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Undefined variable, but it is there

开发者 https://www.devze.com 2023-01-14 10:18 出处:网络
I\'ve got an array, but when I try to use it, I get the Undefined variable notice. Here\'s the relevant lines:

I've got an array, but when I try to use it, I get the Undefined variable notice.

Here's the relevant lines:

    $varEvents = array();
...
    if ($selectedResult) {
while ($row = mysql_fetch_assoc($selectedResult)) {
    array_push($varEvents, $row['eventID']);
}
mysql_free_result($selectedResult);

}

...
            print_r($varEvents);

        if (is_array($varEvents)) {
            if (count($varEvents) > 0) {
            if (in_array($id, $varEvents)) {
                $varRegistered = 1;
            }
        }
        unset($varEvents);
    }

and the re开发者_如何学JAVAsult shows as:

Array ( [0] => 4 ) Notice: Undefined variable: varEvents in /home/.../www/registration.php on line 143 Notice: Undefined variable: varEvents in /home/.../www/registration.php on line 145

line 143: print_r($varEvents); line 145: if (is_array($varEvents)) {

All relevant lines are in the same loop and I do get most of the results I expect, except $varRegistered never changes to 1 and that messes up my result.


It is most likely because of this line:

unset($varEvents);

You are unsetting the variable within the loop and next iterations don't find it again.

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