PHP error "Notice: Use of undefined constant row - assumed 'row'" MySQL

Question

Error:

Notice: Use of undefined constant row - assumed 'row'

Relevant Code [PHP]:

if ($result->num_rows > 0) {
        while($row = $result->fetch_assoc()) {
            echo BuildDisplay($row['name'], $row['price'], row['link'], $sc);
            if ($sc = 4) {
                $sc = 1;
            } else {
                $sc = ($sc + 1);
            }
        }
    } else {
        echo "<p>No Products Available!<p>";
    }

Yup missing dollar sign. Voting to close as typographic error — Jeff Puckett, Jun 18 '16 at 15:54
Also `$sc = 4` will always evaluate as true, you aren't comparing there. Two `==`s compares, one sets. — chris85, Jun 18 '16 at 15:57
Use two `=`s. e.g. `if ($sc == 4) {`. or three if you want to be strict on the variable type as well. The `=` is an assignment operator, `==` and `===` are comparison operators. You can read about operators here, http://php.net/manual/en/language.operators.php, and specific comparison here, http://php.net/manual/en/language.operators.comparison.php. — chris85, Jun 18 '16 at 16:47

score 7 · Answer 1 · edited Jun 18 '16 at 15:55

7

row['link'] is missing a dollar sign: $row['link']

edited Jun 18 '16 at 15:55

chris85

23,846
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answered Jun 18 '16 at 15:53

bucketman

463
6
15

PHP error "Notice: Use of undefined constant row - assumed 'row'" MySQL

1 Answers1