I face a strange problem including php files. Let me show you the code:
// constants.php
$MYSQL_HOST_PORT = 'localhost:3306';
// functions.php
include 'constants.php';
function getVar() {
echo $MYSQL_HOST_PORT;
}
// doSth.php
include 'functions.php';
echo $MYSQL_HOST_PORT; // The variable is visible and echoed normally as expected!
echo getVar(); // The variable is not echoed! its "".
Any ideas ?
For one, the echo in echo getVar(); won't ever print anything, because getVar doesn't return a value.
Secondly, if you (for some reason) want getVar() itself to work correctly, you need to add a global $MYSQL_HOST_PORT; line, to make it look for $MYSQL_HOST_PORT in the global scope.
Rather than globalising the $MYSQL_HOST_PORT variable, why not simply make it a constant?
// constants.php
define('MYSQL_HOST_PORT', 'localhost:3306');
Provided constants.php is included, you can reference the MYSQL_HOST_PORT constant anywhere.
As indicated in zerocrate's answer, the issue is a scoping one. The enclosed scope of the getVar() function does not include $MYSQL_HOST_PORT.
One thing that I can see wrong is that with the line echo getVar(); you are not getting a return value from the function so you can simply write getVar(); by itself.
Related
I'm trying to pass a variable into an include file. My host changed PHP version and now whatever solution I try doesn't work.
I think I've tried every option I could find. I'm sure it's the simplest thing!
The variable needs to be set and evaluated from the calling first file (it's actually $_SERVER['PHP_SELF'], and needs to return the path of that file, not the included second.php).
OPTION ONE
In the first file:
global $variable;
$variable = "apple";
include('second.php');
In the second file:
echo $variable;
OPTION TWO
In the first file:
function passvariable(){
$variable = "apple";
return $variable;
}
passvariable();
OPTION THREE
$variable = "apple";
include "myfile.php?var=$variable"; // and I tried with http: and full site address too.
$variable = $_GET["var"]
echo $variable
None of these work for me. PHP version is 5.2.16.
What am I missing?
Thanks!
You can use the extract() function
Drupal use it, in its theme() function.
Here it is a render function with a $variables argument.
function includeWithVariables($filePath, $variables = array(), $print = true)
{
$output = NULL;
if(file_exists($filePath)){
// Extract the variables to a local namespace
extract($variables);
// Start output buffering
ob_start();
// Include the template file
include $filePath;
// End buffering and return its contents
$output = ob_get_clean();
}
if ($print) {
print $output;
}
return $output;
}
./index.php :
includeWithVariables('header.php', array('title' => 'Header Title'));
./header.php :
<h1><?php echo $title; ?></h1>
Option 3 is impossible - you'd get the rendered output of the .php file, exactly as you would if you hit that url in your browser. If you got raw PHP code instead (as you'd like), then ALL of your site's source code would be exposed, which is generally not a good thing.
Option 2 doesn't make much sense - you'd be hiding the variable in a function, and be subject to PHP's variable scope. You'ld also have to have $var = passvariable() somewhere to get that 'inside' variable to the 'outside', and you're back to square one.
option 1 is the most practical. include() will basically slurp in the specified file and execute it right there, as if the code in the file was literally part of the parent page. It does look like a global variable, which most people here frown on, but by PHP's parsing semantics, these two are identical:
$x = 'foo';
include('bar.php');
and
$x = 'foo';
// contents of bar.php pasted here
Considering that an include statment in php at the most basic level takes the code from a file and pastes it into where you called it and the fact that the manual on include states the following:
When a file is included, the code it contains inherits the variable scope of the line on which the include occurs. Any variables available at that line in the calling file will be available within the called file, from that point forward.
