I'm trying to read the content of config.json file which is available in the same folder of the file that read it, this file is called settings.php, I actually have this line of code in settings.php:
$jsonData = file('config.json');
the problem's that I get:
Warning: file(config.json): failed to open stream: No such file or directory
this is weird because as you can see from the image:
I'm trying this app on localhost with mamp
This is most likely because your settings.php is included by a index.php that's a directory higher.
Php file opening will always assume from the "parent" php file where execution started as base directory and not of the included files.
Try using __DIR__
$jsonData = file(__DIR__.'/config.json');
__DIR__ The directory of the file. If used inside an include, the directory of the included file is returned. This is equivalent to dirname(FILE). This directory name does not have a trailing slash unless it is the root directory.
Related
I was writing an web app in PHP, when I encountered a strange situation. To illustrate my problem, consider a web app of this structure:
/
index.php
f1/
f1.php
f2/
f2.php
Contents of these files:
index.php:
<?php require_once("f1/f1.php"); ?>
f1.php:
<?php require_once("../f2/f2.php"); ?>
f2.php: blank
now when I try to open index.php in my browser I get this error:
Warning: require_once(../f2/f2.php) [function.require-once]:
failed to open stream: No such file or directory in /var/www/reqtest/f1/f1.php on line 2
Fatal error: require_once() [function.require]:
Failed opening required '../f2/f2.php' (include_path='.:/usr/share/php:/usr/share/pear') in /var/www/reqtest/f1/f1.php on line 2
Is there something obvious I'm missing? how do include paths work in PHP?
Before I asked this question, I attempted to experiment and find out. I set up another test, like so:
/
index.php
f1/
f1.php
f2.php
index.php:
<?php require_once("f1/f1.php"); ?>
f1.php:
<?php require_once("f2.php"); ?>
f2.php: blank
To my surprise (and utter confusion), this worked out fine!
So, what is the secret behind the path resolution?
PS I saw this question, but it still does not answer the second case that I've stated here.
If you include another file, the working directory remains where the including file is.
Your examples are working as intended.
Edit: The second example works because . (actual directory) is in your include path (see your error message).
Edit2:
In your second example, the key point of your interest is this line:
<?php require_once("f2.php"); ?>
At first it will look in the current working dir (/var/www/req_path_test), but does not find f2.php.
As fallback, it will try to find f2.php in your include_path ('.:/usr/share/php:/usr/share/pear'), starting with '.' (which is relative to the actual file, not the including one).
So './f2.php' works and the require does not fail.
When you open index.php, working dir is set to the folder this file resides in. And inside insluded f1.php this working dir does not change.
You can include files by using their absolute paths, relative to the current included file like this:
require_once(dirname(__FILE__).'/../../test/file.php')
But better consider using an autoloader if these files contain classes.
Normaly in you old structure
<?php require_once("f2/f2.php"); ?>
instead of
<?php require_once("../f2/f2.php"); ?>
should work. As far as i know php takes the paths from the initial script
It sounds like your server has the open_basedir setting enabled in the PHP configuration. This makes it impossible to include (and open) files in folders above your in the directory structur (i.e., you can't use ../ to go up in the folder structure).
From the PHP Docs PHP include
Files are included based on the file path given or, if none is given, the include_path specified. If the file isn't found in the include_path, include will finally check in the calling script's own directory and the current working directory before failing.
If the file path is not given then i.e require_once("f2.php");
1st. The include_path is checked
2nd. The calling scripts own directory is checked
3rd. Finally the current working directory is checked
If file not found then PHP throws warning on file include & fatal error on require
If a path is defined — whether absolute (starting with a drive letter or \ on Windows, or / on Unix/Linux systems) or relative to the current directory (starting with . or ..) — the include_path will be ignored altogether. For example, if a filename begins with ../, the parser will look in the parent directory to find the requested file.
If you include/require your file beginning with . or .. or ./ then PHP's parser will look in the parent directory which is the current working directory i.e require_once("../f2/f2.php"), php will check at the root directory as the calling script index.php is in that directory.
Now You have not defined any include path in your PHP script thus it always falls back to the calling script and then into the current working directory.
// Check your default include path, most likely to be C:\xampp\php\PEAR
echo get_include_path();
// To set include path
set_include_path ( string $new_include_path ) : string
The Current Working Directory is derived from your main calling script index.php.
// The Current Working Directory can be checked
echo getcwd();
In the first Example where the required file "../f2/f2.php" is from f1.php
You code does not work because -
The specified path is ignored by PHP as your filename begins with ../
f1/ the calling script's own directory is ignored as well.
The parser directory looks into the parent directory to find the requested file. The current working directory is root directory, this is from where you have initiated the working script index.php. The file is not located at this directory, wrong path given.
Thus you get the Fatal Error
In the Second example you have changed the directory & from f1.php you require_once("f2.php").
Your code works because -
This time you require("f2.php") no leading ../ or ./ This time PHP checks the include_path but does find it there, as you haven't defined it and the file does not reside in the default preset include_path.
This time the calling script f1.php's directory is f1/. and you require file ("f2.php") is located at this directory. PHP This time checks the file in this directory and finds it.
