Please can somebody help me

**Stevis2006** · Apr 30, 2006 13:25

Hi all,

Please can somebody help me out with passing data from one page to another via a url variable?
It is driving me mad.

I want to give the user a drop down which is dynamically filled with categories, a-z, on a page named choice.php. This i have managed to do.

Then, i want the url to display their selection and send it to the next page...eg....www.gfsbersf.com/view.php?id=n.

Then i want the page, (view.php), to display all records under the category of 'n' in the database.

I can't get my head around how the variable gets passed, and received.

If anybody could help me out it would be much appreciated.

Many Thanks,
Steve

**paulgb** · Apr 30, 2006 13:41

If I understand what you are saying, all you need is some HTML to pass the variable, and some PHP to interpret it. I usually use a wysiwig editor for forms so this code might not all be right, but it should look something like this:

HTML Code:

<form action="view.php" method="GET"><select name="id">
<option value="43">option1</option>
<option value="59">option2</option>
</select><input type="submit"></form>

This would give you a url of something like this: /view.php?id=43.

Then you get the ID in php, and you can do whatever you want with it:

PHP Code:


$request_id = _GET["id"];

//Now $requst_id has the number that was passed

I hope I answered the right question, good luck with it.

**7stud** · Apr 30, 2006 16:43

I want to give the user a drop down which is dynamically filled with categories, a-z, on a page named choice.php.

That means choice.php has a form on it that contains your select element.

I can't get my head around how the variable gets passed, and received.

The name of the select along with the value attribute of the selected option is automatically sent to the page specified in the form's action attribute, e.g.

Code:

<form method="get" action="somepage.php">

<select name="id">
<option value="a123">choice 1</option>
<option value="b678">choice 2</option>
</select>

<input type="submit" value="Submit" />
</form>

If the method attribute in the form tag is 'get', then on the page somepage.php the value of the id will be in the $_GET[] array, and you access it like this:

echo $_GET['id'];

If the method attribute of the form tag is 'post', then on the page somepage.php the value of the id will be in the $_POST array, and you access it like this:

echo $_POST['id'];

**Stevis2006** · May 1, 2006 11:24

Thanks u 2...will give it a go

Steve

**stant_98** · May 1, 2006 11:33

Great advice here! Thanks!

**Stevis2006** · May 1, 2006 14:04

Sorry to be a pain, but i cant get it to display anything.

here is the code...

<?php require_once('.htconfig.php'); ?>
<?php
$colname_Recordset1 = "-1";
if (isset($_POST['$request_id'])) {
$colname_Recordset1 = (get_magic_quotes_gpc()) ? $_POST['$request_id'] : addslashes($_POST['$request_id']);
}
mysql_select_db($database_soundflicksdb, $soundflicksdb);
$query_Recordset1 = sprintf("SELECT artist_name, album_title, category FROM upload2 WHERE category = '%s' ORDER BY artist_name DESC", $colname_Recordset1);
$Recordset1 = mysql_query($query_Recordset1, $soundflicksdb) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);
?>
<!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>Untitled Document</title>
</head>
<body>
<p>
<?php
echo $_GET['id'];
?>
<?php
$request_id = _GET['id'];
?>

</p>
<table width="40%" border="1" align="center" cellpadding="0" cellspacing="0">
<tr>
<td><div align="center"><?php echo $row_Recordset1['artist_name']; ?></div></td>
</tr>
<tr>
<td><div align="center"><?php echo $row_Recordset1['album_title']; ?></div></td>
</tr>
</table>
<p align="center"> </p>
</body>
</html>
<?php
mysql_free_result($Recordset1);
?>

Can anyone help?

Many Thanks once more,
Steve

Page above is view.php btw

**themightystephen** · May 1, 2006 14:11

PHP Code:


$request_id = _GET['id'];

You missed out a $ sign before the _GET. Thats the first thing I noticed.

**Stevis2006** · May 1, 2006 14:22

Thanks for that.
I had to cut some out as i left the recordset query in from before when i was trying to do it myself.
I have now ammended it, but it still will not show anything

Code

<!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>Untitled</title>
</head>
<?php require_once('.htconfig.php'); ?>
<body>
<?
$request_id = $_GET["id"];

while($my_data = mysql_fetch_array($request_id))
{
echo "<table align=center width=50% cellpadding=1 cellspacing=0 border=0>";
echo "<tr>";
echo "<td>";

echo $my_data['artist_name'];
echo "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>";
echo $my_data['album_title'];
echo "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>";
echo $my_data['album_info'];
echo "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>";
echo '<img src ="/images/'.$my_data['name'].'">';
echo "</td>";
echo "</tr>";
echo "</table>";
}
?>
</body>
</html>

**blackman890** · May 1, 2006 14:31

is it just me, or have you completely missed some mysql code there?

PHP Code:


$request_id = $_GET["id"];



while($my_data = mysql_fetch_array($request_id))

this would definately always return a null since $request_id is not a mysql array.

**Stevis2006** · May 1, 2006 14:42

Ok, great thanks...It's alive now

Many Thanks all

**Stevis2006** · May 1, 2006 14:43

Could anybody tell me if there is a way of turning the displayed album titles into links to the appropiate page?

Thanks again,
Steve

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