These things make me think that there is a diffrent problem alltogether. Also Option number 3 will never work because you're not redirecting to second.php you're just including it and option number 2 is just a weird work around. The most basic example of the include statment in php is:
vars.php
<?php
$color = 'green';
$fruit = 'apple';
?>
test.php
<?php
echo "A $color $fruit"; // A
include 'vars.php';
echo "A $color $fruit"; // A green apple
?>
Considering that option number one is the closest to this example (even though more complicated then it should be) and it's not working, its making me think that you made a mistake in the include statement (the wrong path relative to the root or a similar issue).
I have the same problem here, you may use the $GLOBALS array.
$GLOBALS["variable"] = "123";
include ("my.php");
It should also run doing this:
$myvar = "123";
include ("my.php");
....
echo $GLOBALS["myvar"];
Have a nice day.
I've run into this issue where I had a file that sets variables based on the GET parameters. And that file could not updated because it worked correctly on another part of a large content management system. Yet I wanted to run that code via an include file without the parameters actually being in the URL string. The simple solution is you can set the GET variables in first file as you would any other variable.
Instead of:
include "myfile.php?var=apple";
It would be:
$_GET['var'] = 'apple';
include "myfile.php";
OPTION 1 worked for me, in PHP 7, and for sure it does in PHP 5 too. And the global scope declaration is not necessary for the included file for variables access, the included - or "required" - files are part of the script, only be sure you make the "include" AFTER the variable declaration. Maybe you have some misconfiguration with variables global scope in your PHP.ini?
Try in first file:
<?php
$myvariable="from first file";
include ("./mysecondfile.php"); // in same folder as first file LOLL
?>
mysecondfile.php
<?php
echo "this is my variable ". $myvariable;
?>
It should work... if it doesn't just try to reinstall PHP.
In regards to the OP's question, specifically "The variable needs to be set and evaluated from the calling first file (it's actually '$_SERVER['PHP_SELF']', and needs to return the path of that file, not the included second.php)."
This will tell you what file included the file. Place this in the included file.
$includer = debug_backtrace();
echo $includer[0]['file'];
I know this is an old question, but stumbled upon it now and saw nobody mentioned this. so writing it.
The Option one if tweaked like this, it should also work.
The Original
Option One
In the first file:
global $variable;
$variable = "apple";
include('second.php');
In the second file:
echo $variable;
TWEAK
In the first file:
$variable = "apple";
include('second.php');
In the second file:
global $variable;
echo $variable;
According to php docs (see $_SERVER) $_SERVER['PHP_SELF'] is the "filename of the currently executing script".
The INCLUDE statement "includes and evaluates the specified" file and "the code it contains inherits the variable scope of the line on which the include occurs" (see INCLUDE).
I believe $_SERVER['PHP_SELF'] will return the filename of the 1st file, even when used by code in the 'second.php'.
I tested this with the following code and it works as expected ($phpSelf is the name of the first file).
// In the first.php file
// get the value of $_SERVER['PHP_SELF'] for the 1st file
$phpSelf = $_SERVER['PHP_SELF'];
// include the second file
// This slurps in the contents of second.php
include_once('second.php');
// execute $phpSelf = $_SERVER['PHP_SELF']; in the secod.php file
// echo the value of $_SERVER['PHP_SELF'] of fist file
echo $phpSelf; // This echos the name of the First.php file.
An alternative to using $GLOBALS is to store the variable value in $_SESSION before the include, then read it in the included file. Like $GLOBALS, $_SESSION is available from everywhere in the script.
Pass a variable to the include file by setting a $_SESSION variable
e.g.
$_SESSION['status'] = 1;
include 'includefile.php';
// then in the include file read the $_SESSION variable
$status = $_SESSION['status'];
You can execute all in "second.php" adding variable with jQuery
<div id="first"></div>
<script>
$("#first").load("second.php?a=<?=$var?>")
</scrpt>
I found that the include parameter needs to be the entire file path, not a relative path or partial path for this to work.
