CODEWITHSUNDEEP
phpcontinue
taabouzeid
taabouzeid

Reputation: 939

PHP continue statement

When reading a PHP book I wanted to try my own (continue) example. I made the following code but it doesn't work although everything seems to be ok

$num2 = 1;

    while ($num2 < 19)
        {
            if ($num2 == 15) { 
            continue; 
            } else {
            echo "Continue at 15 (".$num2.").<br />";
            $num2++;
            }

        }

The output is

Continue at 15 (1).
Continue at 15 (2).
Continue at 15 (3).
Continue at 15 (4).
Continue at 15 (5).
Continue at 15 (6).
Continue at 15 (7).
Continue at 15 (8).
Continue at 15 (9).
Continue at 15 (10).
Continue at 15 (11).
Continue at 15 (12).
Continue at 15 (13).
Continue at 15 (14).

Fatal error: Maximum execution time of 30 seconds exceeded in /var/www/php/continueandbreak.php on line 20

Line 20 is that line

if ($num2 == 15) {

Would you please tell me what's wrong with my example ? I am sorry for such a Noob question

Upvotes: 0

Views: 6464

Answers (3)

Jeffrey Aylesworth
Jeffrey Aylesworth

Reputation: 8510

You don't even need continue there, your code equivalent to;

$num2 = 1;
while ($num2 < 19){
    if ($num2 != 15) { 
        echo "Continue at 15 (".$num2.").<br />";
        $num2++;
    }
}

If that's not what you're trying to achieve, you're using continue wrong.

Upvotes: 2

Marco
Marco

Reputation: 2737

in php, use foreach for arrays and for for looping

for($num = 1; $num < 19; $num++) {
    if ($num != 15) { 
       echo "Continue at 15 (" . $num . ") . <br />";
       break;
    }   
}

Upvotes: 0

Marek Karbarz
Marek Karbarz

Reputation: 29314

if you don't increment $num2 before the continue you will get into an infinite loop;

$num2 = 0;

while ($num2 < 18)
    {
            $num2++;
            if ($num2 == 15) { 
              continue; 
            } else {
              echo "Continue at 15 (".$num2.").<br />";
            }


    }

Upvotes: 10

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