Syntax error in my PHP form?

Question

Getting this error:

Parse error: syntax error, unexpected ';' in D:\EasyPHP-DevServer-14.1VC9\websitestuff\www\php-form-processor.php on line 9

The code:

$value = ($_POST['formEnquiry']) ? ($_POST['formName']) ? ($_POST['formTitle']) : '';

As @Rizier123 says, unless you're trying to nest ternaries (really bad mistake) in which case you're missing a `:` and option — Mark Baker, Apr 14 '16 at 08:31
@DomJackson count the `?` in your line of code, there's 2 of them — Mark Baker, Apr 14 '16 at 08:32
If you don't understand the ternary operator, write it as an `if` statement, and if you don't understand ternary operators, you certainly shouldn't try to nest them — Mark Baker, Apr 14 '16 at 08:34
Well you shouldn't add another `?` you should add another `:` if you want to nest the ternary operator. Take @MarkBaker advice. For just making an `if else` statement, which is more readable too. — Daan, Apr 14 '16 at 08:34

score 0 · Answer 1 · answered Apr 14 '16 at 08:35

You appear to be trying to use two ternary operators, yet are not specifying an "else clause" for the first one. Breaking the code down:

$value = ($_POST['formEnquiry'])
         ? ($_POST['formName'])
             ? ($_POST['formTitle'])
             : ''
         // You're missing a ':' (else) statement here, like
         : ''
;

Writing it as procedural code, it would look like this:

if ($_POST['formEnquiry']) {
    if ($_POST['formName']) {
        $value = $_POST['formTitle'];
    } else {
        $value = '';
    }
} else {
    // Because of the way an assignment through a ternary works,
    // there needs to be an else here, since you already wrote $value =
    // Without an else value, it would end up like $value = ;
    $value = '';
}

score 0 · Answer 2 · answered Apr 14 '16 at 08:38

0

Try this:

$value = ($_POST['formEnquiry']) ? ($_POST['formName']) ? ($_POST['formTitle']) : '' : '';

answered Apr 14 '16 at 08:38

prava

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Syntax error in my PHP form?

2 Answers2