PHP 4.0 Bug #7591 Updated: continue doesn't work inside switch statement

ID: 7591 Updated by: stas Reported By: joel@azursoft.fr Status: Closed Bug Type: Scripting Engine problem Assigned To: Comments: continue relates to switch as a looping structure (just as break does), so what it does is to continue switch. Use "continue 2;" to do what you want. Previous Comments: --------------------------------------------------------------------------- [2000-11-02 09:57:15] joel@azursoft.fr When placed inside a switch statement located inside a for loop, the continue doesn't work as usual in C: the following code display the following output: <? for ( $i=0; $i<10; $i++ ) { switch ( $i ) { case 5: continue; default: echo "$i<br>"; } echo "Should be printed only if i != 5 and now i == $i<br>"; } ?> //output: 0 Should be printed only if i != 5 and now i == 0 1 Should be printed only if i != 5 and now i == 1 2 Should be printed only if i != 5 and now i == 2 3 Should be printed only if i != 5 and now i == 3 4 Should be printed only if i != 5 and now i == 4 Should be printed only if i != 5 and now i == 5 6 Should be printed only if i != 5 and now i == 6 7 Should be printed only if i != 5 and now i == 7 8 Should be printed only if i != 5 and now i == 8 9 Should be printed only if i != 5 and now i == 9 // End of output The line : Should be printed only if i != 5 and now i == 5 should not appear --------------------------------------------------------------------------- Full Bug description available at: http://bugs.php.net/?id=7591