PHP/mySQL: How do a concatenate a variable in a mysql query?
Solution 1
You do not need the .
. PHP parses double quoted ("
) strings and replaces the variables with their values. As such:
$pageID = 3;
echo "UPDATE test SET page$pageID = '$pageProgress' WHERE userID = '$userID'";
http://codepad.viper-7.com/uIdqqH
Solution 2
The problem is that your .'$pageID' is inside the double-quoted string; you don't concatenate this on the MySQL side; it gets parsed long before MySQL ever sees it.
It might be that you were trying to escape the field name for Mysql, in that case, you use backticks.
Try:
'UPDATE test SET `page'.$pageID.'`=\''.$pageProgress.'\' WHERE...'
Or, much easier on the eyes:
"UPDATE test SET `page{$pageID}`='{$pageProgress}' WHERE..."
Solution 3
"UPDATE test SET page".$pageID."='".$pageProgress."' WHERE userID='".$userID."';"
Dots are in the wrong spot to do it with PHP's string functions.
Solution 4
Something like this.
mysql_query("UPDATE test SET page" . $pageID . " = '" . $pageProgress . "' WHERE userID = " . $userID)
Mark Rummel
I'm a modern web designer. I have enjoyed designing and building websites for my clients for over 10 years. I have extensive experience in many web technologies, including HTML5, CSS3, javascript, jQuery, LESS, PHP, MySQL, Wordpress, MailChimp, PayPal, Stripe, and more.
Updated on July 15, 2022Comments
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Mark Rummel almost 2 years
What is the proper way to concatenate text and a variable in PHP inside a mysql_query? Here is my attempt:
page.'$pageID'
I want it to output
page3
.Here is all of the code (simplified to focus on the mysql_query):
if ($_POST['pageProgress']) { $pageProgress = $_POST['pageProgress']; $pageID = 3; $userID = 1; $updateUserProgress = mysql_query("UPDATE test SET page.'$pageID'='$pageProgress' WHERE userID='$userID'") or die(mysql_error()); }
All of the code works perfectly if I simply replace
page.'$pageID'
withpage3
.