Mysql_fetch_array problems

[Drezard]

Is there anything wrong with this? as in code wise?

 

The mysql_fetch_array() is failing.

 

Code:

 

$query = "SELECT * FROM user_profiles WHERE user_id='$user_id'";

$result = mysql_query($query);

$array = mysql_fetch_array($result);

$name = $array[1];

$email = $array[2]; 	

$location = $array[3];	

 

Daniel

[micmania1]

You may have an error in your query.

 

Try

if ($result) {
        $array = mysql_fetch_array($result);
} else {
        echo mysql_error(); exit();
}

[psychowolvesbane]

Is there anything wrong with this? as in code wise?

 

The mysql_fetch_array() is failing.

 

Code:

 

$query = "SELECT * FROM user_profiles WHERE user_id='$user_id'";

$result = mysql_query($query);

$array = mysql_fetch_array($result);

$name = $array[1];

$email = $array[2]; 	

$location = $array[3];	

 

Daniel

 

With the Array you need to use:

$name = $array['Name']; (or whatever field name it is)

$email = $array['Email'];

$location = $array['Location'];

 

BTW with an array it would actually start at 0 instead of 1, so 0,1,2 not 1,2,3.

[phorcon3]

<?php

$a=mysql_query("SELECT `name`,`email`,`location` FROM `user_profiles` WHERE `user_id` = '$user_id'");
$b=mysql_fetch_assoc($a) or mysql_error();

echo 'Name: '.$b['name'].'<br />';
echo 'Email: '.$b['email'].'<br />';
echo 'Location: '.$b['location'];

?>

 

thats how id do it...