A. What does this do?
require ("./file.php");
B. in comparison to this?
require ("file.php");
(Its not up-one-directory.. which would be)
require ("../file.php");
./ is the current directory. It is largely the same as just file.php, but in many cases (this one included) it doesn't check any standard places PHP might look for a file, instead checking only the current directory.
From the PHP documentation (notice the last sentence):
Files for including are first looked for in each include_path entry relative to the current working directory, and then in the directory of current script. E.g. if your include_path is libraries, current working directory is /www/, you included include/a.php and there is include "b.php" in that file, b.php is first looked in /www/libraries/ and then in /www/include/. If filename begins with ./ or ../, it is looked only in the current working directory.
The first version forces the internal mechanism to include files relatively to the... directly executed file. So for example you have
index.php
// directly executed script (php -f index.php or from a browser)
include 'second.php';
second.php
// This is included relatively to index.php
// Actually, it is first searched relatively to include_path, then relatively
// to index.php
include './third.php';
third.php
// This is included relatively to second.php ONLY. It does not search
// include_path
return "foo";
The Short Answer
You're right, it's not up one directory. A . refers to the directory you're in, and .. refers to the parent directory.
Meaning, ./file.php and file.php are functionally equivalent in PHP. Here's the relavent page of documentation:
http://us.php.net/manual/en/wrappers.file.php
The Longer Answer
However, just because they work the same in this context doesn't mean they're always the same.
When you're operating in a *nix shell environment, and you type the name of an executable file, the shell will look in the PATH directories, but it won't look in the CWD, or the directory you're currently in.
So, if you're in a directory that has a file called: myprogram.php (this would be a PHP CLI file) and you just type:
myprogram.php
it doesn't matter if your program is executable or not. The shell will look in /bin/, /usr/bin/, etc. for your file, but it won't look in ./: the directory you're in.
To execute that program without adding your directory to the PATH, you need to type
./myprogram
So really, ./ is more explicit. It means, "the file you're looking for HAS to be right here" and lack of ./ means, "the file should be somewhere the program is looking for files".
The dot-slash forces the file to be found in the current directory only, rather than additionally searching the paths mentioned in the include_path setting.
Edit: I have completely rewritten the answer for the sake of clarity
When including a file, you can either use ./myfile.php or myfile.php.
They are not the same and you should always, preferably, use the first syntax, unless you know what you're doing.
The difference is best illustrated with an example: lets say you have the following files and folder structure:
index.php
inc/inner.php
From index.php, you can include your inner template without the ' ./' and it will work as expected:
# index.php
<?php
include "inc/inner.php";
Now let's say we add a new file, so the folder structure is now like this:
index.php
inc/inner.php
inc/inner-inner.php
To include inner-inner.php in inner.php, we would do this... right?
# src/inner.php
<?php
include "inner-inner.php";
Wrong. inner.php will just look for inner-inner.php in the root folder.
index.php
* inner-inner.php <- Doesn't exist, PHP ERROR.
inc/inner.php
inc/inner-inner.php
Instead, we should use the ./ syntax to include inner-inner. This will instruct that the file we are including must be included relative to the current file, not to the current "entry point" of the PHP script (index.php in the example).
# src/inner.php
<?php
include "./inner-inner.php";
In addition, and as mentioned in other comments, if you have configured a different "load path" for your PHP application, that load path will be looked up first without the ./ syntax. Instead, the ./ syntax is more efficient because it doesn't check the "include path" of php 1.
If you wanna make your includes even more explicit, use the __DIR__ constant. This tells php: "Use the directory of this file" 2.
# src/inner.php
<?php
include __DIR__ . "/inner-inner.php";
TLDR; Always use ./ or __DIR__ 2 because it's relative to the current (working) directory and doesn't depend on the PHP "include path" 1.
References:
[Include - https://www.php.net/manual/en/function.include.php]
Magic Constants - https://www.php.net/manual/en/language.constants.magic.php
Simply you are telling php to include the file in the current directory only or fail if the file is not present.
If you use the format "indexcommon3.php" and the file is not present php will search it into the include_path system variable.
