CODEWITHSUNDEEP
phpforms
Kevin Hamer
Kevin Hamer

Reputation: 69

Remember submitted field in form drop down list

I have a HTML form and PHP script on the same page, which calculates a price based in the input from the HTML form.

It is important that the information provided in the form is stored when the PHP script runs, so that the customer can see what the price is based on.

The following code stores the text input:

<input type="number" min="1" max="420" name="stofprijs" width="100" placeholder="Prijs"  value="<?=( isset( $_POST['stofprijs'] ) ? $_POST['stofprijs'] : '' )?>"> euro.

I had tried to store the data from the drop down list with this code:

On top:

<?php
    $gordijnsoort = $_POST['gordijnsoort'];
?>

In the form:

<select name="gordijnsoort">
    <option value="gordijn" <?php echo $gordijnsoort == 'gordijn' ? 'selected="selected"' : ''; ?>>Gordijn</option>
    <option value="vouwgordijn" <?php echo $gordijnsoort == 'vouwgordijn' ? 'selected="selected"' : ''; ?>>Vouwgordijn</option>
    <option value="ringgordijn" <?php echo $gordijnsoort == 'ringgordijn' ? 'selected="selected"' : ''; ?>>Ringgordijn</option>
</select>

The data is stored, but the empty form keeps displaying the error:

Undefined index: gordijnsoort

The form should only save the variable when the customer submits the form, but now it's already looking for the variable at the start.

Does anyone know how to fix this?

Upvotes: 0

Views: 71

Answers (4)

beniamin
beniamin

Reputation: 81

I think you will have to check if $_POST['gordijnsoort'] isset. Your code should look like:

$gordijnsoort = "";
if(isset($_POST['gordijnsoort'])
    $gordijnsoort = $_POST['gordijnsoort'];

This will prevent callign non existing index before sending the form.

Upvotes: 0

realshadow
realshadow

Reputation: 2585

Thats because $_POST is empty. It will only be filled after you submit your form. So first you have to check with isset, empty if gordijnsoort exists.

E.g.

$gordijnsoort = isset($_POST['gordijnsoort']) ? $_POST['gordijnsoort'] : '';

Upvotes: 1

Debflav
Debflav

Reputation: 1151

Set your variable to NULL if it doesn't exist.

$gordijnsoort = isset($_POST['gordijnsoort']) ? $_POST['gordijnsoort'] : NULL;

Upvotes: 0

Tobias Golbs
Tobias Golbs

Reputation: 4616

Try this:

<?php
    $gordijnsoort = array_key_exists('gordijnsoort', $_POST) ? $_POST['gordijnsoort'] : "";
?>

You will receive the error because at first call the $_POST array is empty and you try to access a undefined key.

Upvotes: 1

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