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 Manualis still very comprehensive. One sentence solves all the problems in my title.

The code is as follows:

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

Variableis 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 }

FunctionUse 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.

function_exists determines whether the function exists

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.

The output is. Input type exists

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