2/2010 100 2 1/2/2010 103 1 The database is SQL Server Compact, so it doesn't support all the features of the full SQL Server. The query I have written so far is SELECT DATEADD(dd, DATEDIFF(dd, 0, Timestamp), 0) as Date, Event, Count(Event) as EventCount FROM Log GROUP BY Timestamp, Event This almost works, but EventCount is always 1. How can I get SQL Server to return the correct counts? All fields are mandatory.

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  • PHP Multiple User Login Form - Navigation to Different Pages Based on Login Credentials

    - by Zulu Irminger
    I am trying to create a login page that will send the user to a different index.php page based on their login credentials. For example, should a user with the "IT Technician" role log in, they will be sent to "index.php", and if a user with the "Student" role log in, they will be sent to the "student/index.php" page. I can't see what's wrong with my code, but it's not working... I'm getting the "wrong login credentials" message every time I press the login button. My code for the user login page is here: <?php session_start(); if (isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'IT Technician')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'Student')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/student/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> And the form that the data is pulled from is shown here: <form id="LoginForm" name="LoginForm" method="post" action="http://www.zuluirminger.com/SchoolAdmin/user_login.php"> User Name:<br /> <input type="text" name="username" id="username" size="50" /><br /> <br /> Password:<br /> <input type="password" name="password" id="password" size="50" /><br /> <br /> Log in as: <select name="role" id="role"> <option value="">...</option> <option value="Head">Head</option> <option value="Deputy Head">Deputy Head</option> <option value="IT Technician">IT Technician</option> <option value="Pastoral Care">Pastoral Care</option> <option value="Bursar">Bursar</option> <option value="Secretary">Secretary</option> <option value="Housemaster">Housemaster</option> <option value="Teacher">Teacher</option> <option value="Tutor">Tutor</option> <option value="Sanatorium Staff">Sanatorium Staff</option> <option value="Kitchen Staff">Kitchen Staff</option> <option value="Parent">Parent</option> <option value="Student">Student</option> </select><br /> <br /> <input type="submit" name = "button" id="button" value="Log In" onclick="javascript:return validateLoginForm();" /> </h3> </form> Once logged in (and should the correct page be loaded, the validation code I have at the top of the script looks like this: <?php session_start(); if (!isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/user_login.php"); exit(); } $managerID = preg_replace('#[^0-9]#i', '', $_SESSION["id"]); $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["manager"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if ($existCount == 0) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> Just so you're aware, the database table has the following fields: id, username, password and role. Any help would be greatly appreciated! Many thanks, Zulu

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