isset()

(PHP 4, PHP 5, PHP 7)

检测变量是否已设置并且非NULL

说明

检测变量是否设置，并且不是NULL。

如果已经使用unset()释放了一个变量之后，它将不再是isset()。若使用isset()测试一个被设置成NULL的变量，将返回FALSE。同时要注意的是 null 字符（"0"）并不等同于 PHP 的NULL常量。

如果一次传入多个参数，那么isset()只有在全部参数都以被设置时返回TRUE计算过程从左至右，中途遇到没有设置的变量时就会立即停止。

参数

要检查的变量。

其他变量。

返回值

如果$var存在并且值不是NULL则返回TRUE，否则返回FALSE。

更新日志

范例

Example #1isset()例子

这对于数组中的元素也同样有效：

在字符串位移中使用isset()

PHP 5.4 改变了传入字符串位移时isset()的行为。

以上例程在PHP 5.3中的输出：

bool(true)
bool(true)
bool(true)
bool(true)
bool(true)
bool(true)

以上例程在PHP 5.4中的输出：

bool(false)
bool(true)
bool(true)
bool(true)
bool(false)
bool(false)

注释

isset()只能用于变量，因为传递任何其它参数都将造成解析错误。若想检测常量是否已设置，可使用defined()函数。

Note:

如果使用isset()来检查对象无法访问的属性，如果__isset()方法已经定义则会调用这个重载方法。

参见

• empty() 检查一个变量是否为空
• __isset()
• unset() 释放给定的变量
• defined() 检查某个名称的常量是否存在
• the type comparison tables
• array_key_exists() 检查数组里是否有指定的键名或索引
• is_null() 检测变量是否为 NULL
• 错误控制@运算符。

I, too, was dismayed to find that isset($foo) returns false if ($foo == null). Here's an (awkward) way around it.
unset($foo);
if (compact('foo') != array()) {
 do_your_thing();
}
Of course, that is very non-intuitive, long, hard-to-understand, and kludgy. Better to design your code so you don't depend on the difference between an unset variable and a variable with the value null. But "better" only because PHP has made this weird development choice.
In my thinking this was a mistake in the development of PHP. The name ("isset") should describe the function and not have the desciption be "is set AND is not null". If it was done properly a programmer could very easily do (isset($var) || is_null($var)) if they wanted to check for this!
A variable set to null is a different state than a variable not set - there should be some easy way to differentiate. Just my (pointless) $0.02.

"empty() is the opposite of (boolean) var, except that no warning is generated when the variable is not set."
So essentially

is the same as

doesn't it? :)
!empty() mimics the chk() function posted before.

You can safely use isset to check properties and subproperties of objects directly. So instead of writing
  isset($abc) && isset($abc->def) && isset($abc->def->ghi)
or in a shorter form
  isset($abc, $abc->def, $abc->def->ghi)
you can just write
  isset ($abc->def->ghi)
without raising any errors, warnings or notices.
Examples 

How to test for a variable actually existing, including being set to null. This will prevent errors when passing to functions.

Note: you can't turn this into a function (e.g. is_defined($myvar)) because get_defined_vars() only gets the variables in the current scope and entering a function changes the scope.

in PHP5, if you have 

and

will always echo 'false'. because the isset() accepts VARIABLES as it parameters, but in this case, $foo->bar is NOT a VARIABLE. it is a VALUE returned from the __get() method of the class Foo. thus the isset($foo->bar) expreesion will always equal 'false'.

The new (as of PHP7) 'null coalesce operator' allows shorthand isset. You can use it like so:

Quoted from http://php.net/manual/en/migration70.new-features.php#migration70.new-features.null-coalesce-op

Careful with this function "ifsetfor" by soapergem, passing by reference means that if, like the example $_GET['id'], the argument is an array index, it will be created in the original array (with a null value), thus causing posible trouble with the following code. At least in PHP 5.
For example:

will print 
Array
(
)
Array
(
  [unsetindex] => 
)
Any foreach or similar will be different before and after the call.

Return Values :
Returns TRUE if var exists and has value other than NULL, FALSE otherwise.

Could any one explain me in clarity.

I tried the example posted previously by Slawek:
$foo = 'a little string';
echo isset($foo)?'yes ':'no ', isset($foo['aaaa'])?'yes ':'no ';
He got yes yes, but he didn't say what version of PHP he was using.
I tried this on PHP 5.0.5 and got: yes no
But on PHP 4.3.5 I got: yes yes
Apparently, PHP4 converts the the string 'aaaa' to zero and then returns the string character at that position within the string $foo, when $foo is not an array. That means you can't assume you are dealing with an array, even if you used an expression such as isset($foo['aaaa']['bbb']['cc']['d']), because it will return true also if any part is a string.
PHP5 does not do this. If $foo is a string, the index must actually be numeric (e.g. $foo[0]) for it to return the indexed character.

1) Note that isset($var) doesn't distinguish the two cases when $var is undefined, or is null. Evidence is in the following code.

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