CODEWITHSUNDEEP
phphtml
SG_Rowin
SG_Rowin

Reputation: 622

Display different content based on Select option value - PHP

I am trying to display info based on what the user selects in drop down box and seems I cant get it working:

Here is what I did:

PHP:

if(isset($_POST['order-type'][0])){
$OrderType =  "small format black and white";
}

if(isset($_POST['order-type'][1])){
$OrderType =  "small format color";
}

HTML:

<select name="order-type">
  <option value="SmallFormatBW">Small Format Black & White</option>
  <option value="SmallFormatColor">Small Format Color</option>
</select>

Any help is high appreciated! Thanks,

Upvotes: 0

Views: 2188

Answers (4)

BCM
BCM

Reputation: 675

If you use $_POST['order-type'] as array you must collect array from form. In your case you can try this.

<select name="order-type[]">
  <option value="SmallFormatBW">Small Format Black & White</option>
  <option value="SmallFormatColor">Small Format Color</option>
</select>

Upvotes: 1

Sumit
Sumit

Reputation: 1639

Try this 

if (isset($_POST['order-type'])) {
    if($_POST['order-type'] == 'SmallFormatBW'){
        $OrderType =  "small format black and white";
    } else if($_POST['order-type'] == 'SmallFormatColor'){
        $OrderType =  "small format color";
    }
}

Upvotes: 0

frz3993
frz3993

Reputation: 1635

When you select an option with the value="SmallFormatBW" from a select with name="order-type" and POST it to PHP. You will access it in PHP as $_POST['order-type'] carrying the value SmallFormatBW.

$_POST['order-type'] == 'SmallFormatBW'

That means if you want to test the value, you would do it like this

if(!empty($_POST['order-type']) && ($_POST['order-type'] == 'SmallFormatBW')){
    $OrderType =  "small format black and white";
}

Upvotes: 0

funnyproffy
funnyproffy

Reputation: 1

you are missing a '$_GET' in the HTML I think...

look here for more info about submitting info from a PHP page to a HTML page: http://www.w3schools.com/php/php_forms.asp

Upvotes: 0

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