PHP is_resource() Function
The PHP is_resource() function checks whether a variable is a resource or not. The function returns true if the variable is a resource, otherwise it returns false.
Note: This function is not a strict type-checking function. It will return false if value is a resource variable that has been closed.
Syntax
is_resource(variable)
Parameters
variable |
Required. Specify the variable being evaluated. |
Return Value
Returns true if variable is a resource, false otherwise.
Example:
The example below shows the usage of is_resource() function. Lets assume that we have a file called test.txt in the current working directory.
<?php $file = fopen("test.txt","r"); if (is_resource($file)) { echo "File is opened successfully."; } else { echo "Error in opening the file."; } ?>
The output of the above code will be:
File is opened successfully.
Example:
Consider one more example, where this function is used with a resource variable that has been closed.
<?php $file = fopen("test.txt","r"); fclose($file); if (is_resource($file)) { echo "File is opened successfully."; } else { echo "Error in opening the file."; } ?>
The output of the above code will be:
Error in opening the file.
❮ PHP Variable Handling Reference