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回复讨论(解决方案)

array小问题

Jun 20, 2016 pm 12:37 PM

<?php$roomlist=array('0'=>array('id' => '11'),'1'=>array('id' => '22'),);$rand_keys = array_rand($roomlist, 1);var_dump($rand_keys);echo gettype($rand_keys);echo ("\n");echo ("\n");$rand_keys = array_rand($roomlist, 2);var_dump($rand_keys);echo gettype($rand_keys);?>
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输出为:
int(0)
integer

array(2) {
[0]=>
int(0)
[1]=>
int(1)
}
array

一个小问题,对于第一个测试,array_rand($roomlist, 1); 怎么返回的是 integer,而不是array呢?


回复讨论(解决方案)

array_rand() 函数返回数组中的随机键名,或者如果您规定函数返回不只一个键名,则返回包含随机键名的数组。

貌似只是返回的key

array_rand() 函数返回数组中的随机键名,或者如果您规定函数返回不只一个键名,则返回包含随机键名的数组。



我疑惑的是,为什么array_rand 的第二个参数为 1时,返回的不是 array,而是 integer

如果我这样使用就会出bug
<?php$roomlist=array('0'=>array('id' => '11'),);$array_size = count($roomlist)$rand_keys = array_rand($roomlist, $array_size);   //size为1时,返回为integer,size>=2时,返回的是array,下面的操作无法统一var_dump($rand_keys);echo gettype($rand_keys);echo ("\n");echo ("\n"); $val = $roomlist[$rand_keys[0]];
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看到了不??  由于 $rand_keys为integer时存在bug

貌似只是返回的key



是只返回key,,在c++里,返回array就是array,返回integer时返回就是integer,,不会因为函数的参数改变而改变。。

不一致时需要特殊处理。。而且也不合理,一个函数两种返回类型。。。看我的测试

定义和用法
array_rand() 函数返回数组中的随机键名,或者如果您规定函数返回不只一个键名,则返回包含随机键名的数组。
说明
array_rand() 函数从数组中随机选出一个或多个元素,并返回。
第二个参数用来确定要选出几个元素。如果选出的元素不止一个,则返回包含随机键名的数组,否则返回该元素的键名。

如果看不懂手册,或者非要质疑手册上的描述。那么神仙也帮不到你

手册上已经写的很清楚了:
------
返回值 :
如果你只取出一个,array_rand()返回一个随机单元的键名,否则就返回一个包含随机键名的数组。这样你就可以随机从数组中取出键名和值。 

bug果然是bug。。

脚本语言好奇妙。。。。。啦啦啦啦啦啦啦啦啦

c++er不懂不懂啦啦啦啦啦

另外楼上的回答,难道没看出来我是新手,如果找到了正确/很好的解释我也不开贴了。。。。。。。。。。



自己找到了官网的说法

http://php.net/manual/zh/function.array-rand.php

以后遇到像我一样的新手们,这是php的官方手册

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