PHP does not have to check the working directory as the file was found.
Thus Your Code Works Fine!
I have the following directory structure.
/var/www/base/controller/detail.php
/var/www/base/validate/edit.json
/var/www/html
Within /var/www/base/controller/detail.php, how do I use file_get_contents() with a relative path to read /var/www/base/validate/edit.json? I've tried the following:
//failed to open stream: No such file or directory (error no: 2)
$json=file_get_contents('detail.php');
//No error, but I don't want this file and was just testing
$json=file_get_contents('detail.php', FILE_USE_INCLUDE_PATH);
//failed to open stream: No such file or directory (error no: 2)
$json=file_get_contents('./validate/edit.json', FILE_USE_INCLUDE_PATH);
//failed to open stream: No such file or directory (error no: 2)
$json=file_get_contents('../validate/edit.json', FILE_USE_INCLUDE_PATH);
//failed to open stream: No such file or directory (error no: 2)
$json=file_get_contents('././validate/edit.json', FILE_USE_INCLUDE_PATH);
//failed to open stream: No such file or directory (error no: 2)
$json=file_get_contents('../../validate/edit.json', FILE_USE_INCLUDE_PATH);
//This works, but I want to use a relative path
$json=file_get_contents(dirname(dirname(__FILE__)).'/validate/edit.json');
Have you tried:
$json = file_get_contents(__DIR__ . '/../validate/edit.json');
__DIR__ is a useful magic constant.
For reasons why, see http://yagudaev.com/posts/resolving-php-relative-path-problem/.
When a PHP file includes another PHP file which itself includes yet another file — all being in separate directories — using relative paths to include them may raise a problem.
PHP will often report that it is unable to find the third file, but why?
Well the answer lies in the fact that when including files in PHP the interpreter tries to find the file in the current working directory.
In other words, if you run the script in a directory called A and you include a script that is found in directory B, then the relative path will be resolved relative to A when executing a script found in directory B.
So, if the script inside directory B includes another file that is in a different directory, the path will still be calculated relative to A not relative to B as you might expect.
try using this
$json = file_get_contents("/path/to/your/file/edit.json", true);
As of PHP 5 the FILE_USE_INCLUDE_PATH constant can be used to trigger include path search. This is not possible if strict typing is enabled since FILE_USE_INCLUDE_PATH is an int. Use TRUE instead.
Of cause there is the possibility to use the include_path parameter. Set this parameter to '1' then a search for the file in the include_path (in php.ini.) will be done. You have to do editing in the php.ini!
When using $json=file_get_contents('detail.php'); a call is done from a php file.
Use file_get_contents('detail.php'); to have the detail.php get executed. the file have to be in the root directory (same as the calling php file in which the file_get_contents() is situated). If detail.php is in a subdirectory i can not see any workaround than using the include_path parameter
You can always try to do the opposite first to find out:
file_put_contents('FINDME', 'smth');
exit();
$json=file_get_contents('detail.php');
...
Then run something like this from your terminal on *nix system (or search for a file named FINDME using GUI on Windows)
find <root-dir-of-the-project> -name 'FINDME'
The command will output something like this:
<root-dir-of-the-project>/<directories>/FINDME
Now you know the root dir (from which the relative path is being taken) for the file where you are attempting to read the other files and you can construct the relative path easily
Warning: include_once(../settings/settings.php): failed to open stream: No such file or directory in /Applications/MAMP/htdocs/Project/classes/user.class.php on line 2
Warning: include_once(): Failed opening '../settings/settings.php' for inclusion (include_path='.:/Applications/MAMP/bin/php/php7.2.1/lib/php') in /Applications/MAMP/htdocs/Snapshot/classes/user.class.php on line 2
But the file location is actually right? It even gives me auto path fill suggestions (visual code plugin) to that path!
Why does this happen is this something to do with the / or the second warning line?
So I'm trying to include the settings.php file from inside the User.class.php file with the following code.
include_once("../settings/settings.php");
../ -> goes back into the root folder
settings/ -> enters settings folder
settings.php -> should locate the needed .php file
Replace
include_once("../settings/settings.php");
with
include_once(__DIR__ . "/../settings/settings.php");
__DIR__ will resolve in the absolute path of the file where this constant is used. You can navigate your folders from that.
Use an absolute path to the file. You can use $_SERVER['DOCUMENT_ROOT'] or create a const variable BASEPATH that you can easily set root directory value to.
include_once($_SERVER['DOCUMENT_ROOT']."/settings/settings.php");
OR
define('BASEPATH', "/Applications/MAMP/htdocs");
include_once(BASEPATH."/settings/settings.php");
Depending on where you call your class. Better yet is to use configuration file to fulfil the require/include path like every PHP framework does.
I am trying to move my jpg image file from one directory to another using the rename() function. However it keeps on giving the error saying No such file or directory. I have changed it to the copy() function with the following error failed to open stream: No such file or directory.
I am trying this as following:
rename('/_upload/1.image.jpg', '/_accepted/test/1.image.jpg');
The file is orginally already in my htdocs/_upload folder. This PHP file is already in my htdocs folder. All permissions are set to 777 but giving the same error.
