PHP Mysqli UPDATE error

Question

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I have done a little bit of coding and withdrawed this problem:

Java

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '? = ? WHERE `Username` = ?' at line 1

I don't understand whats wrong with this prepare:

SQL

require 'mysqli_data/mysqli_connection.php';
        $qString = $gConnection->prepare("UPDATE `accounts` SET ? = ? WHERE `Username` = ?");
        printf($gConnection->error);
        $qString->bind_param('sss', $key, $data, $gConnection->real_escape_string($user));
        $qString->execute();
        $qString->close();
        return true;

Thanks for any help!

Posted 7-May-13 10:00am

NextGenDeveloper

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Comments

Mohibur Rashid 8-May-13 1:16am

what is ? = ?

NextGenDeveloper 8-May-13 3:58am

Read up on the prepare statment
http://php.net/manual/en/mysqli.prepare.php

Bernhard Hiller 8-May-13 4:21am

what do you want to achieve? Change a username? Then you must also provide the previous username or userid. And of course, you need some more basics for SQL:
UPDATE `MYTABLE`
SET `MYCOLUMN`=`NEWVALUE`
WHERE `OTHERCOLUMN`=`OTHERVALUE`

NextGenDeveloper 8-May-13 5:58am

Its a general function called UpdateUserInfo($user, $key, $data) at my side.
I'm doing it regulery so I can't find the problem.
'UPDATE `accounts` SET ? = ? WHERE `Username` = ?' is like 'UPDATE `accounts` SET Username = RANDOM_RANDOM WHERE `Username` = NON_RANDOM'

Its the same thing, but with a statement, and I can't figure out why it dosen't work.

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NextGenDeveloper · Accepted Answer · 2013-05-08T00:46:00

Solution 1

Fixed, just implemented some PHP into it:
$qString = $gConnection->prepare("UPDATE `accounts` SET ".$key." = ? WHERE `Username` = ?");

Posted 8-May-13 0:46am

NextGenDeveloper