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关于回调函数无法显示的help

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Release: 2016-06-23 14:12:55
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回调函数

function filter(){
for($i=0;$i if(call_user_func_array('one',array($i))){
continue;
echo $i."
";
}
}
}

function one($num){
return $num%3==0;
}

echo filter();



为什么这段代码无法显示的,我注释掉一些代码,循环都是正常的,求高手!

回复讨论(解决方案)

把 continue; 去掉就可以显示了

但是去掉之后,3的倍数就显示出来了啊 ,我要它不显示3的倍数啊

$num%3==0 不就是 3 的倍数吗?

你最好先描述一下你的需求

我想循环出0-100的数字,然后利用回调函数过滤掉3的倍数

function filter(){  for($i=0;$i<=100;$i++){    if(call_user_func_array('one',array($i))){      echo $i."<br>";	    }  }}function one($num){  return $num % 3;}echo filter();
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为什么不用contiune的呢 亲

你想用 contiune ?那么

function filter(){  for($i=0;$i<=100;$i++){    if(call_user_func_array('one',array($i))){      contiune;    }    echo $i."<br>";      }} function one($num){  return $num % 3 == 0;} echo filter();
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