For reference you can use http://www.php.net/manual/en/function.include.php
It's explicitly naming the current directory.
Related
A. What does this do?
require ("./file.php");
B. in comparison to this?
require ("file.php");
(Its not up-one-directory.. which would be)
require ("../file.php");
./ is the current directory. It is largely the same as just file.php, but in many cases (this one included) it doesn't check any standard places PHP might look for a file, instead checking only the current directory.
From the PHP documentation (notice the last sentence):
Files for including are first looked for in each include_path entry relative to the current working directory, and then in the directory of current script. E.g. if your include_path is libraries, current working directory is /www/, you included include/a.php and there is include "b.php" in that file, b.php is first looked in /www/libraries/ and then in /www/include/. If filename begins with ./ or ../, it is looked only in the current working directory.
The first version forces the internal mechanism to include files relatively to the... directly executed file. So for example you have
index.php
// directly executed script (php -f index.php or from a browser)
include 'second.php';
second.php
// This is included relatively to index.php
// Actually, it is first searched relatively to include_path, then relatively
// to index.php
include './third.php';
third.php
// This is included relatively to second.php ONLY. It does not search
// include_path
return "foo";
The Short Answer
You're right, it's not up one directory. A . refers to the directory you're in, and .. refers to the parent directory.
Meaning, ./file.php and file.php are functionally equivalent in PHP. Here's the relavent page of documentation:
http://us.php.net/manual/en/wrappers.file.php
The Longer Answer
However, just because they work the same in this context doesn't mean they're always the same.
When you're operating in a *nix shell environment, and you type the name of an executable file, the shell will look in the PATH directories, but it won't look in the CWD, or the directory you're currently in.
So, if you're in a directory that has a file called: myprogram.php (this would be a PHP CLI file) and you just type:
myprogram.php
it doesn't matter if your program is executable or not. The shell will look in /bin/, /usr/bin/, etc. for your file, but it won't look in ./: the directory you're in.
To execute that program without adding your directory to the PATH, you need to type
./myprogram
So really, ./ is more explicit. It means, "the file you're looking for HAS to be right here" and lack of ./ means, "the file should be somewhere the program is looking for files".
The dot-slash forces the file to be found in the current directory only, rather than additionally searching the paths mentioned in the include_path setting.
Edit: I have completely rewritten the answer for the sake of clarity
When including a file, you can either use ./myfile.php or myfile.php.
They are not the same and you should always, preferably, use the first syntax, unless you know what you're doing.
The difference is best illustrated with an example: lets say you have the following files and folder structure:
index.php
inc/inner.php
From index.php, you can include your inner template without the ' ./' and it will work as expected:
# index.php
<?php
include "inc/inner.php";
Now let's say we add a new file, so the folder structure is now like this:
index.php
inc/inner.php
inc/inner-inner.php
To include inner-inner.php in inner.php, we would do this... right?
# src/inner.php
<?php
include "inner-inner.php";
Wrong. inner.php will just look for inner-inner.php in the root folder.
index.php
* inner-inner.php <- Doesn't exist, PHP ERROR.
inc/inner.php
inc/inner-inner.php
Instead, we should use the ./ syntax to include inner-inner. This will instruct that the file we are including must be included relative to the current file, not to the current "entry point" of the PHP script (index.php in the example).
# src/inner.php
<?php
include "./inner-inner.php";
In addition, and as mentioned in other comments, if you have configured a different "load path" for your PHP application, that load path will be looked up first without the ./ syntax. Instead, the ./ syntax is more efficient because it doesn't check the "include path" of php 1.
If you wanna make your includes even more explicit, use the __DIR__ constant. This tells php: "Use the directory of this file" 2.
# src/inner.php
<?php
include __DIR__ . "/inner-inner.php";
TLDR; Always use ./ or __DIR__ 2 because it's relative to the current (working) directory and doesn't depend on the PHP "include path" 1.
References:
[Include - https://www.php.net/manual/en/function.include.php]
Magic Constants - https://www.php.net/manual/en/language.constants.magic.php
Simply you are telling php to include the file in the current directory only or fail if the file is not present.
