PHP - 检查变量是否未定义

PHP - check if variable is undefined

提问by Timothy Coetzee

Consider this jquery statment

考虑这个 jquery 语句

isTouch = document.createTouch !== undefined

I would like to know if we have a similar statement in PHP, not being isset() but literally checking for an undefined value, something like:

我想知道我们在 PHP 中是否有类似的语句，不是 isset() 而是字面上检查未定义的值，例如：

$isTouch != ""

$isTouch != ""

Is there something similar as the above in PHP?

PHP中是否有与上述类似的东西？

回答by Sougata Bose

You can use -

您可以使用 -

$isTouch = isset($variable);

It will store trueif $variableis defined else false. If need the invers the simply remove the !.

它将存储trueif$variable定义为 else false。如果需要反转，只需删除!.

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

注意：如果 var 存在且具有非 NULL 值，则返回 TRUE，否则返回 FALSE。

Or if you want to check for false, 0etc also then use empty()-

或者，如果你想检查false，0等等也然后使用empty()-

$isTouch = empty($variable);

empty()works for -

empty()效劳于 -

• ""(an empty string)
• 0(0 as an integer)
• 0.0(0 as a float)
• "0"(0 as a string)
• NULL
• FALSE
• array()(an empty array)
• $var;(a variable declared, but without a value)

• "" （空字符串）
• 0 （0 作为整数）
• 0.0 （0 作为浮点数）
• “0” （0 作为字符串）
• 空值
• 错误的
• array() （空数组）
• $var; （声明的变量，但没有值）

回答by TiDJ

An another way is simply :

另一种方法很简单：

if($test){
    echo "Yes 1";
}
if(!is_null($test)){
    echo "Yes 2";
}

$test = "hello";

if($test){
    echo "Yes 3";
}

Will return :

将返回 ：

"Yes 3"

The best way is to use isset(), otherwise you can have an error like "undefined $test".

最好的方法是使用isset()，否则会出现“undefined $test”之类的错误。

You can do it like this :

你可以这样做：

if( isset($test) && ($test!==null) )

You'll not have any error because the first condition isn't accepted.

你不会有任何错误，因为第一个条件不被接受。

回答by ErasmoOliveira

To check is variable is set you need to use isset function.

要检查是否设置了变量，您需要使用 isset 函数。

$lorem = 'potato';

if(isset($lorem)){
    echo 'isset true' . '<br />';
}else{
    echo 'isset false' . '<br />';
}

if(isset($ipsum)){
    echo 'isset true' . '<br />';
}else{
    echo 'isset false' . '<br />';
}

this code will print:

此代码将打印：

isset true
isset false

read more in https://php.net/manual/en/function.isset.php

在https://php.net/manual/en/function.isset.php 中阅读更多内容

回答by Sumit Thakur

You can use -

您可以使用 -

Ternary oprator to check wheather value set by POST/GET or not somthing like this

三元运算符检查 POST/GET 设置的值是否是这样的

$value1 = $_POST['value1'] = isset($_POST['value1']) ? $_POST['value1'] : '';
$value2 = $_POST['value2'] = isset($_POST['value2']) ? $_POST['value2'] : '';
$value3 = $_POST['value3'] = isset($_POST['value3']) ? $_POST['value3'] : '';
$value4 = $_POST['value4'] = isset($_POST['value4']) ? $_POST['value4'] : '';

回答by Quasimodo's clone

JavaScript's 'strict not equal' operator (!==) on comparison with undefineddoes notresult in falseon nullvalues.

JavaScript的“严格不等于”运算符（!==）与比较undefined并不会导致false对null值。

var createTouch = null;
isTouch = createTouch !== undefined  // true

To achieve an equivalent behaviour in PHP, you can check whether the variable name exists in the keys of the result of get_defined_vars().

为了在 PHP 中实现等价的行为，您可以检查变量名是否存在于结果的键中get_defined_vars()。

// just to simplify output format
const BR = '<br>' . PHP_EOL;

// set a global variable to test independence in local scope
$test = 1;

// test in local scope (what is working in global scope as well)
function test()
{
  // is global variable found?
  echo '$test ' . ( array_key_exists('test', get_defined_vars())
                    ? 'exists.' : 'does not exist.' ) . BR;
  // $test does not exist.

  // is local variable found?
  $test = null;
  echo '$test ' . ( array_key_exists('test', get_defined_vars())
                    ? 'exists.' : 'does not exist.' ) . BR;
  // $test exists.

  // try same non-null variable value as globally defined as well
  $test = 1;
  echo '$test ' . ( array_key_exists('test', get_defined_vars())
                    ? 'exists.' : 'does not exist.' ) . BR;
  // $test exists.

  // repeat test after variable is unset
  unset($test);
  echo '$test ' . ( array_key_exists('test', get_defined_vars())
                    ? 'exists.' : 'does not exist.') . BR;
  // $test does not exist.
}

test();

In most cases, isset($variable)is appropriate. That is aquivalent to array_key_exists('variable', get_defined_vars()) && null !== $variable. If you just use null !== $variablewithout prechecking for existence, you will mess up your logs with warnings because that is an attempt to read the valueof an undefined variable.

大多数情况下，isset($variable)是合适的。这相当于array_key_exists('variable', get_defined_vars()) && null !== $variable. 如果你只是使用null !== $variable而不预先检查是否存在，你会用警告弄乱你的日志，因为这是尝试读取未定义变量的值。

However, you can apply an undefined variable to a reference without any warning:

但是，您可以在没有任何警告的情况下将未定义的变量应用于引用：

// write our own isset() function
function my_isset(&$var)
{
  // here $var is defined
  // and initialized to null if the given argument was not defined
  return null === $var;
}

// passing an undefined variable by reference does not log any warning
$is_set = my_isset($undefined_variable);   // $is_set is false

回答by Waruna Manjula

if(isset($variable)){
    $isTouch = $variable;
}

OR

或者

if(!isset($variable)){
    $isTouch = "";// 
}