检查值是否为多维数组[重复]

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这个问题在这里已有答案：

Checking if array is multidimensional or not? 21 answers
检查数组是否是多维的？ 21个答案

How come when I run this code, I get an output of I am a multidimensional array! (the first block). I thought it would go into the second block, but it doesn't. What am I missing here?

为什么我运行这段代码，我得到一个输出，我是一个多维数组！（第一块）。我以为它会进入第二块，但事实并非如此。我在这里想念的是什么？

$values = array('1','2');
if(isset($values[0][0])){
    echo "I am a multidimensional array!";
}else{
    echo "I am not a multidimensional array.";
}

3 个解决方案

#1

$values = array(1,array(1,2));


$multi = false;
if(is_array($values)){
    foreach($values as $k=>$v){
        if(is_array($v)){
            $multi = true;
            break;
        }
    }
}

echo $multi ? "multi" : "not multi";

#2

Try this:

尝试这个：

if(is_array($values[0]))

Edit: This will check the first element of the array only. You should loop through each element to check if its truly multidimensional.

编辑：这将仅检查数组的第一个元素。您应该遍历每个元素以检查它是否真正是多维的。

This code checks to see if the first element of the array is also an array. isset just checks whether or not a variable is NULL.

此代码检查数组的第一个元素是否也是一个数组。 isset只检查变量是否为NULL。

isset in your example is not working as expected. Perhaps there is a slight difference in functionality between PHP versions or setups. I didn't see anything in the manual but maybe you can:

您的示例中的isset未按预期工作。也许PHP版本或设置之间的功能略有不同。我手册中没有看到任何内容，但也许您可以：

http://php.net/manual/en/function.isset.php

Using is_array is more semantic, so in my opinion is a much better choice.

使用is_array更具语义性，所以在我看来是一个更好的选择。

#3

This code only goes into the if-branch for me, if the first value in the array is explicitly declared as a string,

如果数组中的第一个值显式声明为字符串，则此代码仅进入if分支，

$values = array('1',2);

– and with that the behavior is nothing but logical, because $values[0] is that text literal '1', and that has a first character that can be access using a zero based index.

- 并且这种行为只不过是合乎逻辑的，因为$ values [0]是文本文字'1'，并且具有可以使用基于零的索引访问的第一个字符。

So I guess either your real data is of a string type – or it maybe depends in the PHP version (I tested under 5.3.16).

所以我猜你的真实数据是字符串类型 - 或者它可能取决于PHP版本（我在5.3.16下测试）。

Anyway, using is_array as the other answers already suggested is the right way to go here.

无论如何，使用is_array已经建议的其他答案是正确的方法。

#1

$values = array(1,array(1,2));


$multi = false;
if(is_array($values)){
    foreach($values as $k=>$v){
        if(is_array($v)){
            $multi = true;
            break;
        }
    }
}

echo $multi ? "multi" : "not multi";

#2