If you use the format "indexcommon3.php" and the file is not present php will search it into the include_path system variable.
For reference you can use http://www.php.net/manual/en/function.include.php
It's explicitly naming the current directory.
I'm having trouble understanding the ruleset regarding PHP relative include paths. If I run file A.PHP- and file A.PHP includes file B.PHP which includes file C.PHP, should the relative path to C.PHP be in relation to the location of B.PHP, or to the location of A.PHP? That is, does it matter which file the include is called from, or only what the current working directory is- and what determines the current working directory?
It's relative to the main script, in this case A.php. Remember that include() just inserts code into the currently running script.
That is, does it matter which file the include is called from
No.
If you want to make it matter, and do an include relative to B.php, use the __FILE__ constant (or __DIR__ since PHP 5.2 IIRC) which will always point to the literal current file that the line of code is located in.
include(dirname(__FILE__)."/C.PHP");
#Pekka got me there, but just want to share what I learned:
getcwd() returns the directory where the file you started executing resides.
dirname(__FILE__) returns the directory of the file containing the currently executing code.
Using these two functions, you can always build an include path relative to what you need.
e.g., if b.php and c.php share a directory, b.php can include c.php like:
include(dirname(__FILE__).'/c.php');
no matter where b.php was called from.
In fact, this is the preferred way of establishing relative paths, as the extra code frees PHP from having to iterate through the include_path in the attempt to locate the target file.
Sources:
Difference Between getcwd() and dirname(__FILE__) ? Which should I use?
Why you should use dirname(__FILE__)
The accepted answer of Pekka is incomplete and, in a general context, misleading. If the file is provided as a relative path, the called language construct include will search for it in the following way.
First, it will go through the paths of the environment variable include_path, which can be set with ini_set. If this fails, it will search in the calling script's own directory dirname(__FILE__) (__DIR__ with php >= 5.3.) If this also fails, only then it will search in the working directory ! It just turns out that, by default, the environment variable include_path begins with ., which is the current working directory. That is the only reason why it searches first in the current working directory. See http://php.net/manual/en/function.include.php.
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.
So, the correct answer to the first part of the question is that it does matter where is located the included calling script. The answer to the last part of the question is that the initial working directory, in a web server context, is the directory of the called script, the script that includes all the others while being handled by PHP. In a command line context, the initial working directory is whatever it is when php is invoked at the prompt, not necessarily the directory where the called script is located. The current working directory, however, can be changed at run time with the PHP function chdir. See http://php.net/manual/en/function.chdir.php.
This paragraph is added to comment on other answers. Some have mentioned that relying on include_path is less robust and thus it is preferable to use full paths such as ./path or __DIR__ . /path. Some went as far as saying that relying on the working directory . itself is not safe, because it can be changed. However, some times, you need to rely on environment values. For example, you might want set include_path empty, so that the directory of the calling script is the first place that it will search, even before the current working directory. The code might be already written and updated regularly from external sources and you do not want to reinsert the prefix __DIR__ each time the code is updated.
If include path doesn't start with ./ or ../, e.g.:
include 'C.php'; // precedence: include_path (which include '.' at first),
// then path of current `.php` file (i.e. `B.php`), then `.`.
If include path starts with ./ or ../, e.g.:
include './C.php'; // relative to '.'
include '../C.php'; // also relative to '.'
The . or .. above is relative to getcwd(), which defaults to the path of the entry .php file (i.e. A.php).
Tested on PHP 5.4.3 (Build Date : May 8 2012 00:47:34).
(Also note that chdir() can change the output of getcwd().)
Short answer: it's relative to the including script.
TFM explains it correctly:
If the file isn't found in the include_path, include will check in the calling script's directory and the current working directory
So, if /app/main.php says include("./inc.php") that will find /app/inc.php.
The ./ is not strictly necessary but removes any dependency on include_path.
