break
The continue keyword can skip division by zero:
<?php
$i = 100;
while ($i > -100)
{
$i--;
if ($i == 0)
{
continue;
}
echo (200 / $i) . "<br />";
}
?>
continue
continueは、ループ構造において現在の繰り返しループ の残りの処理をスキップし、条件式を評価した後に 次の繰り返しの最初から実行を続けるために使用されます、
注意: switch 命令はループ構造におけるcontinueの効果 を考慮してつくられていることに留意してください。
continueでは、オプションの引数で 処理をスキップするループ構造のレベルの数を指定できます。
while (list($key,$value) = each ($arr)) {
if (!($key % 2)) { // キーが偶数の組をスキップ
continue;
}
do_something_odd($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";
}
continue の後のセミコロンを省略すると混乱を生じる ことがあります。以下にすべきでないことの例を示します。
<?php
for ($i = 0; $i < 5; ++$i) {
if ($i == 2)
continue
print "$i\n";
}
?>
これは、以下のような結果になることを期待していたのでしょう。
0 1 3 4
しかし、このスクリプトの出力は以下のようになります。
2
これは、print()コールの返り値が int(1) であり、前記のオプションの数値引数のように 見えてしまうためです。
add a note
User Contributed Notescontinue
Anonymous
12-Sep-2008 02:04
12-Sep-2008 02:04
Geekman
28-Dec-2007 10:01
28-Dec-2007 10:01
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.
tufan dot oezduman at gmail dot com
21-Dec-2006 09:28
21-Dec-2006 09:28
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.
szrrya at yahoo dot com
18-Jul-2006 06:18
18-Jul-2006 06:18
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 $i1 . '<br>';
}
?>
This outputs the following:
[0]
[1]
0
[0]
[1]
1
Switch is documented as a block of if...elseif... statements, so you might expect the following output:
[1]
0
[1]
1
This output requires you to either change the switch to an if or use the numerical argument and treat the switch as one loop.
Rene
19-Feb-2006 05:24
19-Feb-2006 05:24
(only) the reason that is given on the "Continue with missing semikolon" example is wrong.
the script will output "2" because the missing semikolon causes that the "print"-call is executed only if the "if" statement is true. It has nothing to to with "what" the "print"-call would return or not return, but the returning value can cause to skip to the end of higher level Loops if any call is used that will return a bigger number than 1.
<?php
continue print "$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.
net_navard at yahoo dot com
26-Dec-2005 02:01
26-Dec-2005 02:01
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
dedlfix gives me a hint
28-Jan-2005 11:47
28-Jan-2005 11:47
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
greg AT laundrymat.tv
15-Jan-2005 01:58
15-Jan-2005 01:58
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 $x;
it should print
"012346789" no five, but it produces an error:
Cannot break/continue 1 level in etc.
www.derosetechnologies.com
11-May-2004 12:58
11-May-2004 12:58
In the same way that one can append a number to the end of a break statement to indicate the "loop" level upon which one wishes to 'break' , one can append a number to the end of a 'continue' statement to acheive the same goal. Here's a quick example:
<?
for ($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.