continue

(PHP 4, PHP 5, PHP 7, PHP 8)

continue wird innerhalb von Schleifen verwendet, um den Rest des aktuellen Schleifendurchlaufs abzubrechen und mit der Auswertung der nächsten Bedingung fortzufahren, um dann den nächsten Durchlauf zu beginnen.

Hinweis: In PHP wird das switch-Statement im Sinne von continue als Schleifenstruktur betrachtet. continue verhält sich wie break (wenn keine Argumente übergeben werden), erzeugt aber eine Warnung, da es sich vermutlich um einen Fehler handelt. Falls sich ein switch innerhalb einer Schleife befindet, wird continue 2 mit der nächsten Iteration der äußeren Schleife fortsetzen.

continue akzeptiert ein optionales numerisches Argument, das angibt, wie viele Ebenen umschließender Schleifen bis zu ihrem Ende übersprungen werden sollen. Die Voreinstellung ist 1, so dass zum Ende der aktuellen Schleife gesprungen wird.

<?php
$arr
= ['zero', 'one', 'two', 'three', 'four', 'five', 'six'];
foreach (
$arr as $key => $value) {
if (
0 === ($key % 2)) { // ignoriere Elemente mit geradem Schlüssel
continue;
}
echo
$value . "\n";
}
?>

Die obigen Bespiele erzeugen folgende Ausgabe:

one
three
five
<?php
$i
= 0;
while (
$i++ < 5) {
echo
"Äußere\n";
while (
1) {
echo
"Mittlere\n";
while (
1) {
echo
"Innere\n";
continue
3;
}
echo
"Das hier wird nie ausgegeben.\n";
}
echo
"Das hier ebenfalls nicht.\n";
}
?>

Die obigen Bespiele erzeugen folgende Ausgabe:

Outer
Middle
Inner
Outer
Middle
Inner
Outer
Middle
Inner
Outer
Middle
Inner
Outer
Middle
Inner

Das Weglassen des Semikolons nach continue kann zu unerwarteten Ergebnissen führen. Hier ist ein Beispiel, das zeigt, was Sie nicht tun sollten.

<?php
for ($i = 0; $i < 5; ++$i) {
if (
$i == 2)
continue
print
"$i\n";
}
?>

Man könnte meinen, die Ausgabe wäre:

0
1
3
4

Changelog für continue
Version Beschreibung
7.3.0 continue innerhalb eines switch, das versucht wie eine break-Anweisung für das switch zu wirken, erzeugt ein E_WARNING.

add a note

User Contributed Notes 20 notes

up
136
jaimthorn at yahoo dot com
14 years ago
The remark "in PHP the switch statement is considered a looping structure for the purposes of continue" near the top of this page threw me off, so I experimented a little using the following code to figure out what the exact semantics of continue inside a switch is:

<?php

for( $i = 0; $i < 3; ++ $i )
{
echo
' [', $i, '] ';
switch(
$i )
{
case
0: echo 'zero'; break;
case
1: echo 'one' ; XXXX;
case
2: echo 'two' ; break;
}
echo
' <' , $i, '> ';
}

?>

For XXXX I filled in

- continue 1
- continue 2
- break 1
- break 2

and observed the different results. This made me come up with the following one-liner that describes the difference between break and continue:

continue resumes execution just before the closing curly bracket ( } ), and break resumes execution just after the closing curly bracket.

Corollary: since a switch is not (really) a looping structure, resuming execution just before a switch's closing curly bracket has the same effect as using a break statement. In the case of (for, while, do-while) loops, resuming execution just prior their closing curly brackets means that a new iteration is started --which is of course very unlike the behavior of a break statement.

In the one-liner above I ignored the existence of parameters to break/continue, but the one-liner is also valid when parameters are supplied.
up
40
Nikolay Ermolenko
15 years ago
Using continue and break:

<?php
$stack
= array('first', 'second', 'third', 'fourth', 'fifth');

foreach(
$stack AS $v){
if(
$v == 'second')continue;
if(
$v == 'fourth')break;
echo
$v.'<br>';
}
/*

first
third

*/

$stack2 = array('one'=>'first', 'two'=>'second', 'three'=>'third', 'four'=>'fourth', 'five'=>'fifth');
foreach(
$stack2 AS $k=>$v){
if(
$v == 'second')continue;
if(
$k == 'three')continue;
if(
$v == 'fifth')break;
echo
$k.' ::: '.$v.'<br>';
}
/*

one ::: first
four ::: fourth

*/

?>
up
18
Koen
11 years ago
If you use a incrementing value in your loop, be sure to increment it before calling continue; or you might get an infinite loop.
up
17
rjsteinert.com
13 years ago
The most basic example that print "13", skipping over 2.

