Detailed example code of how PHP determines whether constants, variables and functions exist

If you understand the above sentence, then the rest is nonsense. PHP Manual is still very comprehensive. One sentence solves all the problems in my title.

The code is as follows:

if (defined('CONST_NAME')) {
    //do something 
}

Variable is detected using isset. Note that the variable is not declared or is assigned a value of NULL when declared, and isset returns FALSE. , such as:

if (isset($var_name)) {
    //do something
}

Function Use function_exists for detection. Note that the function name to be detected also needs to use quotation marks, such as:

if (function_exists('fun_name')) {
 fun_name();
}

Without further ado, let’s look at an example.

<?php 
/* 判断常量是否存在*/ 
if (defined(&#39;MYCONSTANT&#39;)) { 
echo MYCONSTANT; 
} 
//判断变量是否存在 
if (isset($myvar)) { 
echo "存在变量$myvar."; 
} 
//判断函数是否存在 
if (function_exists(&#39;imap_open&#39;)) { 
echo "存在函数imag_openn"; 
} else { 
echo "函数imag_open不存在n"; 
} 
?>

function_exists determines whether the function exists

<?php
if (function_exists(&#39;test_func&#39;)) {
    echo "函数test_func存在";
} else {
    echo "函数test_func不存在";
}
?>

filter_has_var function

filter_has_var() function checks whether a variable of the specified input type exists.
If successful, return true, otherwise return false.

<?php
if(!filter_has_var(INPUT_GET, "name"))
 {
 echo("Input type does not exist");
 }
else
 {
 echo("Input type exists");
 }
?>

The output is. Input type exists

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