I would not rely on finding include files in the current working directory in case someone changes it with chdir().
dir
-> a.php
-> c.php
- dir2
-> b.php
To include a in b you need to include("../a.php");
To include b in c you need to include("dir2/b.php");
I'm having trouble understanding the ruleset regarding PHP relative include paths. If I run file A.PHP- and file A.PHP includes file B.PHP which includes file C.PHP, should the relative path to C.PHP be in relation to the location of B.PHP, or to the location of A.PHP? That is, does it matter which file the include is called from, or only what the current working directory is- and what determines the current working directory?
It's relative to the main script, in this case A.php. Remember that include() just inserts code into the currently running script.
That is, does it matter which file the include is called from
No.
If you want to make it matter, and do an include relative to B.php, use the __FILE__ constant (or __DIR__ since PHP 5.2 IIRC) which will always point to the literal current file that the line of code is located in.
include(dirname(__FILE__)."/C.PHP");
#Pekka got me there, but just want to share what I learned:
getcwd() returns the directory where the file you started executing resides.
dirname(__FILE__) returns the directory of the file containing the currently executing code.
Using these two functions, you can always build an include path relative to what you need.
e.g., if b.php and c.php share a directory, b.php can include c.php like:
include(dirname(__FILE__).'/c.php');
no matter where b.php was called from.
In fact, this is the preferred way of establishing relative paths, as the extra code frees PHP from having to iterate through the include_path in the attempt to locate the target file.
Sources:
Difference Between getcwd() and dirname(__FILE__) ? Which should I use?
Why you should use dirname(__FILE__)
The accepted answer of Pekka is incomplete and, in a general context, misleading. If the file is provided as a relative path, the called language construct include will search for it in the following way.
First, it will go through the paths of the environment variable include_path, which can be set with ini_set. If this fails, it will search in the calling script's own directory dirname(__FILE__) (__DIR__ with php >= 5.3.) If this also fails, only then it will search in the working directory ! It just turns out that, by default, the environment variable include_path begins with ., which is the current working directory. That is the only reason why it searches first in the current working directory. See http://php.net/manual/en/function.include.php.
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.
So, the correct answer to the first part of the question is that it does matter where is located the included calling script. The answer to the last part of the question is that the initial working directory, in a web server context, is the directory of the called script, the script that includes all the others while being handled by PHP. In a command line context, the initial working directory is whatever it is when php is invoked at the prompt, not necessarily the directory where the called script is located. The current working directory, however, can be changed at run time with the PHP function chdir. See http://php.net/manual/en/function.chdir.php.
This paragraph is added to comment on other answers. Some have mentioned that relying on include_path is less robust and thus it is preferable to use full paths such as ./path or __DIR__ . /path. Some went as far as saying that relying on the working directory . itself is not safe, because it can be changed. However, some times, you need to rely on environment values. For example, you might want set include_path empty, so that the directory of the calling script is the first place that it will search, even before the current working directory. The code might be already written and updated regularly from external sources and you do not want to reinsert the prefix __DIR__ each time the code is updated.
If include path doesn't start with ./ or ../, e.g.:
include 'C.php'; // precedence: include_path (which include '.' at first),
// then path of current `.php` file (i.e. `B.php`), then `.`.
If include path starts with ./ or ../, e.g.:
include './C.php'; // relative to '.'
include '../C.php'; // also relative to '.'
The . or .. above is relative to getcwd(), which defaults to the path of the entry .php file (i.e. A.php).
Tested on PHP 5.4.3 (Build Date : May 8 2012 00:47:34).
(Also note that chdir() can change the output of getcwd().)
Short answer: it's relative to the including script.
TFM explains it correctly:
If the file isn't found in the include_path, include will check in the calling script's directory and the current working directory
So, if /app/main.php says include("./inc.php") that will find /app/inc.php.
The ./ is not strictly necessary but removes any dependency on include_path.
I would not rely on finding include files in the current working directory in case someone changes it with chdir().
dir
-> a.php
-> c.php
- dir2
-> b.php
To include a in b you need to include("../a.php");
To include b in c you need to include("dir2/b.php");
I have seen this:
<?php
include( dirname(__FILE__) . DIRECTORY_SEPARATOR . 'my_file.php');
?>
Why would I ever need to do this? Why would I go to the trouble of getting the dirname and then concatenating that with a directory separator, and a new filename?
