mysql_query() problem

Anon

this is my code.

// here if no ID then adding else we're editing

if ($id) {

$sql = "UPDATE $table_name 
	SET coname='$coname', acctnum='$acctnum', username='$username', contact='$contact', phone='$phone', url='$url' 
	WHERE id=$id"; 

$sql2 = "UPDATE $table_name2
	SET custid='$custid', faq='$faq', education='$education', sitelinks='$sitelinks', footer='$footer', employment='$employment', pressrel='$pressrel', whatsnew='$whatsnew', groupware='$groupware', webcalendar='$webcalendar', personalcalendar='$personalcalendar', groupcalendar='$groupcalendar', projectcalendar='$projectcalendar', giffiles='$giffiles', pdffiles='$pdffiles', audiofiles='$audiofiles', videofiles='$videofiles', ppfiles='$ppfiles', inventory='$inventory'
	WHERE custid = $custid";

}

// run SQL against the DB

$result = mysql_query($sql,$connection);
$result2 = mysql_query($sql2,$connection) or die(mysql_errno());
echo "Record updated/edited!<p>";

First query works perfectly, second doesn't, same database, different table. It is giving me this error: 1064, mysql_error() shows: You have an error in your SQL syntax near '' at line 3

Any ideas will be appreciated, I am getting bleary eyed looking at it.

moonie 😉

Anon

SET custid='$custid'
WHERE custid = $custid"

So what's the type of custid? 'cause if it's a number, you don't need quotes, but if it's a string, you need quotes. You're using both with and without quotes, so which one is it?

Diego

theslinky

Okay two things:

1) Like diego said: make sure custid is an integer if it is outside quotes. You can also put integers inside quotes.

2) In your second query, you have "UPDATE $table_name2 ..."

PHP will replace $table_name with it's value and then append the 2. If you want to use the variable $table_name2, either put it in braces {} or append it using the string addition operator:

"UPDATE " . $table_name2 . " ...";

-sridhar

Anon

yep, found it before you answered, custid is text the ' ' did it, thanks.

moonie 😉