What should I use in the following statement? Include or required.
if(a ==b){
require 'requiredfile.php';
} else {
require 'requiredfile_2.php'
}
If in a function, I know that one, either include or require, only includes the file when called, the other one will include the file regardless. Am I correct?
The difference between include and require is that include will only emit a warning when the file is not found, and require will terminate with a fatal error.
If you are loading vital program parts, you probably want to go with require.
Manual
require() is identical to include() except upon failure it will also produce a fatal E_ERROR level error. In other words, it will halt the script whereas include() only emits a warning (E_WARNING) which allows the script to continue.
Related
Here is the code I have been struggling for several hours:
if ((require $_SESSION['ROOT_PATH'] . '/templates/core/menu_js.php') == 'OK') {
echo 'OK';
} else {
echo 'KO';
}
If I understand the PHP documentation on the "require" directive correctly, the "KO" should never be written because, if the require doesn't work, an error is raised.
In my case, the "KO" is always displayed even with error tunning :
error_reporting(E_ERROR | E_WARNING | E_PARSE | E_COMPILE_ERROR);
Note that the same require (or include) works perfectly in other pages of the site.
EDIT
The if structure has been added after watching the documentation.
At first, I had a single line :
require $_SESSION['ROOT_PATH'] . '/templates/core/menu_js.php';
As I checked that this line was not working, I added the if.
By the way. Required page (when it works) adds a script tag to calling page that I never see on this unique page. On any other page where this require is used, the script appears on output.
So my question should be "if the output of the required php file is not displayed, why is there no error raised ?"
According to the documentation
Successful includes, unless overridden by the included file, return 1... Also, it's possible to return values from included files. You can take the value of the include call as you would for a normal function.
So your file isn't returning 'OK'. It's returning either 1 (for success) or a custom value.
require is a language construct, not a standard function. By using require you're already indicating the code should fail if the file isn't found. Most likely you do not need to check the return value.
This only works the way you expect if menu_js.php contains return 'OK';. Otherwise the require works just fine, but the returned value is not "OK", which is why your "KO" condition is triggered. require doesn't return 'OK'. If the require does not work, the program is halted immediately, your false condition will never be hit the way you think it will.
There are some mistakes here:
According to the PHP documentation:
require is identical to include except upon failure it will also produce a fatal E_COMPILE_ERROR level error.
A fatal error will stop your script execution.
require returns what the included file returns:
// a.php
return 'test';
// b.php
$result = require 'a.php';
echo $result; // will display 'test'
So, don't test the return value of require! Use is_file(), then if true, require that file!
You cannot use require in if
Use this
if(file_exists($_SESSION['ROOT_PATH'] . '/templates/core/menu_js.php'))
{
echo 'OK';
} else {
echo 'KO';
}
While it's true that the require will not return anything, making the conditional always return 'KO' since the left hand value in the parenthesis will never equal 'OK', you can check to see if the file has failed by setting the display_errors ini value in this script temporarily On.
Since you have your error reporting set to catch this, your display_errors is probably off, therefore not showing you this. Set it for this script like this:
ini_set('display_errors', 'on');
I've read that one of the differences between include and require is that
while include only includes the included file if the include statement is
encountered, require includes the file into the file hosting the require
statement even when the require code is not reached by the execution flow.
How can this have any implications. After all, if I have a file which says:
<?php
echo __LINE__;
?>
the output will always be 3, instead of printing the line inside at the
position inside the file which includes such include file.
This is simply not true, as far as I know. If the require line is not reached in the code, the file will certainly not be included.
The PHP Manual on require states:
require is identical to include except upon failure it will also
produce a fatal E_COMPILE_ERROR level error.
(emphasis on "except" added by me)
As far as I know, that's not how require works. The only practical difference is that require will cause your script to terminate if the file cannot be included, and include will just throw a warning and keep executing.
A very important idiom depends on your stated behavior not being true: conditional required files.
if(is_secure_page()) {
require "security.php";
}
Both require and include include their files at the point where they are reached in the execution flow.
A simple test case for this is:
if (false) {
require_once("Nonexistent.php");
}
echo "I'm doing SCIENCE and I'm still alive.\n";
Let's say you are running a mission critical application and will lose $1 per second. And some libraries are only used in certain case and are not affecting your business flow. You should probably use include_once for these optional libraries. Because if the file is not there require will stop your application from running with a fatal error.
require, or even better, require_once should be used exclusively when you want to include code that is actually required by the application, e.g. any libraries you need. include and its brother, include_once should only be used when the included file is not necessary to the execution of your script, such as when you are loading a sidebar or some tangental content.
The only real difference between the two statements is that is the file is not found, require with throw a fatal error, and include will just throw a warning and go on happily chugging away.
I believe that both require and include are loaded and parsed in exactly the same manner. If require would load a file in a given context, so would include.
I personally always use require_once, I find in most cases it's what you need (either you get the page once or you halt the script). And include throws warnings that you need to hide if you want to "try" including stuff. But that's just bad programming practice, a simple if (file_exists($file)) require_once $file; does the trick and totally beats include.
So in my opinion, require wins by far.
PS: Beaten by a sec !
I have several pages which use the include or require language constructs within PHP. Many of these lie within IF, ELSE statements.
I do realize that a page will not load at all if a require'd file is missing but the main purpose of including this way is to:
1) reduce code clutter on the page
2) not load the file unless a statement is met.
Does issuing an include or require statement load the file regardless (and thus eliminating the benefits I was trying to achieve by placing within the if/else statement?
Brief example:
<?php
$i = 1
if($i ==1) {
require_once('somefile.php');
} else {
require_once('otherfile.php');
}
?>
At page load, are both files checked AND loaded?
