$_POST issues..?

[woolyg]

Hi all,

 

I'm trying to gather data from a dropdown menu to be processed and added into a table

 

Here's the dropdown menu code that displays a list of selectable items:

<select name="prop_area" size="1" class="boxes">
        <?PHP
$places = mysql_query ("SELECT * FROM places order by name");
$ordered = array();
while ($placelist = mysql_fetch_row($places)) {
	$id = $placelist[0];
	$name = $placelist[1];
	$ordered[$id] = $name;
}
natsort($ordered);
while (list($id,$use) = each($ordered)) {
	$placeid = current($ordered); 
	$option = "<option value=\"$id\"";
	if ($id == $area) {
		$option .= " selected";
	}
	$option .= ">";
	$option .= "$use</option>";
	print $option;
}
?>
    </select>

 

 

Then I use

$prop_area = $_POST['prop_area'];

 

 

Then I try to write the data to a table using:

 

$query = "INSERT INTO table (area) ".
"VALUES ('$prop_area')";

 

 

Because there are a few possible entries from the dropdown selection, should I be using some sort of FOREACH statement? If so, how can I implement it? Still nOObing it up here:)

 

Cheers,

Woolyg

 

 

 

 

[Daniel0]

<select name="prop_area" size="1" class="boxes">
<?php
$places = mysql_query("SELECT * FROM places ORDER BY name");
while($place = mysql_fetch_assoc($places))
{
$selected = $place['id']==$area ? " selected='selected'" : null;
echo "<option value='{$place['id']}'{$selected}>{$place['name']}</option>\n";
}
?>
</select>

Is better... I supposed that the fields were named id and name.

 

I believe that if multiple selections are made, then it will be an array. Try to use print_r() and then die(). Then you can see what it does. Try with both a single value and multiple values selected, then you'll see.

[woolyg]

Hi Daniel0

 

Thanks for that - though the information displays fine in the dropdown the query created, what I'm really trying to do is process the information that a user selects from it.

 

In other words:

Say I had a dropdown menu with info about "Number of Rooms"

Selectable options:

1

2

3

4

5

 

...and a user selects '4'

 

How would I tell the PHP file to use this selection and post it to a MySQL table? When I use $_POST, the info simply does not enter onto the table!

 

Hope this is understandable - Woolyg.

[rameshfaj]

You can also try this stuff:

<select name="prop_area" size="1" class="boxes">

<?php

$places = mysql_query("SELECT * FROM places ORDER BY name");

while($place = mysql_fetch_assoc($places))

{?>

$selected = $place['id']==$area ? " selected='selected'" : null;

 

<option value=<?php $place['id']?>><?php $place['id']?></option>

<?php

}

?>

</select>

[rameshfaj]

On the next page do this:

 

$something=$_POST['name_of_selection_list'];

 

What ever the value is assigned to the element of the list ;will be posted to this value.

Note that the action of the firstpage should be set to the next page to recieve the values by POST and GET commands.

I think this helps!

[Daniel0]

I believe that if multiple selections are made, then it will be an array. Try to use print_r() and then die(). Then you can see what it does. Try with both a single value and multiple values selected, then you'll see.

Nevermind this. I confused it with a select with multiple='multiple'... Besides, I found out that it doesn't return arrays anyways (unless you name it something like test[])...

 

Thanks for that - though the information displays fine in the dropdown the query created [...]

I know... I just gave you a shorter (and probably more efficient since your contains multiple loops) way of doing it.

 

[...] what I'm really trying to do is process the information that a user selects from it.

 

In other words:

Say I had a dropdown menu with info about "Number of Rooms"

Selectable options:

1

2

3

4

5

 

...and a user selects '4'

 

How would I tell the PHP file to use this selection and post it to a MySQL table? When I use $_POST, the info simply does not enter onto the table!

 

You can access it using $_POST...

 

You can see in this little test script how the data will be like in $_POST:

<form action='<?php echo $_SERVER['PHP_SELF']; ?>' method='post'>
<select size='4' name='test'>
	<option value='1'>one</option>
	<option value='2'>two</option>
	<option value='3'>three</option>
	<option value='4'>four</option>
</select><br />
<button type='submit'>Submit</button>
</form>

<?php
print_r($_POST);
?>