UPDATE Problem should be easy

[gokeeffe]

can anyone see why the below code is not working. I've echo'd out county and user_id and they come out fine but
when I try to use the UPDATE statement it doesn't work, i keep getting failure. Please help.

Regards


$j = $_SESSION['county'];
echo "county = $j";
$a = $_SESSION['user_id'];
echo "user_id = $a";

$query_two = "UPDATE users SET id_county = '$j',
WHERE user_id = '$a'";
$result_two = mysql_query ($query_two);

if ($result_two)
{
echo'sucess';
}
else
{
echo'failure';
}

 

[sleipnir214]

What do you get when you echo or print $query_two?


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TANSTAAFL!!

 

[gokeeffe]

I get

UPDATE users SET id_county = '1', WHERE user_id = '434'failure

 

[LTeeple]

You should also try some more descriptive error messages such as:

$j = $_SESSION['county'];
echo "county = $j";
$a = $_SESSION['user_id'];
echo "user_id = $a";
        
$query_two = "UPDATE users SET  id_county = '$j',
             WHERE user_id = '$a'";
$result_two = mysql_query ($query_two);
if ($result_two){
  echo 'sucess';
}
else{
  echo 'Failure: Error on query:' . $query _two . '<br>' . mysql_error();
}

Cheers!
Laura

 

[LTeeple]

UPDATE users SET id_county = '1', WHERE user_id = '434'failure"

Eliminate the comma between the '1' and WHERE:

"UPDATE users SET id_county = '1' WHERE user_id = '434'failure"


Cheers!
Laura

 

[gokeeffe]

Tks Laura,

It was the comma great spot, driving me mad for ages

Regards
Graham