Changing Default Value of Dropdown

[Moorcam]

Hi guys,

Before we go further, yes, I know I should use prepared statements. This is just a project that will probably never go live. If I do decide to go live, I will change to prepared statements.

Anyways, I am populating a text box from the selection of an options dropdown and updating MySQL. This all works fine. However, what I want to do is have the newly inserted option display by default in the dropdown. Hope this makes sense.

Here is the dropdown code with the select statement...

                         <div class="row form-group">
                          <div class="col-6">
                        <div class="form-group"><label for="location" class=" form-control-label">location</label>
<?php
$sql = "SELECT location_name, location_phone FROM locations";
$result = $con->query($sql);
?>

 <select name="location" id="location" onchange="myFunction()" class="form-control">
<?php

        while($r = mysqli_fetch_row($result))
{
        echo "<option data-location_name='$r[1]'  data-location_phone='$r[2]'  value='$r[0]'> $r[0] </option>";
}
?>

</select>
<label>Phone</label><input type="text" class="form-control" name="location_phone" id="location_name" value = "<?php echo $row['location_phone']; ?>"/>

I am grabbing the location name and phone number from locations. Then inserting them into tours. So after the form is submitted, I want the location name to stay in the dropdown. This is in the table "tours". How do I implement that in here?

Thanks heaps.

[requinix]

1. Figure out which one was just submitted. Best method would be to create the location, note the location ID (you have an ID, right?), and check $r for a match (after you've added that ID column to the query).
2. As you output <option>s, mark the new one as selected.

<option data-location_name="..." data-location_phone="..." value="..." selected>

 

[Moorcam]

Hi requinix,

Thanks for the reply.

When I add location_id to the query I only get the id number in the dropdown.

<?php

$sql = "SELECT DISTINCT location_id, location_name, location_phone FROM locations";
$result = $con->query($sql);
?>

 <select name="location" id="location" onchange="myFunction()" class="form-control">
<?php

        while($r = mysqli_fetch_row($result))
{
        echo "<option data-location_name='$r[1]'  data-location_phone='$r[2]'  value='$r[0]' selected> $r[0] </option>";
}
?>

Sorry mate. Struggling with this one.

Edited March 18, 2021 by DanEthical

[Phi11W]

1 hour ago, DanEthical said:

When I add location_id to the query I only get the id number in the dropdown.

I'm guessing that's because you told it to? 

while( $r = mysqli_fetch_row( $result ) )
{
   echo "<option data-location_name='$r[1]'  data-location_phone='$r[2]'  value='$r[0]' selected> $r[0] </option>";
   //                                ^                            ^              ^                ID!! 
   //                                |                            |              ID
   //                                |                            Phone 
   //                                Name
   //
}

Trying putting the name ($r['location_id']) inside the option element, not the id or, rather, whatever happens to be the first column that your query retrieves ($r[0]). 

Regards, 
   Phill  W.

 

[Moorcam]

Thanks mate. Still wasn't working for me. So, gave in and reverted back to good old text boxes.