2 database columns displaying the same values?

[Omael]

I have a simple query, but for some reason when I display the value of two columns they both return the value of the first column.
 

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<?php
mysql_connect($theserver,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT column1, column2 FROM questions WHERE id = '$_GET[id]'";

$result1=mysql_query($query);
$num1=mysql_numrows($result1);

mysql_close();

 $colum1=mysql_result($result1,"column1");
 $column2=mysql_result($result1,"column2");

?>

<?php echo $column1; ?> <?php echo $column2; ?>

 

For some reason column 2 displays the same value as column 1 but they have different values.

[adam_bray]

That method is very insecure. Try this - 

<?php
	mysql_connect($theserver,$username,$password);
	@mysql_select_db($database) or die( "Unable to select database");
	
	
	if( preg_match( "/[0-9]/", $_GET['id'] ) )
	{ 
		$question_query = mysql_query( "SELECT column1, column2 FROM questions WHERE id = '$_GET[id]';" ) or die( 'MySQL Error: ' . mysql_error() );
		
		$num1 = mysql_num_rows($question_query);
		
		while( $row = mysql_fetch_object( $question_query ) )
		{
			echo $row->column1 . '<br />';
			echo $row->column2 . '<br />';
		}
	}  
	else
	{
		die( 'Invalid ID' );	
	}
	
	mysql_close();
	
?>

[Omael]

Hmm. I'm getting invalid ID when i try that.

[Omael]

I changed my query to this and it now works:

 

mysql_connect($theserver,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

$query="SELECT column1, column2 FROM questions WHERE questionid = '$_GET[qid]'";

 

$result1=mysql_query($query);

$num1=mysql_numrows($result1);

 

while ($row = mysql_fetch_assoc($result1))

{

 $column1=$row['column1'];

 $column2=$row['column2'];

 

 }

mysql_close();

 

 

[Psycho]

Right, the problem was in the usage of  mysql_result().
 
Per the manual, that function (which is deprecated) takes two parameters. The first is the result resource, which you provided. The second parameter is the 'row' which you did not provide. You provided strings corresponding to the column names. But, the manual states:
 
 

The row number from the result that's being retrieved. Row numbers start at 0.

EDIT: Since you are only expecting 1 row, no need to use a while() loop.

if(mysql_numrows($result1))
{
    $row = mysql_fetch_assoc($result1);
    $column1=$row['column1'];
    $column2=$row['column2'];
}
else
{
    //Error condition
}

[Omael]

Thank you Pcycho