use a while mysql_fetch_array and UPDATE during the loop - PART TWO

Question

I apologize for not knowing how else to do this. I went to add to my other question, and I could not figure out how to append this to it.

I am trying to apply your answer to the previous question in another place.

In this instance I have a bit more going on but thought I could do the same thing as above -with the adjustment of changing id for user_name, since the ids would not match table to table, but the user_name does. I know I am biting off a lot for my first project but I really like this coding stuff. Hope I am not pulling on too many shirt tails. Thanks for your help.

Heres the code. Its placing the same user_name in every row(like the previous question was)

<?php

$Var1 =$_POST['Var1'];

require("connection.php");

mysql_query("UPDATE table1 SET actor = '$Var1'");

$result = mysql_query("SELECT * FROM table2 WHERE subject ='$Var1'");

while($row = mysql_fetch_array($result))
  { $un = $row['user_name'];
     $a =$row['subject'];
     $a_val = $row['subject_val'];

    $sql=mysql_query("UPDATE table3 SET user_name='$un', subject='$', subject_val= $a_val WHERE user_name=".$row['user_name']);     

  }
 mysql_close($connection);
?>

Please pay attention to SQL injections. $Var1 is not sanitized! — Francesco Laurita
– Francesco Laurita, Commented Feb 27, 2011 at 20:10

Ryan Castillo · Accepted Answer · 2011-02-27 20:15:04Z

This is the last time I answer this kind of question. I'm also adding some optimization.

<?php
var expires = (isset($_POST['var1']) ? $_POST['var1'] : die("no VAR");

require_once("connection.php");

mysql_query("UPDATE table1 SET actor = '$var1'") or die("I cannot run , reason : ".mysql_error());

$result = mysql_query("SELECT * FROM table2 WHERE subject ='$var1'") or die("I cannot run , reason : ".mysql_error());

while($row = mysql_fetch_array($result)) {
    extract($row); // field name as variable , content as value , so be sure that the fields have the right name
    /*
          $user_name = $row['user_name'];
          $subject =$row['subject'];
          $subject_val = $row['subject_val'];
    */
        $sql="UPDATE table3 SET user_name='".$user_name."', subject='".$subject."', subject_val= ".$subject_val." WHERE user_name='".$user_name."'";
    mysql_query($sql) or die("I cannot run , reason : ".mysql_error());

  }
 mysql_close();
?>

zdvdla · Accepted Answer · 2011-03-02 02:55:10Z

0

Changed the key of third table to user_name - then

    $sql="UPDATE table3 SET user_name='".$user_name."', subject='".$subject."', subject_val= ".$subject_val." WHERE user_name='".$un."'";

worked!!

thanks for the help!!

answered Mar 2, 2011 at 2:55

zdvdla

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