ahvceo Posted May 23, 2009 Share Posted May 23, 2009 Hi All, 20 years ago I thought I knew how to do a query but I guess I have forgotten. I am trying to update a single value in one row in one table in a database with 5 tables. I thought the following code would work but no joy. $payment = "3.50"; $ref = $_COOKIE['ref']; mysql_connect($server, $db_user, $db_pass) or die ("Database CONNECT Error (line "); $monthlydue = mysql_db_query($database, "select monthlydue from affiliates where memid=" . $ref); $monthlydue = $monthlydue + $payment mysql_db_query($database, "INSERT INTO affiliates (monthlydue) VALUES ('".$monthlydue."') where memid =" . $ref) or die(mysql_error()); Thyere are no errors in the database values or names because I use the same values and names to add rows with no problem. If it makes any difference the value I am trying to return is a double but I have the same problem with a string when I try to access just the email address. Can anyone tell me what I am doing wrong? Thanks ahvceo PS Would you believe I have looked a over 20 examples of this on the net and not a single one of them addressed the issue of accessing only one value in a table. Every one showed how to return or insert or add or update a complete row, but not one showed how to get or insert a single value in a row. Quote Link to comment Share on other sites More sharing options...
papaface Posted May 23, 2009 Share Posted May 23, 2009 Code should be: <?php $payment = "3.50"; $ref = $_COOKIE['ref']; mysql_connect($server, $db_user, $db_pass) or die ("Database CONNECT Error (line "); mysql_select_db($database); //cant see the $database var defined anywhere.... $monthlydue = mysql_query("select `monthlydue` from `affiliates` where `memid`= '".$ref."'"); while ($row = mysql_fetch_array($monthlydue)) { mysql_query("UPDATE `affiliates` SET `monthlydue` ='".$row['monthlydue'] + $payment."' WHERE `memid` ='".$ref."'"); } ?> You may also be able to do this: <?php $payment = "3.50"; $ref = $_COOKIE['ref']; mysql_connect($server, $db_user, $db_pass) or die ("Database CONNECT Error (line "); mysql_select_db($database); //cant see the $database var defined anywhere.... mysql_query("UPDATE `affiliates` SET `monthlydue` = '`monthlydue`+".$payment."' WHERE `memid` ='".$ref."'"); ?> Quote Link to comment Share on other sites More sharing options...
GingerRobot Posted May 23, 2009 Share Posted May 23, 2009 Hold up? The above code is the working bit to insert things into the database, but your having issues with the other end and getting items from it? If so you might like to check out this tutorial, which covers all the basics: http://www.phpfreaks.com/tutorial/php-basic-database-handling Quote Link to comment Share on other sites More sharing options...
kickstart Posted May 23, 2009 Share Posted May 23, 2009 Hi To the OP. You are mixing up the syntax for an insert and an update. The 2nd suggestion from papaface is probably best. Doing the update there is no need to read it first (although you might want to if you are making use of the other columns). <?php $payment = "3.50"; $ref = $_COOKIE['ref']; $conn = mysql_connect($server, $db_user, $db_pass) or die ("Database CONNECT Error (line "); mysql_select_db($database, $conn); //cant see the $database var defined anywhere.... mysql_query("UPDATE `affiliates` SET `monthlydue` = `monthlydue`+ $payment WHERE `memid` ='$ref'", $conn) or die(mysql_error()); ?> All the best Keith Quote Link to comment Share on other sites More sharing options...
ahvceo Posted May 24, 2009 Author Share Posted May 24, 2009 Hi All, Thanks for the help! I am really getting too old for this stuff. What I used to do in 5 minutes now takes me an hour or two. I was really starting to get frustrated but finally got there. Your code works and now so does mine. Thank you all! ahvceo Quote Link to comment Share on other sites More sharing options...
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