[RESOLVED] Select data from Drop Down

[nawrik]

After sleeping on it, my head is already spinning with this one.

I'm new to PHP and to be honest don't really know what I'm doing, just winging it!

I'll do my best to explain my problem ALL help is greatly appreciated.

My Database containsthe fields 'linkname' and 'linklocation'.

Each row of these fields is a Main Menu link on a website.

The idea is to create a drop down menu within the website admin pages.
This drop down will enable the user to select and save the 'linklocation' into the database as 'sitehome'.

'sitehome' will then be accessed as a variable, automatically redirecting the vistor to the admin's desired Home Page or 'linklocation'.

I have managed to populate the drop down menu using:

        <select name="sitehome">

	<?php
    while($row = mysql_fetch_array( $result )) {
echo "<option value=\"".$row['linklocation']."\">".$row['linkname']." </option>"; 

} 
?>
        </select>

This seems to work fine. It loops through the fields 'linklocation' and 'linkname' - filling the drop down with the correct data from the database.

Because I haven't used the code:

selected="selected"

within that 'looped' drop down, naturally the first item in that list is displayed by default.

and obiously if I added the code - they would all be 'selected' - eventually making the last item in the list appear as the 'selected' option.

What I'm hoping to achieve is:

Find out what data is stored in 'sitehome' ($sitehome).

That data should always be ONE of the rows in the field 'linklocation' ($linklocation).

As the LOOP drop down is populating, when you come across the 'linklocation' that is equal to the data in 'sitehome' - echo the text 'selected="selected"'.

This should then make the default starting item in the drop down - the item that is currently the 'sitehome'.

Hope this makes sense!!!

Regards,

[coldwerturkey]

you would add something like below in the loop

if ($current_value == $this) {
  $selected = " selected='selected'";
} else {
  $selected = '';
}

then add the $selected var in the <option>, so if your desired value matches that of the current value $selected becomes selected='selected' otherwise it's not selected,

[nawrik]

coldwerturkey,

thankyou for your swift reply.

That was my initial thoughts on resloving - and I'm sure if I new what I was doing I would have solved it that way.

I did try exactly that...

...but couldn't seem to get it working.

Can you start an IF statement in the middle of a LOOP. I couldn't seeme to get it to work.
You probably can - I just wouldn't know how to correctly mark it up.

Again thanks for your swift reply I have now solved the problem. Below is my work around (thanks to a previous thread).

So for all those in a similar situation:

AFTER selecting the data and storing in variables.

$query = "SELECT linklocation FROM menu";
$result = mysql_query($query) or die('Query failed: ' . mysql_error()); 
$linklocation = mysql_result ( $result, 0, "linklocation" );

Start the Drop Down:

<select name="sitehome"><?php


// query your database for your list box values
$sqlquery     = "SELECT linkname,linklocation FROM menu";
$dataresult = mysql_query($sqlquery);

// loop through the values retrieved storing the data to variables
while ($datarow = mysql_fetch_array($dataresult)) {
    $linkname         = $datarow['linkname'];
    $linklocation     = $datarow['linklocation'];

// if the value of the list box element is the same as the value you retrieved from the database for the user in edit mode ($stored_value_in_db)
if ($linklocation == $sitehome) {
    $showselected = "selected";
} else {
    $showselected = "";
}

// $showselected will contain 'selected' or be empty here...
echo " <option ".$showselected." value=\"".$linklocation."\">".$linkname."</option>\n";

}
mysql_free_result($dataresult);        

?>
</select>

Thankyou!

Now I plow head-on to the NEXT problem!