PHP drop down box not showing selected value

Question

Hi I have a piece of php code which takes php variables from a database and then populates A drop down box. This works fine but once the page updates it doesn't display the selected value and instead just displays the original list again.

I have played around with some different ideas of using isset to displays the selected value, but then it just displays that value and nothing else.

I have attached the original code where it works but always only shows the original list of values from the database.

If any body could provide a solution or even a nudge in the right direction then it would be much appreciated.

Thanks

if (isset($select) && $select != "location") {
    $select = $_POST['location'];
}
?>

<select name="location">
<?php
// Get records from database (table "name_list").
$list = mysql_query("select DISTINCT region_name from masterip_details WHERE country_code='GB' AND TRIM(IFNULL(region_name,'')) <> '' order by region_name asc");

// Show records by while loop.
while ($row_list = mysql_fetch_assoc($list)) {
    $location = $_POST['location']; ?>
    <option value="<?php echo $row_list['region_name']; ?>"
    <?php if  $row_list['region_name'] == $select) {
        echo "selected";
    } ?>><?php echo $row_list['region_name']; ?></option>        
<?php } ?>

Check $_POST['location'] has any of one dropdown box values. — Ranjith
– Ranjith, Commented Jul 18, 2013 at 9:51

alabama · Accepted Answer · 2013-07-18 10:06:09Z

1

You are sure you send that form with a post request?

Your code is unreadable Better do something like this:

<?php

    //database request don't belong in the presentation layer
    $list=mysql_query("select DISTINCT region_name from masterip_details WHERE country_code='GB' AND TRIM(IFNULL(region_name,'')) <> '' order by region_name asc");
    $location = "";
    $isSelect = "";
    if(isset($_POST['location'])){
        $location = $_POST['location']; 
    }
    while($row_list=mysql_fetch_assoc($list)){
        if($row_list['region_name'] == $location ){ 
            $isSelect = "selected"; 
         } 
         echo '<option value="'.$row_list['region_name'].'" '.$isSelect.'  >'. $row_list['region_name'].'</option>';

    }
?>

edited Jul 18, 2013 at 10:06

answered Jul 18, 2013 at 9:57

alabama

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1 Comment

user2516043 Over a year ago

Thanks, I like the code much neater and easier to follow though it should an error around If(if($row ... But i will look at this and get it working as it is much neater than my way.

Chibueze Opata · Accepted Answer · 2013-07-18 10:02:39Z

1

You are not checking the select value properly.

Change this:

if (isset($select) && $select != "location") {
    $select = $_POST['location'];
}

TO

if (isset($_POST['location']) && $_POST['location'] != "location") {
    $select = $_POST['location'];
}

answered Jul 18, 2013 at 10:02

Chibueze Opata

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Comments

Vivek Jain · Accepted Answer · 2013-07-18 10:04:57Z

0

While creating the <option> tag, you may use the following:

<?php if ($row_list['region_name'] == $select) { echo "selected='true'"; } ?>

You may also like to check if the condition $row_list['region_name'] == $select is returning true. You may try to echo their result separately as:

echo ($row_list['region_name'] == $select)

or something similar that would show if you get true or not.

edited Jul 18, 2013 at 10:04

answered Jul 18, 2013 at 9:59

Vivek Jain

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2 Comments

Vivek Jain Over a year ago

@Ranjith, yes it is not necessary to set it like this. The issue could be with $row_list['region_name'] == $select also. But adhering to XHTML conventions, every attribute should have a value.

Ranjith Over a year ago

Yes, you are right. But here its not the major problem. refer this w3.org/community/webed/wiki/HTML/Elements/option#Example_A link of w3.org

Vivek Jain · Accepted Answer · 2013-07-18 10:27:43Z

0

The problem lies in the first line - actually the system will never execute 2nd line to assign $select because $select is never set:

if(isset($_POST['select']) && $select!="location") { $select=$_POST['location']; } ?>

edited Jul 18, 2013 at 10:27

Vivek Jain

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answered Jul 18, 2013 at 10:04

MBaas

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1 Comment

MBaas Over a year ago

and also the bit $location = $_POST['location'] in the while loop is not making sense...

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