PHP：当元素可能不存在时测试数组元素的值
假设我有一个数组 $arr。它可能有一个元素的键名为“music”($arr['music'])，我想测试该值是否等于“classical”：
if($arr['music'] === 'classical'){
//do something cool
}
但是，$ arr 没有包含“music”键的值。为了避免 PHP 错误，我因此执行以下操作：
if($arr['music']){
if($arr['music'] === 'classical'){
//do something cool
}
}
这看起来完全荒谬。在我看来，如果 $arr['music'] 不存在，那么它绝对不等于“古典”。有没有办法避免在测试键的值之前首先测试键是否存在？
Assume I have an array $arr. It's possible that it has an element with a key named 'music' ($arr['music']), and I want to test whether that value equals "classical":
if($arr['music'] === 'classical'){
//do something cool
}
However, it's possible that $arr does not have a value with the key 'music'. In order to avoid a PHP error, I therefore do the following:
if($arr['music']){
if($arr['music'] === 'classical'){
//do something cool
}
}
This seems totally ridiculous. In MY opinion, if $arr['music'] doesn't exist, then it DEFINITELY doesn't equal 'classical'. Is there a way to avoid first testing whether a key exists before testing it's value?
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评论（3）
如果密钥不存在，它将抛出错误（或警告）。
因此，为了实现这一点，您必须检查它是否确实存在。
有两种方法可以做到这一点：
您可以使用
isset()（如果$arr['music'] == null则为 false）：或者使用
array_key_exists()：If the key does not exist it will throw an error (or a warning).
So in order for that to happen, you have to check if it does exist.
Here are two ways to do that:
You can check it using
isset()(which will be false if$arr['music'] == null):Or use
array_key_exists():我认为 array_key_exists() 满足您的需求。
I think array_key_exists() meets your needs.
还有另一种解决方案。有些人不喜欢它，因为它使用“@”符号，但我发现它非常可读，所以我使用它。
就这样。对我来说，这是最好的可读解决方案。
There is another solution. Some people doesn't like it because it uses the '@' sign, but I find it very readable so I use it.
That's all. For me, the best readable solution.