Html Select Box Get value from mysql

Question

Hi I have a stored value in mysql from a select box. When i call a edit page I would like the select box to populate with the value i have stored in mysql

$taskCompany = $mysqli->real_escape_string($_POST['taskCompany']); This returns 1 

<select id="taskCompany" required name="taskCompany" class="form-control">

    <option value="" disabled selected>Select a Company</option>
    <option value="1">building</option>
    <option value="2">maintenace</option>
</select></div>

I would like the select box to show the correct option when loading the edit page. I do not store any select option in the database values or text just the value that is select when job was created. Thanks Jon

Is it me or are I have a stored value in mysql from a select box and I do not store any select option in the database values or text contradicting each other — SuperDJ
– SuperDJ, Commented Aug 20, 2015 at 10:05

DirtyBit · Accepted Answer · 2015-08-20 10:57:26Z

1

I'd I have understood correctly, you have he value/values in your MySQL database and you want them to be shown in your drop down list: You could do it like:

<option value = "1" <?php if($taskCompany == 1) echo "select='selected'";?>>building</option>

Or if you have the names coming from the database too.

<select id="taskCompany" required name="taskCompany" class="form-control">

<option>SELECT</option>
<?php
$res = musql_query("SELECT * FROM yourtablename");
while($row = mysql_fetch_array($res))
{
$taskCompany = $row['taskCompany'];
$co_name = $row['taskCompanyName'];
?>
<option value = "<?php echo $taskCompany; ?>" ><?php echo co_name; ?></option>
<?php } ?>

Here I have used $co_name to be displayed as a name assuming that you have building, maintenance etc stored in your database.

This is for the time being, as I wanted to demonstrate but Please consider mysqli/PDO as mysql_* is deprecated.

edited Aug 20, 2015 at 10:57

answered Aug 20, 2015 at 10:52

DirtyBit

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abdullacm · Accepted Answer · 2015-08-20 10:47:37Z

0

<?php $taskCompany = $mysqli->real_escape_string($_POST['taskCompany']); ?>

<select id="taskCompany" required name="taskCompany" class="form-control">

    <option value="" disabled selected>Select a Company</option>
    <option value="1" <?php if($taskCompany == 1) echo "selected='selected'"; ?>>building</option>
    <option value="2" <?php if($taskCompany == 2) echo "selected='selected'"; ?>>maintenace</option>
</select></div>

edited Aug 20, 2015 at 10:47

answered Aug 20, 2015 at 10:05

abdullacm

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4 Comments

jonathan young Over a year ago

Thanks for your quick reply but it does not seem to work. I have setup a fiidle take a look FIDDLE

abdullacm Over a year ago

add jquery library in your fiddle and remove the script tag from fiddle's script area

jonathan young Over a year ago

Thanks but did not want to use jquery I have answered my own question my adding and is statement <?php if ($row['taskCompany'] == '1') {$select ='selected'; }else{$select2 ='selected';}; ?> Thanks

abdullacm Over a year ago

check the modified answer

Niranjan N Raju · Accepted Answer · 2015-08-20 10:07:37Z

0

You can do this

<?php foreach($taskCompany as $row){ ?>
    <option value="unique id for select options">name for select options</option>
<?php } ?>

answered Aug 20, 2015 at 10:07

Niranjan N Raju

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