CODEWITHSUNDEEP
phpmysqlsql-update
David Reinberger
David Reinberger

Reputation: 540

My SQL UPDATE error

I've been scratching my head for the last hour and just can't get this to work.

Doing this query:

$dbtoken = mysql_real_escape_string($invite_token);
$dbexpire = mysql_real_escape_string($invite_expire);
$dbid = mysql_real_escape_string($uid);
$update_user = mysql_query("UPDATE inviters SET invitetoken='$dbtoken', inviteexpire='$dbexpire' WHERE uid='$dbid'");
$save = mysql_query($update_user) or die(mysql_error());

It gives me this error:

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 '1' at line 1

Upvotes: 0

Views: 69

Answers (3)

chandresh_cool
chandresh_cool

Reputation: 11830

There is a error You are trying to execute your query twice. Remove second execution of it.i.e remove this

 $save = mysql_query($update_user) or die(mysql_error());

it is trying to do this

$save = mysql_query(1);

which is wrong

Upvotes: 1

Borniet
Borniet

Reputation: 3546

You are sending the result of your first query as an sql statement for your second query:

$update_user = mysql_query("UPDATE inviters SET invitetoken='$dbtoken',
inviteexpire='$dbexpire' WHERE uid='$dbid'");
$save = mysql_query($update_user) or die(mysql_error());

$update_user contains the result, and you feed it back to mysql_query. Try this instead:

$update_user = "UPDATE inviters SET invitetoken='$dbtoken',
inviteexpire='$dbexpire' WHERE uid='$dbid'";
$save = mysql_query($update_user) or die(mysql_error());

BTW: Use Mysqli instead of Mysql_ statements, they're deprecated and unsafe.

Upvotes: 1

nl-x
nl-x

Reputation: 11832

You are doing mysql_query(); twice. Once in the last line. And once in the line above it.

The first time it returns 1 (success)

The second time you try to execute 1 as a query

Ps. Scratching your head for an hour can cause skin irritation.

Upvotes: 0

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