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($var_name)) { //do something }
Function Use function_exists for detection. Note that the function name to be detected also needs to use quotation marks, such as:
if (function_exists('fun_name')) { fun_name(); }
Without further ado, let’s look at an example.
<?php /* 判断常量是否存在*/ if (defined('MYCONSTANT')) { echo MYCONSTANT; } //判断变量是否存在 if (isset($myvar)) { echo "存在变量$myvar."; } //判断函数是否存在 if (function_exists('imap_open')) { 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('test_func')) { 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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