php 检查变量是否为空
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check if variable empty
提问by ZeroSuf3r
if ($user_id == NULL || $user_name == NULL || $user_logged == NULL) {
$user_id = '-1';
$user_name = NULL;
$user_logged = NULL;
}
if ($user_admin == NULL) {
$user_admin = NULL;
}
- Is there any shortest way to do it ?
- And if i right, it should be tested with
is_null? - It's possible
$user_id,$user_nameand$user_loggedwrite in one line (maybe array?) without repeatingNULL?
- 有没有最短的方法来做到这一点？
- 如果我是对的，应该用
is_null? - 这是可能的
$user_id，$user_name并且$user_logged写在一行（也许是数组？）而不重复NULL？
回答by Rob W
If you want to test whether a variable is reallyNULL, use the identity operator:
如果要测试变量是否为真的NULL，请使用恒等运算符：
$user_id === NULL // FALSE == NULL is true, FALSE === NULL is false
is_null($user_id)
If you want to check whether a variable is not set:
如果要检查变量是否未设置：
!isset($user_id)
Or if the variable is not empty, an empty string, zero, ..:
或者，如果变量不为空，则为空字符串、零、..：
empty($user_id)
If you want to test whether a variable is not an empty string, !will also be sufficient:
如果要测试变量是否不是空字符串，!也足够了：
!$user_id
回答by Mark
You can check if it's not set (or empty) in a number of ways.
您可以通过多种方式检查它是否未设置（或为空）。
if (!$var){ }
Or:
或者：
if ($var === null){ } // This checks if the variable, by type, IS null.
Or:
或者：
if (empty($var)){ }
You can check if it's declared with:
您可以检查它是否声明为：
if (!isset($var)){ }
Take note that PHP interprets 0 (integer) and "" (empty string) and false as "empty" - and dispite being different types, these specific values are by PHP considered the same. It doesn't matter if $var is never set/declared or if it's declared as $var = 0 or $var = "". So often you compare by using the === operator which compares with respect to data type. If $var is 0 (integer), $var == "" or $var == false will validate, but $var === "" or $var === false will not.
请注意，PHP 将 0（整数）和“”（空字符串）和 false 解释为“空” - 尽管类型不同，但 PHP 认为这些特定值是相同的。$var 是否从未设置/声明或声明为 $var = 0 或 $var = "" 都没有关系。您经常使用 === 运算符进行比较，该运算符根据数据类型进行比较。如果 $var 为 0（整数），则 $var == "" 或 $var == false 将验证，但 $var === "" 或 $var === false 不会。
回答by user3470929
here i have explained how the empty function and isset works please use the one that is appropriate also you can use is_null function also
在这里我已经解释了空函数和 isset 是如何工作的，请使用合适的，你也可以使用 is_null 函数
<?php
$val = 0;
//evaluates to true because $var is empty
if (empty($val)) {
echo '$val is either 0, empty, or not set at all';
}
//evaluates to true because $VAR IS SET
if (isset($val)) {
echo '$val is set even though it is empty';
}
?>
回答by Antony
Its worth noting - and I only found this out after nearly 9 years of PHP coding that the best way of checking any variable exists is using the empty() function. This is because it doesn't generate errors and even though you turn them off - PHP still generates them! empty() however won't return errors if the variable doesn't exist. So I believe the correct answer is not to check if its null but to do the following
值得注意的是 - 我在将近 9 年的 PHP 编码之后才发现这一点，检查任何变量是否存在的最佳方法是使用 empty() 函数。这是因为它不会生成错误，即使您关闭它们 - PHP 仍然会生成它们！但是如果变量不存在，empty() 不会返回错误。所以我相信正确的答案不是检查它是否为空，而是执行以下操作
if (!empty($var) && is_null($var))
Note the PHP manual
注意 PHP 手册
variable is considered empty if it does not exist or if its value equals FALSE
如果变量不存在或其值等于 FALSE，则变量被认为是空的
As opposed to being null which is handy here!
与 null 不同，这在这里很方便！
回答by Michael Berkowski
empty()is a little shorter, as an alternative to checking !$user_idas suggested elsewhere:
empty()有点短，作为检查!$user_id其他地方建议的替代方法：
if (empty($user_id) || empty($user_name) || empty($user_logged)) {
}
回答by Mike Q
Felt compelled to answer this because of the other responses. Use empty() if you can because it covers more bases. Future you will thank me.
由于其他回答，感到不得不回答这个问题。如果可以，请使用 empty() ，因为它涵盖了更多的碱基。未来你会感谢我的。
For example you will have to do things like isset() && strlen() where instead you could use empty(). Think of it like this empty is like !isset($var) || $var==false
例如，您将不得不执行isset() && strlen() 之类的操作，而您可以使用empty()。把它想象成这个空就像 !isset($var) || $var==false
回答by Mike Q
回答by Dmitry Zaytsev
1.
1.
if(!($user_id || $user_name || $user_logged)){
//do your stuff
}
2 . No. I actually did not understand why you write such a construct.
2 . 不，我实际上不明白你为什么要写这样一个结构。
3 . Put all values into array, for example $ar["user_id"], etc.
3 . 将所有值放入数组，例如 $ar["user_id"] 等。
回答by Alnitak
回答by Vijeenrosh P.W
you can use isset() routine .
您可以使用 isset() 例程。
also additionaly you can refer an range of is_type () functions like
此外，您还可以参考一系列 is_type () 函数，例如
is_string(), is_float(),is_int() etc to further specificaly test
is_string(), is_float(),is_int() 等进一步具体测试