<?php
$arr
= array(1, 2, 3);
foreach(
$arr as $number) {
if(
$number == 2) {
continue;
}
print
$number;
}
?>
up
11
www.derosetechnologies.com
19 years ago
In the same way that one can append a number to the end of a break statement to indicate the "loop" level upon which one wishes to 'break' , one can append a number to the end of a 'continue' statement to acheive the same goal. Here's a quick example:

<?
for ($i = 0;$i<3;$i++) {
echo
"Start Of I loop\n";
for (
$j=0;;$j++) {

if (
$j >= 2) continue 2; // This "continue" applies to the "$i" loop
echo "I : $i J : $j"."\n";
}
echo
"End\n";
}
?>

The output here is:
Start Of I loop
I : 0 J : 0
I : 0 J : 1
Start Of I loop
I : 1 J : 0
I : 1 J : 1
Start Of I loop
I : 2 J : 0
I : 2 J : 1

For more information, see the php manual's entry for the 'break' statement.
up
5
greg AT laundrymat.tv
19 years ago
You using continue in a file included in a loop will produce an error. For example:

//page1.php
for($x=0;$x<10;$x++)
{
include('page2.php');
}

//page2.php

if($x==5)
continue;
else
print $x;

it should print

"012346789" no five, but it produces an error:

Cannot break/continue 1 level in etc.
up
2
tufan dot oezduman at gmail dot com
17 years ago
a possible explanation for the behavior of continue in included scripts mentioned by greg and dedlfix above may be the following line of the "return" documentation: "If the current script file was include()ed or require()ed, then control is passed back to the calling file."
The example of greg produces an error since page2.php does not contain any loop-operations.

So the only way to give the control back to the loop-operation in page1.php would be a return.
up
0
clau r jimenez
3 years ago
I've been playing around to see what it does in practice. This is what helped me understand what it does and its difference with using break.

<?php
$i
= 0;
while (
$i++ < 5) {
while (
$i % 2 === 0) {
echo
"$i is even. \n";
####
}
echo
"$i is odd. \n";
}
?>

Where ####:
- break: outputs both '$i is even' and '$i is odd' for even numbers.
- continue: infinite loop as soon as it evaluates true for the first even number.
- break 2: as soon as it runs, it exits from both loops.
- continue 2: outputs numbers correctly.

What I understand from this is that break will exit current looping structure and will keep running outer loop code. Continue will make loop get back to evaluation, and will iterate over itself until it evaluates to false. Break 2 will exit 2 levels, which in this case will stop the iteration altogether. Continue 2 will evaluate not the current loop (level 1 so to speak), but the outer loop in this case.
up
0
Geekman
16 years ago
For clarification, here are some examples of continue used in a while/do-while loop, showing that it has no effect on the conditional evaluation element.

<?php
// Outputs "1 ".
$i = 0;
while (
$i == 0) {
$i++;
echo
"$i ";
if (
$i == 1) continue;
}

// Outputs "1 2 ".
$i = 0;
do {
$i++;
echo
"$i ";
if (
$i == 2) continue;
} while (
$i == 1);
?>

Both code snippets would behave exactly the same without continue.
up
-3
skippychalmers at gmail dot com
12 years ago
To state the obvious, it should be noted, that the optional param defaults to 1 (effectively).
up
-1
mparsa1372 at gmail dot com
3 years ago
The continue statement breaks one iteration (in the loop), if a specified condition occurs, and continues with the next iteration in the loop.

This example skips the value of 4:

<?php
for ($x = 0; $x < 10; $x++) {
if (
$x == 4) {
continue;
}
echo
"The number is: $x <br>";
}
?>
up
-9
maik penz
11 years ago
Please note that with PHP 5.4 continue 0; will fail with

PHP Fatal error: 'continue' operator accepts only positive numbers

(same is true for break).
up
-11
Anonymous
12 years ago
<?php
function print_primes_between($x,$y)
{
for(
$i=$x;$i<=$y;$i++)
{
for(
$j= 2; $j < $i; $j++) if($i%$j==0) continue 2;
echo
$i.",";
}
}
?>

This function, using continue syntax, is to print prime numbers between given numbers, x and y.
For example, print_primes_between(10,20) will output:

11,13,17,19,23,29,
up
-9
send at mail dot 2aj dot net
6 years ago
As of PHP 7.0, instead of code executing up until encountering a continue (or break) call outside of a loop statement, the code will simply not execute.

If you need to correct such error cases as part of an upgrade, you may need to substitute either an exit or return to maintain the existing behavior of such legacy code.

