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php 回调函数中的 continue 报错

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2016-06-06 20:33:211571browse

回调函数功能：遇到是3的倍数，不输出

下面不用回调函数的代码正常执行：

<code>function number($n,$m=''){
    for($i=0;$i</code>

报错内容：: Cannot break/continue 1 level in xxx.php

回复内容：

回调函数功能：遇到是3的倍数，不输出

下面不用回调函数的代码正常执行：

<code>function number($n,$m=''){
    for($i=0;$i</code>

报错内容：: Cannot break/continue 1 level in xxx.php

continue;的上一层没循环当然跳不了啊

php是块作用域，你要跳出的话直接return

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