检查数组键是否存在 [英] Checking if an array key exists

问题描述

我有一个由 json_decode()生成的多维数组. json是动态生成的，这意味着一些密钥将随机出现.

I have a multidimensional array produced by json_decode(). The json is dynamically generated, that means some keys will be present randomly.

我想避免未定义的索引:通知，所以我将对数组的调用封装在这样的函数中:

I would like to avoid Undefined index: notice, so i encapsulated the calls to the array in a function like this:

function exists($value) {
    if (isset($value)) {
        return $value;
    }
}

然后我调用数据:

$something = exists($json_array['foo']['bar']['baz']);

但是我仍然得到未定义索引:baz 的通知.有什么建议吗?

But i still get the Undefined index: baz notice. Any suggestions?

推荐答案

看来您是PHP的新手，所以我给出的答案比平时长一些.

It seems you are new to PHP, so I'll give a bit lengthier answer than normal.

$something = exists($json_array['foo']['bar']['baz']);

这等同于您写的内容:

$baz = $json_array['foo']['bar']['baz'];
$something = exists($baz);

您可能已经注意到，这意味着$json_array['foo']['bar']['baz']在传递给exists()之前先经过评估.这是未定义索引的来源.

As you may have noticed, this means that $json_array['foo']['bar']['baz'] is evaluated before it's passed to exists(). This is where the undefined index is coming from.

正确的习惯用法更像这样:

The correct idiom would be more like this:

$something = NULL;
if (isset($json_array['foo']['bar']['baz'])) {
    $something = $json_array['foo']['bar']['baz'];
}

以下内容也与上述几行相同:

The following is also identical to the above lines:

$something = isset($json_array['foo']['bar']['baz'])
    ? $json_array['foo']['bar']['baz']
    : NULL;

这篇关于检查数组键是否存在的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！