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How to extract the value from a select field type, not the label?

  • I’m trying using the_sub_field and $value = get_sub_field(‘choice’); and both display “Choice 1”, I want to get “choice_1” instead.

    This is what I have in choices:

    choice_1 : Choice 1
    choice_2 : Choice 2

    Regards

  • Nevermind, it seems to work now.

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