drop down select form in a block?

Regular Joined: Jan 16, 2008 Posts: 75

I have php script that has a drop down select form which returns the user selection as a variable in a query to display records from a mysql table.

I want the dropdown form to be in a block and the output to show in main window.

Currently I have learned how to place my scrip inside a module and that works great...time to step it up a notch Smile

Any advice, would be most appreciated ty

Included the php code I am using currently.

Code:

//Database Class Connect


mysql_select_db($database_Lw, $Lw);


$query_Class = "SELECT `Class` FROM class_list ORDER BY id ASC";


$Class = mysql_query($query_Class, $Lw) or die(mysql_error());


$row_Class = mysql_fetch_assoc($Class);


$totalRows_Class = mysql_num_rows($Class);








//Database Cards Connect


$name = $_POST['select'];


mysql_select_db($database_Lw, $Lw);


$query_Cards = "SELECT Name, `Class`, Drops, Details FROM skill_cards WHERE skill_cards.`Class` LIKE '$name%' ORDER BY Name ASC";


$Cards = mysql_query($query_Cards, $Lw) or die(mysql_error());


$row_Cards = mysql_fetch_assoc($Cards);


$totalRows_Cards = mysql_num_rows($Cards);





?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">


<html xmlns="http://www.w3.org/1999/xhtml">


<head>


<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />


<title>Loot Whore</title>


</head>





<body>


<form id="form1" name="form1" method="post" action="">


  <label>Class


  <select name="select">


   <option value="">ALL</option>


    <?php


do {  


?>


    <option value="<?php echo $row_Class['Class']?>"><?php echo $row_Class['Class']?></option>


    <?php


} while ($row_Class = mysql_fetch_assoc($Class));


  $rows = mysql_num_rows($Class);


  if($rows > 0) {


      mysql_data_seek($Class, 0);


     $row_Class = mysql_fetch_assoc($Class);


  }


?>


  </select>


  </label>


<input type="submit" name="submit" value="submit">


</form>


<table border="1" cellpadding="5" cellspacing="5">


  <tr>


    <td>Name</td>


    <td>Class</td>


    <td>Drops</td>


    <td>Details</td>


  </tr>


  <?php do { ?>


    <tr>


      <td><?php echo $row_Cards['Name']; ?></td>


      <td><?php echo $row_Cards['Class']; ?></td>


      <td><?php echo $row_Cards['Drops']; ?></td>


      <td><?php echo $row_Cards['Details']; ?></td>


    </tr>


    <?php } while ($row_Cards = mysql_fetch_assoc($Cards)); ?>


</table>


<br />


</body>


</html>


<?php


mysql_free_result($Class);





mysql_free_result($Cards);








CloseTable(); 


include_once 'footer.php'; 


?>

registered · Posted: Tue Jan 22, 2008 3:02 pm

Well, in a brief nutshell, you'll have to code up a form in a block, and have the post result (the "action") get directed to a module so the results can be displayed. See the PHP-Nuke How-To on what you need to do to code a block, and/or look at some of the supplied blocks.

Posted: Tue Jan 22, 2008 9:48 pm

I will admit I could not figure out how to work the content string into my script.
however using the editor I made a block by pasting the code. which works great!

But Where is this block saved I looked in the blocks directory thinking it would be named the name I gave it in the editor...but it was not.

so where is the file for it?

Thanks for all the help I have recived here.

Posted: Wed Jan 23, 2008 8:32 am

If you use the editor to make a block, the contents are stored in the database (I think in the _blocks table).

Posted: Wed Jan 23, 2008 2:39 pm

Thank you