<?php

class ok {

function
foo() {
echo
"start\n";

for (
$i = 0; $i < 5; $i++) {
echo
"before\n";
$this->bar($i);
echo
"after\n";
}

echo
"finish\n";
}

function
bar($i) {
echo
"inside iteration $i\n";

if (
$i == 3) {
echo
"continuing\n";
continue;
}

echo
"inside after $i\n";
}
}

$ex = new ok();

$ex->foo();

?>

sh> php56 continue.php
start
before
inside iteration 0
inside after 0
after
before
inside iteration 1
inside after 1
after
before
inside iteration 2
inside after 2
after
before
inside iteration 3
continuing
PHP Fatal error: Cannot break/continue 1 level in continue.php on line 22
PHP Stack trace:
PHP 1. {main}() continue.php:0
PHP 2. ok->foo() continue.php:31
PHP 3. ok->bar() continue.php:10

sh> php70 continue.php
PHP Fatal error: 'continue' not in the 'loop' or 'switch' context in continue.php on line 22

Fatal error: 'continue' not in the 'loop' or 'switch' context in continue.php on line 22
up
-12
net_navard at yahoo dot com
18 years ago
Hello firends

It is said in manually:
continue also accepts an optional numeric argument which tells it how many levels of enclosing loops it should .

In order to understand better this,An example for that:
<?php

/*continue also accepts an optional numeric argument which
tells it how many levels of enclosing loops it should skip.*/

for($k=0;$k<2;$k++)
{
//First loop


for($j=0;$j<2;$j++)
{
//Second loop

for($i=0;$i<4;$i++)
{
//Third loop
if($i>2)
continue
2;// If $i >2 ,Then it skips to the Second loop(level 2),And starts the next step,
echo "$i\n";
}

}

}

?>

Merry's christmas :)

With regards,Hossein
up
-14
Anonymous
15 years ago
The continue keyword can skip division by zero:
<?php
$i
= 100;
while (
$i > -100)
{
$i--;
if (
$i == 0)
{
continue;
}
echo (
200 / $i) . "<br />";
}
?>
up
-10
dedlfix gives me a hint
19 years ago
a possible solution for
greg AT laundrymat.tv

I've got the same problem as Greg
and now it works very fine by using
return() instead of continue.

It seems, that you have to use return()
if you have a file included and
you want to continue with the next loop
up
-18
John
10 years ago
<?php

echo"\n";
echo
"\n";

for (
$i = 0; $i < 5; $i++ ) {

switch (
$i)
{

case
0:
echo
$i . "b";
continue;
echo
$i . "a";
case
1:
echo
$i . "b";
continue
2;
echo
$i . "a";
case
2:
echo
$i . "b";
break;
echo
$i . "a";
case
3:
echo
$i . "b";
break
2;
echo
$i . "a";
case
4:
echo
$i;

}

echo
9;

}

echo
"\n";
echo
"\n";

?>

This results in: 0b91b2b93b

It goes to show that in a switch statement break and continue are the same. But in loops break stops the loop completely and continue just stops executing the current iterations code and moves onto the next loop iteration.
up
-20
szrrya at yahoo dot com
17 years ago
Documentation states:

"continue is used within looping structures to skip the rest of the current loop iteration"

Current functionality treats switch structures as looping in regards to continue. It has the same effect as break.

The following code is an example:

<?php
for ($i1 = 0; $i1 < 2; $i1++) {
// Loop 1.
for ($i2 = 0; $i2 < 2; $i2++) {
// Loop 2.
switch ($i2 % 2) {
case
0:
continue;
break;
}
print
'[' . $i2 . ']<br>';
}
print
$i1 . '<br>';
}
?>

This outputs the following:
[0]
[1]
0
[0]
[1]
1

Switch is documented as a block of if...elseif... statements, so you might expect the following output:
[1]
0
[1]
1

This output requires you to either change the switch to an if or use the numerical argument and treat the switch as one loop.
up
-32
Rene
18 years ago
(only) the reason that is given on the "Continue with missing semikolon" example is wrong.

the script will output "2" because the missing semikolon causes that the "print"-call is executed only if the "if" statement is true. It has nothing to to with "what" the "print"-call would return or not return, but the returning value can cause to skip to the end of higher level Loops if any call is used that will return a bigger number than 1.

<?php
continue print "$i\n";
?>

because of the optional argument, the script will not run into a "unexpected T_PRINT" error. It will not run into an error, too, if the call after continue does return anything but a number.

i suggest to change it from:
because the return value of the print() call is int(1), and it will look like the optional numeric argument mentioned above.

to
because the print() call will look like the optional numeric argument mentioned above.
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