PHP How to determine constants, variables and functions exist in the instance code of the detailed

If you understand the above sentence, then the next is nonsense, the PHP manual is written in a very full. In a word, I solved all the problems in my title.

The code is as follows:

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

Variable detection is the use of isset, note that variables are not declared or declared when the assignment value is Null,isset return false, such as:

if (Isset ($var _name)) {    //do something}

function detection with function_exists, note that the function name to be detected also needs to use quotation marks, such as:

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

Let's just say we're going to see an example.

<?php/* Determines if the constant exists */if (defined (' myconstant ')) {echo myconstant;}//Determines if the variable exists if (Isset ($myvar)) {echo "exists variable $myvar." ; }//Determine if the function exists if (function_exists (' Imap_open ')) {echo "exists function Imag_openn";} else {echo "function Imag_open does not exist n";}?>

Function_exists determine if a function exists

<?phpif (function_exists (' Test_func ') {    echo "function Test_func exists";} else {    echo "function Test_func does not exist";}? >

Filter_has_var function

The Filter_has_var () function checks for the existence of a variable of the specified input type.
Returns true if successful, otherwise false is returned.

<?phpif (!filter_has_var (input_get, "name") {Echo ("INPUT type does not exist");} else {echo ("Input type exists");}? >

Output is. Input type exists