If you place an include/require statement in the body of an if (or else), it'll be executed if the if's condition is true.
if ($a == 10) {
// your_first_file.php will only be loaded if $a == 10
require 'your_first_file.php';
} else {
// your_second_file.php will only be loaded if $a != 10
require 'your_second_file.php';
}
And, if you want, you can test this pretty easily.
This first example :
if (true) {
require 'file_that_doesnt_exist';
}
will get you :
Warning: require(file_that_doesnt_exist) [function.require]: failed to open stream: No such file or directory
Fatal error: require() [function.require]: Failed opening required 'file_that_doesnt_exist'
i.e. the require is executed -- and fails, as the file doesn't exist.
While this second example :
if (false) {
require 'file_that_doesnt_exist';
}
Will not get you any error : the require is not executed.
At page load, are both files checked AND loaded?
No, at least not since (IIRC) PHP 4.0.1.
If you want to reduce include clutter, and you are working with mainly object-oriented code, also take a look at PHP's autoloading.
No, only one of those files would be loaded.
The include and require constructs are only evaluated when passed. The files are only read when your expression is met.
It's simple to explain by considering that the constructs might contain variables:
require_once("otherfile-{$i}.php");
That's supported. But it could not possibly work before PHP runs over that particular line, because it needs to know the state of $i to load the right file.
Is there any difference between them? Is using them a matter of preference? Does using one over the other produce any advantages? Which is better for security?
require will throw a PHP Fatal Error if the file cannot be loaded. (Execution stops)
include produces a Warning if the file cannot be loaded. (Execution continues)
Here is a nice illustration of include and require difference:
From: Difference require vs. include php (by Robert; Nov 2012)
You find the differences explained in the detailed PHP manual on the page of require:
require is identical to include except upon failure it will also produce a fatal E_COMPILE_ERROR level error. In other words, it will halt the script whereas include only emits a warning (E_WARNING) which allows the script to continue.
See #efritz's answer for an example
Use include if you don't mind your script continuing without loading the file (in case it doesn't exist etc) and you can (although you shouldn't) live with a Warning error message being displayed.
Using require means your script will halt if it can't load the specified file, and throw a Fatal error.
The difference between include() and require() arises when the file being included cannot be found: include() will release a warning (E_WARNING) and the script will continue, whereas require() will release a fatal error (E_COMPILE_ERROR) and terminate the script. If the file being included is critical to the rest of the script running correctly then you need to use require().
For more details : Difference between Include and Require in PHP
As others pointed out, the only difference is that require throws a fatal error, and include - a catchable warning. As for which one to use, my advice is to stick to include. Why? because you can catch a warning and produce a meaningful feedback to end users. Consider
// Example 1.
// users see a standard php error message or a blank screen
// depending on your display_errors setting
require 'not_there';
// Example 2.
// users see a meaningful error message
try {
include 'not_there';
} catch(Exception $e) {
echo "something strange happened!";
}
NB: for example 2 to work you need to install an errors-to-exceptions handler, as described here http://www.php.net/manual/en/class.errorexception.php
function exception_error_handler($errno, $errstr, $errfile, $errline ) {
throw new ErrorException($errstr, 0, $errno, $errfile, $errline);
}
set_error_handler("exception_error_handler");
<?PHP
echo "Firstline";
include('classes/connection.php');
echo "I will run if include but not on Require";
?>
A very simple Practical example with code.
The first echo will be displayed. No matter you use include or require because its runs before include or required.
To check the result, In second line of a code intentionally provide the wrong path to the file or make error in file name. Thus the second echo to be displayed or not will be totally dependent on whether you use require or include.
If you use require the second echo will not execute but if you use include not matter what error comes you will see the result of second echo too.
In case of Include Program will not terminate and display warning on browser,On the other hand Require program will terminate and display fatal error in case of file not found.
file1.php and file2.php with die(); function.
include.php:
<? include 'file1.php';
include 'file2.php' ?>
file1.php
<? echo 'included'; die(); ?>
file2.php
<? echo 'not included'; die(); ?>
How can I include both files with die(); function?
Non-English Speakers:
You can provide your question in your native language as well, and somebody here may be able to translate it for you. Just make your best effort to ask in English, and add your native tongue below.
If you would like to test whether the includes happened successfully, you can test the return value of the include function itself:
// http://us3.php.net/manual/en/function.include.php Example #4
if ((include 'file1.php') != 'OK') {
die();
}
You may also consider require() instead of include() depending on your needs:
require() is identical to include() except upon failure it will also produce a fatal E_ERROR level error. In other words, it will halt the script whereas include() only emits a warning (E_WARNING) which allows the script to continue.
If I understand correctly what you are trying to do then unfortunately it isn't possible.
die(); will stop the script from executing at the point from where it is called.
Here is how to include a file, or die with a message if the include fails code sample.
(include('file.php')) || die('Failed to include file!');
if (!condition){
include_once('./inc/header.inc.php');
echo "Errormessage";
include_once('./inc/footer.inc.php');
die();
}
I hope this is what you wanted.
Just a minor improvement to LorenzoP's method:
(#include("file.php")) or die("Failed to include!");
// notice the # sign!
This way, you save yourself 2 ugly lines of php warning when inclusion fails. Otherwise I think this is truly the best way to handle failed includes. (Also, his answer should be the accepted one.)
If the execution of your included file is not dependent on the current file (no shared variables, functions, etc.), then use
file_get_contents('link_to_file.php');
instead of an include method. When you force the file to execute independently it will not make a effect in the script execution.
die() is just an exit with an error, you can't include files with it and I don't really understand why you want to. Could you provide more details as to what you're trying to accomplish?