如何检查 JSON 对象在 PHP 中是否为空？
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How to check if JSON object is empty in PHP?
提问by vava
I'm reading JSON data with PHP and that data contains empty objects (like {}). So the problem is, I have to handle the case when object is empty in different manner but I can't find good enough way to do the check. empty(get_object_vars(object))looks too scary and very inefficient. Is there good way to do the check?
我正在使用 PHP 读取 JSON 数据，该数据包含空对象（如{}）。所以问题是，我必须以不同的方式处理对象为空的情况，但我找不到足够好的方法来进行检查。empty(get_object_vars(object))看起来太可怕了，而且效率很低。有什么好办法进行检查吗？
回答by ?ystein Riiser Gundersen
How many objects are you unserializing? Unless empty(get_object_vars($object))or casting to array proves to be a major slowdown/bottleneck, I wouldn't worry about it – Greg's solution is just fine.
你反序列化了多少个对象？除非empty(get_object_vars($object))或强制转换到 array 被证明是一个主要的减速/瓶颈，否则我不会担心 - Greg 的解决方案很好。
I'd suggest using the the $associativeflag when decoding the JSON data, though:
不过，我建议$associative在解码 JSON 数据时使用该标志：
json_decode($data, true)
This decodes JSON objects as plain old PHP arrays instead of as stdClassobjects. Then you can check for empty objects using empty()and create objects of a user-defined class instead of using stdClass, which is probably a good idea in the long run.
这将 JSON 对象解码为普通的旧 PHP 数组而不是stdClass对象。然后您可以使用empty()和创建用户定义类的对象来检查空对象，而不是使用stdClass，从长远来看，这可能是一个好主意。
回答by Greg
You could cast it to an array (unfortunately you can't do this within a call to empty():
您可以将其转换为数组（不幸的是，您不能在调用中执行此操作empty()：
$x = (array)$obj;
if (empty($x))
...
Or cast to an array and count():
或转换为数组和count()：
if (count((array)$obj))
...
回答by kenorb
Try without using empty()which is:
尝试不使用empty()which 是：
get_object_vars($obj) ? TRUE : FALSE;
On PHP docswe can read the note:
在PHP 文档上，我们可以阅读注释：
When using
empty()on inaccessible object properties, the__isset()overloading method will be called, if declared.
当
empty()在不可访问的对象属性上使用时，__isset()重载方法将被调用（如果已声明）。
Which means when using empty()on an object which is having __get()method, it will always return True.
这意味着在empty()具有__get()方法的对象上使用时，它将始终返回 True。
回答by Parziphal
I had to tell if an object was empty or not, but I also had to ignore private and protected properties, so I made this little function with which you can do this.
我必须判断一个对象是否为空，但我也不得不忽略私有和受保护的属性，所以我制作了这个小函数，你可以用它来做到这一点。
function empty_obj(&$object, $ignore_private = true, $ignore_protected = true) {
$obj_name = get_class($object);
$obj = (array)$object;
foreach(array_keys($obj) as $prop) {
$is_private = $is_protected = false;
$prop = preg_replace("/[^\w*]/", '', $prop);
$prop_name = str_replace(array($obj_name, '*'), '', $prop);
if(preg_match("~^$obj_name$prop_name$~", $prop))
$is_private = true;
if(preg_match("~^\*$prop_name$~", $prop))
$is_protected = true;
if(!$is_private || !$is_protected || ($is_private && !$ignore_private) || ($is_protected && !$ignore_protected))
return;
}
return true;
}
回答by Adam
I am not sure if this is more or less effective than casting to an array but I would guess more. You could just start to loop the object and as soon as you find something you have an answer and stop looping.
我不确定这是否比转换到数组更有效，但我猜更多。您可以开始循环对象，一旦找到某些内容，您就会有答案并停止循环。
function is_obj_empty($obj){
if( is_null($obj) ){
return true;
}
foreach( $obj as $key => $val ){
return false;
}
return true;
}
