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:" . $var ."br /" ; 10 ? js 1 function fn
php
<span> 1</span> <span>php
</span><span> 2</span> <span>function</span><span> fn(){
</span><span> 3</span> <span>echo</span> "inside the function:" .<span>$var</span> ."<br>"<span>;
</span><strong><span> 4 $var = "con 2";
5 echo "inside the function:" .$var ."<br>";
</span></strong><span> 6</span> <span> }
</span><span> 7</span> <span>$var</span> = "con 1"<span>;
</span><span> 8</span> <span> fn();
</span><span> 9</span> <span>echo</span> "outside the function:" .<span>$var</span> ."<br>"<span>;
</span><span>10</span> ?>js
<span>1</span> <span>function</span><span> fn2(){
</span><span>2</span> <span>//</span><span>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？</span>
<span>3</span> <span>alert</span>("inside the function:" + dada +"<br>"<span>);
</span><span><strong>4 var dada = "con 2";
5 <span>alert</span>("inside the function:" + dada +"<br>");
</strong></span><span>6</span> <span> }
</span><span>7</span> <span>var</span> dada = "con 1"<span>;
</span><span>8</span> <span> fn2();
</span><span>9</span> <span>alert</span>("outside the function:" + dada +"<br>");
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？
