php empty,isset,is_null judgment comparison (differences and similarities)_PHP tutorial

1. Examples
A. A variable is not defined, how should we judge it?

Copy code The code is as follows:

#$test variable does not exist

$isset= isset($test)?"test is define!":"test is undefine!" ;
echo "isset:$issetrn";

$empty=!empty($test)?"test is define!":"test is undefine!";
echo "empty:$ emptyrn";

$is_null=is_null($test)?"test is define!":"test is undefine!";
echo "is_null:$is_nullrn";

The test result is:

The result is out: empty and isset will first check whether the variable exists, and then detect the variable value. And is_null just checks the variable value directly to see if it is null, so if the variable is not defined, an error will occur!

B. See what parameters each receives?

isset function parameters:

$test=100;
echo isset($test),isset(100),$isset($b=100);

Parse error: parse error, unexpected T_LNUMBER, expecting T_STRING or T_VARIABLE or '$' in PHPDocument3 on line < ;b>3

empty function parameters:

$test=100;

echo empty($test),empty(100),empty($b=100);

Parse error: parse error, unexpected T_LNUMBER, expecting T_STRING or T_VARIABLE or '$' in PHPDocument3 on line < ;b>3

is_null function parameters:

$test=100;

echo is_null($test),is_null(100),is_null($b=100);

Running results: No errors.

The comparison results are out: empty, isset input parameters must be a variable (php variables start with the $ character), and is_null input parameters only need to have a return value. (constants, variables, expressions, etc.). In the PHP manual, their analysis is: empty, isset is a language structure rather than a function, so it cannot be called by variable functions.

2. Summarize the differences between isset, empty, and is_null:
What I just introduced: checking variables and parameter types is the basis of the differences between these three functions, and also The most easily overlooked. I saw many articles comparing these three functions on the Internet. These are rarely covered. What I want to talk about next is the difference when both check existing variables.

Example:

Copy code The code is as follows:

$a=100;
$b="";
$c=null;
//isset check
echo "isset","$a=$a",isset($a) ?"define":"undefine","rn";
echo "isset","$b=$b",isset($b)?"define":"undefine","rn";
echo "isset","$c=$c",isset($c)?"define":"undefine","rn";
unset($b);
echo "isset","$ b",isset($b)?"define":"undefine","rn";
$b=0;
echo "rnrn";

//empty check
echo "empty","$a=$a",!empty($a)?"no empty":"empty","rn";
echo "empty","$b=$b",! empty($b)?"no empty":"empty","rn";
echo "empty","$c=$c",!empty($c)?"no empty":"empty" ,"rn";
unset($b);
echo "empty","$b",!empty($b)?"no empty":"empty","rn";
$b=0;
echo "rnrn";

//is_null check
echo "is_null","$a=$a",!is_null($a)?"no null" :"null","rn";
echo "is_null","$b=$b",!is_null($b)?"no null":"null","rn";
echo " is_null","$c=$c",!is_null($c)?"no null":"null","rn";
unset($b);
echo "is_null","$ b",is_null($b)?"no null":"null","rn";

Through the above simple test, we can generally know that when a variable exists: isset, empty, is_null detection, and get the value. There are more variables not exemplified above. In fact, the test found:

empty

If the variable is a non-empty or non-zero value, empty() returns FALSE. In other words, "", 0, "0", NULL, FALSE, array(), var $var, undefined; and objects without any properties will be considered empty, and TRUE will be returned if var is empty. .

isset

Returns TRUE if the variable exists (not NULL), otherwise returns FALSE (including undefined). The variable value is set to: null, and the return value is also false; after unsetting a variable, the variable is canceled. Note that isset handles NULL value variables specially.

is_null

Check whether the incoming value [value, variable, expression] is null. Only one variable is defined and its value is null. Returns TRUE. Others return FALSE [An error will occur after undefined variables are passed in! 】.


Question: How to determine if a variable is set and its value is NULL?

Through the above comparison, I guess everyone, like me, will have this question in their mind. To detect whether a variable is null, you can use: is_null, but if the variable is not defined, it will cause an error. Therefore, we thought that to detect whether a variable is defined, we can use: isset, but if a variable value is: null, it will return false. Haha, how to solve this problem? Waiting for everyone to share...

Check that the variable exists and the value is NULL.

Copy the code The code is as follows:

function checkNull($a)
{
if(array_key_exists($a,$GLOBALS))
{
global $$a;
if( is_null($$a))
return true;
}
return false;
}
$test=null;
var_dump(checkNull("test"));
var_dump(checkNull("test1"));

http://www.bkjia.com/PHPjc/322515.htmlwww.bkjia.comtruehttp: //www.bkjia.com/PHPjc/322515.htmlTechArticle1. Example A. A variable is not defined, how should we judge it? Copy the code as follows: ? php #$test variable does not exist $isset= isset($test)?"test is define!":"test i...