Is the code above not equivalent to this:
<?php
include( 'my_file.php' );
?>
??
The PHP doc says,
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. The include() construct will emit a warning if it cannot find a file; this is different behavior from require(), which will emit a fatal error.
Let's say I have a (fake) directory structure like:
.../root/
/app
bootstrap.php
/scripts
something/
somescript.php
/public
index.php
Now assume that bootstrap.php has some code included for setting up database connections or some other kind of boostrapping stuff.
Assume you want to include a file in boostrap.php's folder called init.php. Now, to avoid scanning the entire include path with include 'init.php', you could use include './init.php'.
There's a problem though. That ./ will be relative to the script that included bootstrap.php, not bootstrap.php. (Technically speaking, it will be relative to the working directory.)
dirname(__FILE__) allows you to get an absolute path (and thus avoid an include path search) without relying on the working directory being the directory in which bootstrap.php resides.
(Note: since PHP 5.3, you can use __DIR__ in place of dirname(__FILE__).)
Now, why not just use include 'init.php';?
As odd as it is at first though, . is not guaranteed to be in the include path. Sometimes to avoid useless stat()'s people remove it from the include path when they are rarely include files in the same directory (why search the current directory when you know includes are never going to be there?).
Note: About half of this answer is address in a rather old post: What's better of require(dirname(__FILE__).'/'.'myParent.php') than just require('myParent.php')?
I might have even a simpler explanation to this question compared to the accepted answer so I'm going to give it a go: Assume this is the structure of the files and directories of a project:
Project root directory:
file1.php
file3.php
dir1/
file2.php
(dir1 is a directory and file2.php is inside it)
And this is the content of each of the three files above:
//file1.php:
<?php include "dir1/file2.php"
//file2.php:
<?php include "../file3.php"
//file3.php:
<?php echo "Hello, Test!";
Now run file1.php and try to guess what should happen. You might expect to see "Hello, Test!", however, it won't be shown! What you'll get instead will be an error indicating that the file you have requested(file3.php) does not exist!
The reason is that, inside file1.php when you include file2.php, the content of it is getting copied and then pasted back directly into file1.php which is inside the root directory, thus this part "../file3.php" runs from the root directory and thus goes one directory up the root! (and obviously it won't find the file3.php).
Now, what should we do ?!
Relative paths of course have the problem above, so we have to use absolute paths. However, absolute paths have also one problem. If you (for example) copy the root folder (containing your whole project) and paste it in anywhere else on your computer, the paths will be invalid from that point on! And that'll be a REAL MESS!
So we kind of need paths that are both absolute and dynamic(Each file dynamically finds the absolute path of itself wherever we place it)!
The way we do that is by getting help from PHP, and dirname() is the function to go for, which gives the absolute path to the directory in which a file exists in. And each file name could also be easily accessed using the __FILE__ constant. So dirname(__FILE__) would easily give you the absolute (while dynamic!) path to the file we're typing in the above code. Now move your whole project to a new place, or even a new system, and tada! it works!
So now if we turn the project above to this:
//file1.php:
<?php include(dirname(__FILE__)."/dir1/file2.php");
//file2.php:
<?php include(dirname(__FILE__)."/../file3.php");
//file3.php:
<?php echo "Hello, Test!";
if you run it, you'll see the almighty Hello, Test!! (hopefully, if you've not done anything else wrong).
It's also worth mentioning that from PHP5, a nicer way(with regards to readability and preventing eye boilage!) has been provided by PHP as well which is the constant __DIR__ which does exactly the same thing as dirname(__FILE__)!
Hope that helps.
I used this below if this is what you are thinking. It it worked well for me.
<?php
include $_SERVER['DOCUMENT_ROOT']."/head_lib.php";
?>
What I was trying to do was pulla file called /head_lib.php from the root folder. It would not pull anything to build the webpage. The header, footer and other key features in sub directories would never show up. Until I did above it worked like a champ.
If you want code is running on multiple servers with different environments,then we have need
to use dirname(FILE) in an include or include_once statement.
reason is follows.
