How can I fix this?

[Faracus]

I'm trying to make this code work

$sql3= "UPDATE lottomax SET " . $query1 . "='" . $number1 . "'";if (!mysql_query($sql3,$con)){die('Error: ' . mysql_error());}

and I keep getting this error: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 '5='1'' at line 1

[birbal]

i assume $number1 is int field. if it is like so remove the quotes around it.

$sql3= "UPDATE lottomax SET $query1=$number1";

and you dont need to use concatenate oprator inside double quotes value of variable wil be evaluated itself

[Faracus]

sadly, I'm still getting the error "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 '5=1' at line 1"

[birbal]

it looks like $query1 is evaluating a number . print out $sql3 to see what the query looks like and to be sure.

Edited April 19, 2012 by birbal

[Faracus]

if I put in

"You have chosen " . $query1 . ", and updated the total to " . $number1 . ".<br />

it prints out the right values. and if I print out sql3 I get UPDATE lottomax SET 5=1

Edited April 19, 2012 by Faracus

[birbal]

use backtick around $query1. its better though to avoid integer column name. use a descriptive name instead.

[Faracus]

Thanks for all the help.