How to display two while loop variables

Question

I am trying to display multiple columns with multiple rows from a database. It works to display one row, but when I put in the second one it just displays two of the same. Like an echo. How do I get it to display the two different numbers?

  $result = mysql_query("SELECT plea, COUNT(plea) as cee FROM tee WHERE section='d' GROUP BY     
plea"      , $c) or die("two");  

$number=mysql_num_rows($result);    
  if($number>0)   
        {
            $i=0;
        while($row_result = mysql_fetch_array($result))
            {
        $plea.$i = $row_result['plea'];
        $cee.$i = $row_result['cee'];
        echo $plea.$i."&nbsp;&nbsp;&nbsp;&nbsp;".$cee.$i."<br><br>";

            $i++;
           }

    }

Do you get the same result when running the query in mysql/phpmyadmin? — kero
– kero, Commented Feb 26, 2014 at 19:07
Did you check your sql in something like mysql workbench to ensure your call is not getting "mirrored" column values? — SpYk3HH
– SpYk3HH, Commented Feb 26, 2014 at 19:09
No, it works perfectly in the admin, well, thats obviously not using the while stuff, just the query works perfectly — Tim
– Tim, Commented Feb 26, 2014 at 19:10

Marc B · Accepted Answer · 2014-02-26 19:10:44Z

1

This is VERY strange code:

 $plea.$i = $row_result['plea'];

and NOT doing what you want. It's parsed as:

 $plea . ($i = $row_result['plea']);

and boils down to $i getting your value from the query, and then being further executed down to just

 TRUE

without EVER modifying the value in $plea. You probably want an array:

$plea[$i] = $row_result['plea'];

answered Feb 26, 2014 at 19:10

Marc B

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