Instead of starting / in your path, use ./:
rename('./_upload/1.image.jpg', './_accepted/test/1.image.jpg');
/ means that you are giving path starting the server's root directory.
./ starts with your current directory.
In your example, the arguments to rename are file names. /_upload/1.image.jpg is an absolute file name. That means, it is relative to the root directory. It is not relative to your server root, document root or current dierctory.
try this to give complete abosolute path
rename($_SERVER['DOCUMENT_ROOT'].'/_upload/1.image.jpg', $_SERVER['DOCUMENT_ROOT'].'/_accepted/test/1.image.jpg');
I was writing an web app in PHP, when I encountered a strange situation. To illustrate my problem, consider a web app of this structure:
/
index.php
f1/
f1.php
f2/
f2.php
Contents of these files:
index.php:
<?php require_once("f1/f1.php"); ?>
f1.php:
<?php require_once("../f2/f2.php"); ?>
f2.php: blank
now when I try to open index.php in my browser I get this error:
Warning: require_once(../f2/f2.php) [function.require-once]:
failed to open stream: No such file or directory in /var/www/reqtest/f1/f1.php on line 2
Fatal error: require_once() [function.require]:
Failed opening required '../f2/f2.php' (include_path='.:/usr/share/php:/usr/share/pear') in /var/www/reqtest/f1/f1.php on line 2
Is there something obvious I'm missing? how do include paths work in PHP?
Before I asked this question, I attempted to experiment and find out. I set up another test, like so:
/
index.php
f1/
f1.php
f2.php
index.php:
<?php require_once("f1/f1.php"); ?>
f1.php:
<?php require_once("f2.php"); ?>
f2.php: blank
To my surprise (and utter confusion), this worked out fine!
So, what is the secret behind the path resolution?
PS I saw this question, but it still does not answer the second case that I've stated here.
If you include another file, the working directory remains where the including file is.
Your examples are working as intended.
Edit: The second example works because . (actual directory) is in your include path (see your error message).
Edit2:
In your second example, the key point of your interest is this line:
<?php require_once("f2.php"); ?>
At first it will look in the current working dir (/var/www/req_path_test), but does not find f2.php.
As fallback, it will try to find f2.php in your include_path ('.:/usr/share/php:/usr/share/pear'), starting with '.' (which is relative to the actual file, not the including one).
So './f2.php' works and the require does not fail.
When you open index.php, working dir is set to the folder this file resides in. And inside insluded f1.php this working dir does not change.
You can include files by using their absolute paths, relative to the current included file like this:
require_once(dirname(__FILE__).'/../../test/file.php')
But better consider using an autoloader if these files contain classes.
Normaly in you old structure
<?php require_once("f2/f2.php"); ?>
instead of
<?php require_once("../f2/f2.php"); ?>
should work. As far as i know php takes the paths from the initial script
It sounds like your server has the open_basedir setting enabled in the PHP configuration. This makes it impossible to include (and open) files in folders above your in the directory structur (i.e., you can't use ../ to go up in the folder structure).
From the PHP Docs PHP include
Files are included based on the file path given or, if none is given, the include_path specified. If the file isn't found in the include_path, include will finally check in the calling script's own directory and the current working directory before failing.
If the file path is not given then i.e require_once("f2.php");
1st. The include_path is checked
2nd. The calling scripts own directory is checked
3rd. Finally the current working directory is checked
If file not found then PHP throws warning on file include & fatal error on require
If a path is defined — whether absolute (starting with a drive letter or \ on Windows, or / on Unix/Linux systems) or relative to the current directory (starting with . or ..) — the include_path will be ignored altogether. For example, if a filename begins with ../, the parser will look in the parent directory to find the requested file.
If you include/require your file beginning with . or .. or ./ then PHP's parser will look in the parent directory which is the current working directory i.e require_once("../f2/f2.php"), php will check at the root directory as the calling script index.php is in that directory.
Now You have not defined any include path in your PHP script thus it always falls back to the calling script and then into the current working directory.
// Check your default include path, most likely to be C:\xampp\php\PEAR
echo get_include_path();
// To set include path
set_include_path ( string $new_include_path ) : string
The Current Working Directory is derived from your main calling script index.php.
// The Current Working Directory can be checked
echo getcwd();
In the first Example where the required file "../f2/f2.php" is from f1.php
You code does not work because -
The specified path is ignored by PHP as your filename begins with ../
f1/ the calling script's own directory is ignored as well.
The parser directory looks into the parent directory to find the requested file. The current working directory is root directory, this is from where you have initiated the working script index.php. The file is not located at this directory, wrong path given.
Thus you get the Fatal Error
In the Second example you have changed the directory & from f1.php you require_once("f2.php").
Your code works because -
This time you require("f2.php") no leading ../ or ./ This time PHP checks the include_path but does find it there, as you haven't defined it and the file does not reside in the default preset include_path.
This time the calling script f1.php's directory is f1/. and you require file ("f2.php") is located at this directory. PHP This time checks the file in this directory and finds it.
PHP does not have to check the working directory as the file was found.
Thus Your Code Works Fine!