How to check if value is not NaN in PHP?

4 Views

I’m trying to make a vote script in PHP but I have some syntax error. I’m a real rocky when it comes to PHP so please be kind…

My “doft” value is NaN if not set?

if ( ! isset(int) $_POST["doft"] ){
    $new_value_doft = $prev_value_doft + $doft;
    $new_num_votes_doft = $prev_num_votes_doft + 1;
}
else{
    $new_value_doft = $prev_value_doft + $doft;
    $new_num_votes_doft = $prev_num_votes_doft; 
}
Post Views: 4

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3 comments

  1. The isset method checks if a variable or index is actually set. You want to use it like this

    if( isset( $_POST["doft"] ) ) { }

    or, to check if it is not set

    if( !isset( $_POST["doft"] ) ) { }

    At the same time, you can make sure/check wether the posted value is an integer like using the is_int method:

    if( isset( $_POST["doft"] )  && is_int( $_POST["doft"] ) ) { }

    Or simply use the is_numeric method

    if( isset( $_POST["doft"] )  && is_numeric( $_POST["doft"] ) ) { }

    Your given script should probably look like this:

    if ( !isset( $_POST["doft"] ) ){
    $new_value_doft = $prev_value_doft + $doft;
    $new_num_votes_doft = $prev_num_votes_doft + 1;
}
else{
    $new_value_doft = $prev_value_doft + $doft;
    $new_num_votes_doft = $prev_num_votes_doft; 
}

  3. Your if statement isn’t working.

    It should look something like this.

    if (!isset($_POST['doft']))

    What you have now isn’t valid, thus the syntax error.

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