PHP if($_POST) vs if(isset($_POST))
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Php if($_POST) vs if(isset($_POST))
提问by Aycan Ya??t
I have a simple form as demonstrated below:
我有一个简单的表格，如下所示：
<form action="" method="post">
<input type="text" />
<input type="submit" value="SEND" />
</form>
When I try to receive data sent from this form via if($_POST), I fail, but when try with isset, I success.
当我尝试通过 接收从此表单发送的数据时if($_POST)，我失败了，但是当尝试使用 时isset，我成功了。
if($_POST){
echo 'a'; //Doesn't print anything.
}
if(isset($_POST)){
echo 'b'; //Prints 'b'
}
I guess the reason behind it is missing name attribute in my form input, but I can't understand why if($_POST)and isset($_POST)react different ways in this case.
我想背后的原因是缺少name属性在我的形式输入，但我不明白为什么if($_POST)和isset($_POST)在这种情况下的反应不同的方式。
回答by Quentin
issetdetermine if a variable is set and is not NULL. $_POSTwill always be set and will always be an array.
isset确定变量是否已设置并且不是 NULL。$_POST将始终被设置，并将始终是一个数组。
Without issetyou are just testing if the value is truthy. An empty array (which $_POSTwill be if you aren't posting any data) will not be truthy.
如果没有，isset您只是在测试该值是否为真。空数组（$_POST如果您不发布任何数据，则为空数组）不会是真实的。
回答by Daniele D
isset determines if a variable is set and not NULL, see the manual: http://php.net/manual/en/function.isset.php
isset 确定是否设置了变量而不是 NULL，请参阅手册：http: //php.net/manual/en/function.isset.php
while if($_POST)checks $_POSTfor being true.
whileif($_POST)检查是否$_POST为真。
in your case, $_POSTwill always be set. If doing that with other variables not related to a form, keep in mind that checking for if($var)without knowing if it is set or not, will throw a notice. Checking if(isset($var))will not throw a notice.
在您的情况下，$_POST将始终设置。如果使用与表单无关的其他变量执行此操作，请记住，在if($var)不知道是否已设置的情况下进行检查会抛出通知。检查if(isset($var))不会抛出通知。
Unrelated to your question: if you want to know if there is data inside your $_POSTarray you could try working with count($_POST), see:
http://php.net/manual/en/function.count.php
与您的问题无关：如果您想知道您的$_POST数组中是否有数据，您可以尝试使用count($_POST)，请参阅：http:
//php.net/manual/en/function.count.php
回答by n-dru
It is because the $_POSTis an array of inputs names/values pairs, and in your form no input has any name, therefore it is an empty array (evaluating to false). You can verify it by var_dump($_POST).
这是因为这$_POST是一个输入名称/值对的数组，并且在您的表单中没有输入具有任何名称，因此它是一个空数组（计算为 false）。您可以通过 验证它var_dump($_POST)。
Try to add a name to text input to access its value:
尝试向文本输入添加名称以访问其值：
<form action="" method="post">
<input type="text" name="somename" />
<input type="submit" value="SEND" />
</form>
回答by Akash Tyagi
The major difference is issetdetermine variable is set and is not null for $_POSTis not work here because you are not define here input name. The $_POSTconsider an array of inputs name/values pairs.
主要区别是isset确定变量已设置并且不为空，因为$_POST在这里不起作用，因为您没有在此处定义输入名称。该$ _ POST考虑输入名称/值对的数组。
