(PHP 4, PHP 5, PHP 7)
continue is used within looping structures to skip the rest of the current loop iteration and continue execution at the condition evaluation and then the beginning of the next iteration.
Note: In PHP the switch statement is considered a looping structure for the purposes of continue. continue behaves like break (when no arguments are passed). If a switch is inside a loop, continue 2 will continue with the next iteration of the outer loop.
continue accepts an optional numeric argument which tells it how many levels of enclosing loops it should skip to the end of. The default value is 1, thus skipping to the end of the current loop.
<?php
foreach ($arr as $key => $value) {
if (!($key % 2)) { // skip even members
continue;
}
do_something_odd($value);
}
$i = 0;
while ($i++ < 5) {
echo "Outer<br />\n";
while (1) {
echo "Middle<br />\n";
while (1) {
echo "Inner<br />\n";
continue 3;
}
echo "This never gets output.<br />\n";
}
echo "Neither does this.<br />\n";
}
?>
Omitting the semicolon after continue can lead to confusion. Here's an example of what you shouldn't do.
<?php
for ($i = 0; $i < 5; ++$i) {
if ($i == 2)
continue
print "$i\n";
}
?>
One can expect the result to be:
0 1 3 4
but, in PHP versions below 5.4.0, this script will output:
2
because the entire continue print "$i\n"; is evaluated as a single expression, and so print is called only when $i == 2 is true. (The return value of print is passed to continue as the numeric argument.)
Note:
As of PHP 5.4.0, the above example will raise an
E_COMPILE_ERRORerror.
| Version | Description |
|---|---|
| 7.0.0 |
continue outside of a loop or switch
control structure is now detected at compile-time instead of run-time as
before, and triggers an E_COMPILE_ERROR.
|
| 5.4.0 | continue 0; is no longer valid. In previous versions it was interpreted the same as continue 1;. |
| 5.4.0 | Removed the ability to pass in variables (e.g., $num = 2; continue $num;) as the numerical argument. |