PHP in_array未找到一个值，该值是有 [英] PHP in_array isn't finding a value that is there

问题描述

我有一个名为$ friend_array的数组。当我的print_r（$ friend_array），它看起来是这样的：

I have an array called $friend_array. When I print_r($friend_array) it looks like this:

Array ( [0] => 3,2,5 ) 

我也有正在从网址拉出可变叫做$流体

I also have a variable called $uid that is being pulled from the url.

在我测试的页面，$ uid具有3的值，因此它是数组中

On the page I'm testing, $uid has a value of 3 so it is in the array.

但是，以下是说，它不存在：

However, the following is saying that it isn't there:

if(in_array($uid, $friend_array)){
  $is_friend = true;
}else{
  $is_friend = false;

这始终返回false。我呼应$ uid和值为3我打印阵列和3是存在的。

This always returns false. I echo the $uid and it is 3. I print the array and 3 is there.

我是什么做错了吗？任何帮助将大大AP preciated！

What am I doing wrong? Any help would be greatly appreciated!

推荐答案

阵列（[0] =＆GT; -3,2,5-）表示数组元素 0 是一个字符串 -3,2,5- ，这样，你做一个 is_array 检查 $ UID 所以你必须使用先打破字符串到一个数组，作为一个分离器，然后再检查 $ UID ：

Array ( [0] => 3,2,5 ) means that the array element 0 is a string 3,2,5, so, before you do an is_array check for the $uid so you have to first break that string into an array using , as a separator and then check for$uid:

// $friend_array contains as its first element a string that
// you want to make into the "real" friend array:
$friend_array = explode(',', $friend_array[0]);

if(in_array($uid, $friend_array)){
  $is_friend = true;
}else{
  $is_friend = false;
}

的工作的例子

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