PHP: Hypertext Preprocessor

`continue`

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

continue se utiliza dentro de las estructuras iterativas para saltar el resto de la iteración actual del bucle y continuar la ejecución en la evaluación de la condición, para luego comenzar la siguiente iteración.

Nota: En PHP, la sentencia switch se considera una estructura iterativa para los propósitos de continue. continue se comporta igual que break (cuando no se proporcionan argumentos). Si un switch está dentro de un bucle, continue 2 continurá con la siguiente iteración del bucle externo.

continue acepta un argumento numérico opcional, que indica a cuántos niveles de bucles encerrados se ha de saltar al final. El valor por omisión es 1, por lo que salta al final del bucle actual.

<?php
while (list($clave, $valor) = each($arr)) {
    if (!($clave % 2)) { // saltar los miembros pares
        continue;
    }
    hacer_algo($valor);
}

$i = 0;
while ($i++ < 5) {
    echo "Exterior<br />\n";
    while (1) {
        echo "Medio<br />\n";
        while (1) {
            echo "Interior<br />\n";
            continue 3;
        }
        echo "Esto nunca se imprimirá.<br />\n";
    }
    echo "Ni esto tampoco.<br />\n";
}
?>

Omitir el punto y coma después del continue puede llevar a confusión. He aquí un ejemplo de lo que no se debe hacer.

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

Se esperaría que el resultado fuera:

pero, en versiones de PHP anteriores a la 5.4.0, la salida de este script será:

debido a que continue print "$i\n"; se evalúa completo como una sola expresión, y así print se llama solamente cuando $i == 2 es verdadero. (El valor de retorno de print es pasado a continue como el argumento numérico.)

Nota:
A partir de PHP 5.4.0, el ejemplo anterior emitirá un error E_COMPILE_ERROR.

**Registro de cambios para `continue`**
Versión	Descripción
5.4.0	`continue 0;` ya no es válido. En versiones anteriores era interpretado de la misma manera que `continue 1;`.
5.4.0	Se eliminó la posibilidad de pasar variables (por ejemplo, `$num = 2; continue $num;`) como el argumento numérico.

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User Contributed Notes 7 notes

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140

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.

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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



*/



?>

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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.

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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;
}
?>

down

www.derosetechnologies.com ¶

20 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.

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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.

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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.

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