Question

SOLVED – USER ERROR – helps if one declares a variable doesn’t it…

Help,

I have the following PHP Function code:

function new_respondent() {
    global $link;
    $proc = mysqli_prepare($link, "INSERT INTO trespondent_bps (code) VALUES (uuid());");
    mysqli_stmt_execute($proc);
    $respondent_id = mysqli_insert_id($link);
    mysqli_stmt_fetch($proc);
    mysqli_stmt_close($proc);
    mysqli_clean_connection($link);
    return($code);
}

For the life of me, it’s adding the UUID into the database no problems at all – but it’s not returning it so I can use it in the code – what am I missing/ doing wrong!!!

Thanks in advance,

Homer.

Answer 1

You return a variable $code that is undefined.

function new_respondent() {
    global $link;
    $proc = mysqli_prepare($link, "INSERT INTO trespondent_bps (code) VALUES (uuid());");
    mysqli_stmt_execute($proc);
    $respondent_id = mysqli_insert_id($link);
    mysqli_stmt_fetch($proc);
    mysqli_stmt_close($proc);
    mysqli_clean_connection($link);
    return($respondent_id);
}