$a = mysql_result( $b, 0, $c); - does not work

[jasonc]

i am wanting to use the following code but it is not working as i thought it would.

 

instead of hard coding the field name in the script i wish to have it insert using a variable like below.

 

what is the correct way i should do this.

 

$c= "fieldname";

$a = mysql_result( $results, 0, $c);

[Adam]

Your code is syntactically correct provided that $results is a valid MySQL result resource. Without seeing more of the code we can't really say why it's not working.

[jasonc]

how do i print the result on the screen so i can paste it here

 

i am using the following to get the data

 

$sql = "SELECT * FROM `users` WHERE `user` = 'me' LIMIT 1";

 

 

[Yesideez]

As you're only getting one row back you can use something like this:

$query=mysql_query($sql);
$row=mysql_fetch_assoc($query);
echo $row['name'];

That's presuming the field "name" appears in your table.

[jasonc]

ok but can i use something like this?

 

$fieldname = "name";

$query=mysql_query($sql);
$row=mysql_fetch_assoc($query);
echo $row[$fieldname];

[Yesideez]

The first thing to do is give it a go but yes - you can do that.

[jasonc]

but i tried that as in my first post, and it did not work

$c= "fieldname";

$a = mysql_result( $results, 0, $c);

 

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in line

$a = mysql_result( $results, 0, $c);

 

 

$c = "username";

 

and there is a field in the mysql database called "username"

[Yesideez]

$sql="SELECT * FROM users WHERE user='me' LIMIT 1";
$query=mysql_query($sql);
$row=mysql_fetch_assoc($query);
echo $row[$col];

[Adam]

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in line

 

Would have been helpful to quote this error earlier. $results 'is not a valid MySQL result resource'. Try amending to the end of your mysql_query() call:

 

or trigger_error('MySQL error: ' . mysql_error());

 

Make sure you don't have a semi-colon before 'or'.. i.e. "; or trigger(...)"