PHP for loop not displaying correct data

Question

I have this PHP code/form:

<form method="post" action="tickets_report2.php">
<table width="800" border="0" cellspacing="5" cellpadding="5">
  <tr>
    <td><strong>Select</strong></td>
    <td><strong>Company</strong></td>
  </tr>
  <?php
  $sql="SELECT * from customer ";
  $rs=mysql_query($sql,$conn) or die(mysql_error());
  $counter=0;
  while($result=mysql_fetch_array($rs)) {
    $counter++;
    echo '<tr>
       <td><input type="checkbox" value="'.$result["sequence"].'" name="checkbox'.$counter.'" /></td>
       <td>'.$result["company"].'</td>
    </tr>';
  }
  echo '<input type="hidden" name="counter" value="'.$counter.'" />';
  ?>
  </table>
  <input type="submit" name="submit" value="Next" />
</form>

and on the form action page:

<table width="800" border="0" cellspacing="5" cellpadding="5">
  <tr>
    <td><strong>Company</strong></td>
  </tr>
  <?php
  for($i=1; $i<=$_POST["counter"]; $i++) {
    if($_POST["checkbox$i"]) {
      $sql="SELECT * from customer where sequence = '".$i."' ";
      $rs=mysql_query($sql,$conn) or die(mysql_error());
      $result=mysql_fetch_array($rs);       
      echo '<tr>
                    <td>'.$result["company"].'</td>
      </tr>';
    }
  }
  ?>
</table>

lets say i check the checkbox on the form for the row in the database with sequence of 278, the SQL Query should say SELECT * from customer where sequence = '278' but its showing SELECT * from customer where sequence = '1'

i think its using the counter rather than the customer sequence

what do i need to change to get this working correctly so it selected the correct customer(s)

Just a little question back: Why do you think that code should display that data at all? — hakre
– hakre, Commented Aug 31, 2013 at 14:40
can you post the result of print_r($_REQUEST) from the form action page? — DrewP84
– DrewP84, Commented Aug 31, 2013 at 14:43
This could be your solution: add every checkbox to an array by doing so: <input type="checkbox" name="chknames[]" value="'.$result["sequence"].'"/. after that you can just do if(isset($_POST['chknames'])) — Maxim
– Maxim, Commented Aug 31, 2013 at 14:44

guido · Accepted Answer · 2013-08-31 14:48:19Z

1

On the form action page it should read:

if($_POST["checkbox$i"]) {
    $sql="SELECT * from customer where sequence = '".$_POST["checkbox$i"]."' ";

Note: sanitize your inputs.

answered Aug 31, 2013 at 14:48

guido

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Comments

Fu Xu · Accepted Answer · 2013-08-31 14:48:32Z

form page:

<form method="post" action="tickets_report2.php">
<table width="800" border="0" cellspacing="5" cellpadding="5">
  <tr>
    <td><strong>Select</strong></td>
    <td><strong>Company</strong></td>
  </tr>
<?php
$sql="SELECT * from customer ";
$rs=mysql_query($sql,$conn) or die(mysql_error());
$counter=0;
while($result=mysql_fetch_array($rs))
{
    $counter++;
    echo '<tr>
         <td><input type="checkbox" value="'.$result["sequence"].'" name="checkbox['.$counter.']" /></td>
         <td>'.$result["company"].'</td>
         </tr>';
}
?>
</table>
<input type="submit" name="submit" value="Next" />
</form>

action page:

<table width="800" border="0" cellspacing="5" cellpadding="5">
  <tr>
    <td><strong>Company</strong></td>
  </tr>
<?php
foreach($_POST["checkbox"] as $value)
{
        $sql="SELECT * from customer where sequence = '".$value."' ";
        $rs=mysql_query($sql,$conn) or die(mysql_error());
        $result=mysql_fetch_array($rs);     
        echo '<tr>
                    <td>'.$result["company"].'</td>
                  </tr>';
}
?>
</table>

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