Copy code The code is as follows:
if (defined('CONST_NAME')) {
//do something
}
Variable detection uses isset. Note that if the variable is not declared or is assigned a value of NULL when declared, isset returns FALSE, such as:
Copy code The code is as follows:
if (isset($var_name)) {
//do something
}
Use function_exists for function detection. Note that the function name to be detected also needs to use quotation marks, such as:
Copy code The code is as follows:
if (function_exists('fun_name')) {
fun_name();
}
Without further ado, let’s look at an example
Copy code The code is as follows:
/* Determine whether the constant exists*/
if (defined('MYCONSTANT')) {
echo MYCONSTANT;
}
//Determine whether the variable exists
if (isset($myvar)) {
echo "The variable $myvar exists .";
}
//Determine whether the function exists
if (function_exists('imap_open')) {
echo "Function imag_openn exists";
} else {
echo " Function imag_open does not existn";
}
?>
function_exists determines whether the function exists
Copy code The code is as follows:
if (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 whether a variable of the specified input type exists.
If successful, return true, otherwise return false.
Copy code The code is as follows:
if(!filter_has_var(INPUT_GET, " name"))
{
echo("Input type does not exist");
}
else
{
echo("Input type exists");
}
?>
The output is. Input type exists
http://www.bkjia.com/PHPjc/319967.htmlwww.bkjia.comtruehttp: //www.bkjia.com/PHPjc/319967.htmlTechArticleCopy the code as follows: if (defined('CONST_NAME')) { //do something } Variable detection is When using isset, note that if the variable is not declared or is assigned a value of NULL when declared, isset returns FALSE, such as:...