1. Do not give absolute path to include files on your server.
2. Dynamically calculate the full path like absolute path.
Use a combination of dirname(FILE) and subsequent calls to itself until you reach to the home of your '/myfile.php'.
Then attach this variable that contains the path to your included files.
I'm having trouble understanding the ruleset regarding PHP relative include paths. If I run file A.PHP- and file A.PHP includes file B.PHP which includes file C.PHP, should the relative path to C.PHP be in relation to the location of B.PHP, or to the location of A.PHP? That is, does it matter which file the include is called from, or only what the current working directory is- and what determines the current working directory?
It's relative to the main script, in this case A.php. Remember that include() just inserts code into the currently running script.
That is, does it matter which file the include is called from
No.
If you want to make it matter, and do an include relative to B.php, use the __FILE__ constant (or __DIR__ since PHP 5.2 IIRC) which will always point to the literal current file that the line of code is located in.
include(dirname(__FILE__)."/C.PHP");
#Pekka got me there, but just want to share what I learned:
getcwd() returns the directory where the file you started executing resides.
dirname(__FILE__) returns the directory of the file containing the currently executing code.
Using these two functions, you can always build an include path relative to what you need.
e.g., if b.php and c.php share a directory, b.php can include c.php like:
include(dirname(__FILE__).'/c.php');
no matter where b.php was called from.
In fact, this is the preferred way of establishing relative paths, as the extra code frees PHP from having to iterate through the include_path in the attempt to locate the target file.
Sources:
Difference Between getcwd() and dirname(__FILE__) ? Which should I use?
Why you should use dirname(__FILE__)
The accepted answer of Pekka is incomplete and, in a general context, misleading. If the file is provided as a relative path, the called language construct include will search for it in the following way.
First, it will go through the paths of the environment variable include_path, which can be set with ini_set. If this fails, it will search in the calling script's own directory dirname(__FILE__) (__DIR__ with php >= 5.3.) If this also fails, only then it will search in the working directory ! It just turns out that, by default, the environment variable include_path begins with ., which is the current working directory. That is the only reason why it searches first in the current working directory. See http://php.net/manual/en/function.include.php.
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.
So, the correct answer to the first part of the question is that it does matter where is located the included calling script. The answer to the last part of the question is that the initial working directory, in a web server context, is the directory of the called script, the script that includes all the others while being handled by PHP. In a command line context, the initial working directory is whatever it is when php is invoked at the prompt, not necessarily the directory where the called script is located. The current working directory, however, can be changed at run time with the PHP function chdir. See http://php.net/manual/en/function.chdir.php.
This paragraph is added to comment on other answers. Some have mentioned that relying on include_path is less robust and thus it is preferable to use full paths such as ./path or __DIR__ . /path. Some went as far as saying that relying on the working directory . itself is not safe, because it can be changed. However, some times, you need to rely on environment values. For example, you might want set include_path empty, so that the directory of the calling script is the first place that it will search, even before the current working directory. The code might be already written and updated regularly from external sources and you do not want to reinsert the prefix __DIR__ each time the code is updated.
If include path doesn't start with ./ or ../, e.g.:
include 'C.php'; // precedence: include_path (which include '.' at first),
// then path of current `.php` file (i.e. `B.php`), then `.`.
If include path starts with ./ or ../, e.g.:
include './C.php'; // relative to '.'
include '../C.php'; // also relative to '.'
The . or .. above is relative to getcwd(), which defaults to the path of the entry .php file (i.e. A.php).
Tested on PHP 5.4.3 (Build Date : May 8 2012 00:47:34).
(Also note that chdir() can change the output of getcwd().)
Short answer: it's relative to the including script.
TFM explains it correctly:
If the file isn't found in the include_path, include will check in the calling script's directory and the current working directory
So, if /app/main.php says include("./inc.php") that will find /app/inc.php.
The ./ is not strictly necessary but removes any dependency on include_path.
I would not rely on finding include files in the current working directory in case someone changes it with chdir().
dir
-> a.php
-> c.php
- dir2
-> b.php
To include a in b you need to include("../a.php");
To include b in c you need to include("dir2/b.php");