Get the name of running script from a PHP extension
-
26-06-2021 - |
Question
I'm writing a small extenstion for PHP. Is there way to know at runtime, the name of the script file (e.g.: test.php
) that is running? Maybe some global or environment variables?
La solution
You can fetch $_SERVER['PHP_SELF']
(or any other $_SERVER variable if you need to), like this:
// This code makes sure $_SERVER has been initialized
if (!zend_hash_exists(&EG(symbol_table), "_SERVER", 8)) {
zend_auto_global* auto_global;
if (zend_hash_find(CG(auto_globals), "_SERVER", 8, (void **)&auto_global) != FAILURE) {
auto_global->armed = auto_global->auto_global_callback(auto_global->name, auto_global->name_len TSRMLS_CC);
}
}
// This fetches $_SERVER['PHP_SELF']
zval** arr;
char* script_name;
if (zend_hash_find(&EG(symbol_table), "_SERVER", 8, (void**)&arr) != FAILURE) {
HashTable* ht = Z_ARRVAL_P(*arr);
zval** val;
if (zend_hash_find(ht, "PHP_SELF", 9, (void**)&val) != FAILURE) {
script_name = Z_STRVAL_PP(val);
}
}
The script_name
variable will contain the name of the script.
In case you're wondering, the first block, that initializes $_SERVER
, is necessary because some SAPIs (e.g.: the Apache handler) will initialize $_SERVER
only when the user script accesses it (just-in-time). Without that block of code, if you try to read $_SERVER['PHP_SELF']
before the script tried accessing $_SERVER
, you'd end up with an empty value.
Obviously, you should add error handling in the above code in case anything fails, so that you don't invoke undefined behavior when trying to access script_name
.
Autres conseils
I tried this script, but it didn't work for me. The first statement:
if (!zend_hash_exists(&EG(symbol_table), "_SERVER", 8)
fails. I'm running PHP from the CLI. However, I did set variables through my PHP script and when I use print_r($_SERVER) through the same script I get a full array of values.
I think the negation in before the zend_hash_exists()
is not necessary in this context.