php和js的问题?
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2022-04-27 15:41:30
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php 1 ? php 2 function fn(){ 3 echo "inside the function:" . $var ."br /" ; 4 $var = "con 2"; 5 echo "inside the function:" .$var ."br /"; 6 } 7 $var = "con 1" ; 8 fn(); 9 echo "outside the function:" . $var ."br /" ; 10 ? js 1 function fn
php
1 php 2 function fn(){ 3 echo "inside the function:" .$var ."
"; 4 $var = "con 2"; 5 echo "inside the function:" .$var ."
"; 6 } 7 $var = "con 1"; 8 fn(); 9 echo "outside the function:" .$var ."
"; 10 ?>
js
1 function fn2(){ 2 //因为dada已经存在fn2函数内部了。当它在函数内部找到这个变量之后就不往外找? 3 alert("inside the function:" + dada +"
"); 4 var dada = "con 2"; 5 alert("inside the function:" + dada +"
"); 6 } 7 var dada = "con 1"; 8 fn2(); 9 alert("outside the function:" + dada +"
");
如上两个代码片段,
php输出:
inside the function:
inside the function:con 2
outside the function:con 1
js输出:
inside the function:undefined
inside the function:con 2
outside the function:con 1
但是如果把红色的代码删除了,
php输出:
inside the function:
outside the function:con 1
js输出:
inside the function:con 1
outside the function:con 1
分析:
js代码中看到dada 后就在fn中找有没有定义,看到有定义了,所以就不往上面找?
但是因为定义在使用下面,所以值还是空。
删除了fn内部定义的变量后,因为在fn中没有找到dada的定义所以往外找,找到了所以为1?
php是怎么回事?