Calling a custom field's contents in PHP

Problem:
Calling a custom field's contents in PHP

Solution:
you should use the Types fields API PHP function types_render_field() to get the value of your custom field.

$value = types_render_field("is-the-offer-a-gambling-offer", array("output"=>"raw"));

Relevant Documentation:
=> https://toolset.com/documentation/customizing-sites-using-php/functions/#select
=> https://toolset.com/documentation/customizing-sites-using-php/

This support ticket is created vor 5 Jahren, 8 Monaten. There's a good chance that you are reading advice that it now obsolete.

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Dieses Thema enthält 3 Antworten, hat 2 Stimmen.

Assistiert von: Minesh.

Author

Artikel

Juni 9, 2019 at 1:59 pm #1263481

matthewL-7

Hi,

I have a couple of questions about using a custom field and conditional formatting but in PHP code as opposed to using the visual editor to do it.

I have a field with the field slug "is-the-offer-a-gambling-offer", it is a select field with the contents of the field set to 1 or 2.

The code I am trying to use is relatively simple but its not working:

if ($wpcf-is-the-offer-a-gambling-offer = 2)
  echo "Some text";

Do I need to do something else to be able to call the field? Am I missing something.

Juni 10, 2019 at 5:19 am #1263647

Minesh

Supporter

Sprachen: Englisch (English )

Zeitzone: Asia/Kolkata (GMT+05:30)

Hello. Thank you for contacting the Toolset support.

Well - you should use the Types fields API PHP function types_render_field() to get the value of your custom field.

For example:

$value = types_render_field("is-the-offer-a-gambling-offer", array("output"=>"raw"));

if ($value == 2){
   // add something here
}

Juni 10, 2019 at 3:02 pm #1264283

matthewL-7

Great thanks. 🙂

Juni 10, 2019 at 3:05 pm #1264293

matthewL-7

All sorted.