[SOLVED] Passing info from forms for deletion

Esqulax · November 30, 2007

Hiya, if anyone can help, id be really appreciative...

Basically, im trying to get info from the checkboxes on delete.php, pass it to del.php and run a delete script.

It wasnt working, so in del.php i put in a count() to see if any info is actually being passed into it.. i get a big fat 0

This is so vexing, its been annoying me for days

Code:

delete.php:

<html>
<head>
<title>DELETED!</title>
</head>

<?php
//Standard connection scripts  
$conn=@mysql_connect("localhost","root","")
or die("error connecting to database");

$rs=@mysql_select_db("hotel", $conn)
or die("Error connecting to Hotel");

  
  $sql       = "SELECT * FROM guest_data ORDER BY guest_id, name, surname";
  $result    = mysql_query($sql,$conn) or die("error using result query");

echo ("<table border= 1>");

echo ('<form method="post" action="'.$self.'">');

  while ($row = mysql_fetch_array($result, MYSQL_ASSOC))   
  {
    echo "<tr>\n";
echo "<td>". $row['guest_id'] . '<input name="guestid[]" type="checkbox" value="' . $row['guest_id'] . '"></td>';						
    echo "<td>" . $row['name'] ;
    echo "<td>" . $row['surname'];
  }



?>
    <tr>
      <td>
     
      <input type="submit" name="submit" value="submit"></td>
    </form>
    </tr>
</table>

<?php
  
  echo('<form method="post" action="del.php">');
	$delid = $_POST["guestid"];
	$how_many=count($delid);


        if ($how_many>0) 
	{
            echo ("You chose the following records:<br>");
        
        	for ($i=0; $i<$how_many; $i++) 
		{
                echo (($i+1) . " - " . $delid[$i] . "<br>");
        	}

		echo("Ok to delete ".$how_many." record(s)?");
		echo('<input type="submit" name="submit" value="Yes">');
		echo('</form>');
	}
?>

del.php:

<html>
<head>
<title>Sucessfully Deleted</title>
</head>
<?php
$delvalue=$_POST['guestid'];

	//Stanatrd connection scripts  
$conn=@mysql_connect("localhost","root","")
or die("error connecting to database");

$rs=@mysql_select_db("hotel", $conn)
or die("Error connecting to Hotel");



$amount=count($delvalue);
echo(''.$amount);

for($i=0;$i<$amount; $i++)
{
echo(''.$delvalue[$i]);
}



/*	 while ($row = mysql_fetch_array($delval))  
{
$sql2="DELETE FROM guest_data WHERE guest_id=".$delval;
$gone= mysql_query($sql2,$conn);
}
*/

?><!--
<script language="javascript"><!--
location.replace(delete.php")
//-->
</script>-->
<a href="delete.php">Return</a>
<body>
</body>
</html>

MadTechie · November 30, 2007

thats because your 2nd form doesn't have $guestid

Esqulax · November 30, 2007

My bad...

That was a frustrated experiment, the same issue appears if i have (in del.php):

$delvalue=$_POST['guestid'];

MadTechie · November 30, 2007

still doesn't help if you don't put it in the form..

for example


for ($i=0; $i<$how_many; $i++) 
{
        echo (($i+1) . " - " . $delid[$i] . "<br>");
        echo "<input type=\"hidden\" name=\"guestid[]\" value=\"{$delid[$i]}\"> "; //<--add
}

Esqulax · November 30, 2007

Fantastic! So the form diddn't have any info to store, i see.

one last syntax Q

Is it ok to put the variable on the end of a SELECT or DELETE statement like this?

        $sql= "SELECT * FROM guest_data WHERE guest_id=".$delvalue;
  	$result    = mysql_query($sql,$conn) or die("error using result query");

 while ($row = mysql_fetch_array($result))  
{
$sql2="DELETE FROM guest_data WHERE guest_id=".$delvalue;
$gone= mysql_query($sql2,$conn);
}
[/code

MadTechie · November 30, 2007

yes that fine..

Techie Tip:

you have an array of ID's

ie

$_POST['guestid'];

array(1,2,3,6,8,10)

etc

you can do this

<?php
$delvalue=implode(",",$_POST['guestid']);
//returns "1,2,3,6,8,10"

$sql2="DELETE FROM guest_data WHERE guest_id IN ($delvalue)";
//SQL statment to deletes all the records in one hit 
?>

Esqulax · November 30, 2007

Sorted!

Massive weight is herby lifted!!

Thanks Loads!

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[SOLVED] Passing info from forms for deletion

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