[SOLVED] PHP drop list value

[monty187]

Hi Guys, Im relitively new to php but I am currently trying to create a drop menu that will decide what contents are displayed on screen. so i need a drop menu and then when the user selects an option from the list an sql query uses the selection value to get details from the db,

The problem is I cant seem to get the value out of the list to use in the query. Here is my code so far, any suggestions much appreciated

 

<form action="<? echo $HTTP_SERVER_VARS['PHP_SELF']; ?>" method="post"> 
			<select name="siteselect">
				<? $result = mysql_query("SELECT site_name FROM communitypage");

				while($row = mysql_fetch_array($result))
				{
				?>												
				<option value="<? $row['site_name'] ?>"><? echo $row['site_name'] ?></option>
				<? } ?>
			</select>
			<input type="submit" border="0" name="reg" value="GO">

		</form>

 

and then I attempt to get the data out as below

if(isset($_POST["siteselect"]))
	{		
		echo "select :".$_POST["siteselect"];
		$result = mysql_query("SELECT * FROM communitypage WHERE site_name='richmond court'");

		while($row = mysql_fetch_array($result))
		{
			$site_name=$row[0]; // 42
			$uname=$row[1]; // 
			$site_details=$row[3]; // 
			$site_notice=$row[4]; // 
			$site_files=$row[5]; // 
		}

		?> <br /> <?
		echo $site_name;
		?> <br /> <?
		echo $site_details;

 

I was trying to make it so the user could just click on the selection but i think it may be easier to use a submit button.

 

Thanks a million in advance:)

 

Michael

[trq]

You never actually use the posted variable within your query. eg;

 

$result = mysql_query("SELECT * FROM communitypage WHERE site_name='{$_POST['siteselect']}'");

[monty187]

You never actually use the posted variable within your query. eg;

 

$result = mysql_query("SELECT * FROM communitypage WHERE site_name='{$_POST['siteselect']}'");

 

appologies, I forgot to say I am just trying to print out the value on the previous line to that, cheers for looking:)

[Wildbug]

Did you print_r($_POST)?  Check the HTML source on the form to make sure the action is pointing to the correct script.  You're not echoing the value in the value="" section.

 

Easier to read:

<select name="siteselect">
<?php
$result = mysql_query("SELECT site_name FROM communitypage");
while($row = mysql_fetch_array($result))
	echo "<option value=\"$row[site_name]\">$row[site_name]</option>";
?>
</select>

[monty187]

Did you print_r($_POST)?  Check the HTML source on the form to make sure the action is pointing to the correct script.  You're not echoing the value in the value="" section.

 

Easier to read:

<select name="siteselect">
<?php
$result = mysql_query("SELECT site_name FROM communitypage");
while($row = mysql_fetch_array($result))
	echo "<option value=\"$row[site_name]\">$row[site_name]</option>";
?>
</select>

 

Cool, I have never used that print_r before but this is what it displays

 

Array ( [siteselect] => [reg] => GO )

 

How would I echo the value?

 

Thanks

[redarrow]

example.

<select name="siteselect">
<?php

$row[]="i";
$row[]="love";
$row[]="php";

foreach($row as $r){

echo "<option value='$r'>$r</option>";

}
?>
</select>

 

edited

 

<select name="siteselect">
<?php
$result = mysql_query("SELECT site_name FROM communitypage");
while($row = mysql_fetch_array($result))
	echo "<option value='$row[site_name]'>$row[site_name]</option>";
?>
</select>

[dbillings]

The loop is left open, maybe that is the problem ??? In order to simply click the drop down box you will need to use AJAX else you will need a submit button.

 

<select name="siteselect">
<?php
$result = mysql_query("SELECT site_name FROM communitypage");
while($row = mysql_fetch_array($result)){
	echo "<option value=\"$row[site_name]\">$row[site_name]</option>";
            }
?>
</select>

[monty187]

The loop is left open, maybe that is the problem ??? In order to simply click the drop down box you will need to use AJAX else you will need a submit button.

 

<select name="siteselect">
<?php
$result = mysql_query("SELECT site_name FROM communitypage");
while($row = mysql_fetch_array($result)){
	echo "<option value=\"$row[site_name]\">$row[site_name]</option>";
             }
?>
</select>

 

Thank you so much guys, it is working now, cheers for all the advice.

 

monty

[Wildbug]

The loop is left open, maybe that is the problem ???

 

The loop wasn't left open; you can use while/for/foreach/if, etc as a single line if there's only one line.

 

I.e.,

 

while ($row = mysql_fetch_row($result)) echo "$row[0]\n";

[dbillings]

But the loop wasn't on one line. 

[Wildbug]

It's still "one line" despite the whitespace (newline).  The line is terminated by a semi-colon, not a newline.  I used a carriage return since it was too long to fit on the screen.  Legibility, you know!