This worked for me: To wrap the contents of the second file into a function, as follows:
firstFile.php
<?php
include("secondFile.php");
echoFunction("message");
secondFile.php
<?php
function echoFunction($variable)
{
echo $variable;
}
Do this:
$checksum = "my value";
header("Location: recordupdated.php?checksum=$checksum");
i code the following
<?php
if ($id = mysql_real_escape_string(#$_GET['pid'])
&& $uid = mysql_real_escape_string(#$_GET['file']))
echo include "foo.php";
else
echo include "bar.php";
?>
When I use the include function in conjunction with a function that's designed to output to the page (e.g., or echo include 'foo.php'), it returns the include but with a "1" after the content that has been included.
echo include "foo.php"
should be
include 'foo.php';
Note that this can also happen when using include with shorthand echo:
<?= include 'foo.php'; ?>
This will also print out the return value of 1 when used inside a script. To get rid of this you need to use the regular PHP opening tag like so:
<?php include 'foo.php'; ?>
PHP will now include the contents of the file without printing the return value.
Okey so the answers here are actually not entirely correct; in some sense even misleading.
include takes the contents of the file and places them in context. One of the more common uses is to pass variable scope around, ie. passing scoped variables in your view by including them in the handler and using include on the view. Common, but there are also other uses; you can also return inside a included file.
Say you have a file like this:
<?php return array
(
'some',
'php'
'based'
'configuration',
'file'
); # config
Doing $config = include 'example-config-above.php'; is perfectly fine and you will get the array above in the $config variable.
If you try to include a file that doesn't have a return statement then you will get 1.
Gotcha Time
You might think that include 'example-config-above.php'; is actually searching for the file in the directory where the file calling the include is located, well it is, but it's also searching for the file in various other paths and those other paths have precedence over the local path!
So if you know you had a file like the above with a return inside it, but are getting 1 and potentially something like weird PEAR errors or such, then you've likely done something like this:
// on a lot of server setups this will load a random pear class
include 'system.php'
Since it's loading a file with out a return you will get 1 instead of (in the case of our example) the configuration array we would be expecting.
Easy fix is of course:
include __DIR__.'/system.php'
That is because the include function returns 1 on success. You get, as you say, 'my name is earl1' because the code inside the included file runs first, printing 'my name is earl' and then you local echo runs printing the return value of include() which is 1.
Let the file.txt contain xxx
echo include("file.txt");
This returns,
xxx1
The '1' is the return value of the include function, denoting the success that the file is accessed. Otherwise it returns nothing, in this case parse error is thrown.
echo print "hello";
This too returns,
hello1
Same as above; '1' denoting the success,it's printed.
echo echo "hello";
print echo "hello";
Both the above cases produces an error.
Since the echo function has no return value, hence undefined.
echo echo "hello"; print echo "hello";
(1st) (2nd)
Now the second 'echo' in the both cases produces an undefined.The first 'echo' or 'print' can't take in the hello output with the undefined (produced by the second echo).
A verification:
if((print "hello")==1)
echo "hey!";
output: hellohey! ('echo' in the 2nd line can be a print, it doesn't matter)
Similarly,
if((include ("file.txt"))==1)
echo "hey!";
output: xxxhey!
Other hand,
if((echo "hello")==1)
echo "hey!";
output: an error
In the first two cases the functions (print and include) returned 1, in the third case 'echo' produces no return value (undefined) hence the third case produces an error.
Well... I am using Codeigniter(php frame work). And I encountered the same problem. What I concluded is that when we try to print/echo the include method then it prints 1 on screen and when we just simply write the include command(example given below) it will only do what it is supposed to do.
<?php include('file/path'); ?> // this works fine for me
<?= include('file/path'); ?> // this works fine but prints "1" on screen
Hope my explaination will be helful to someone
= is assigning operator
== is for checking equal to
check for php operators
I have solved it returning nothing at the end of the included file:
$data = include "data-row.php";
return $data;
Inside data-row.php:
<div>etc</div>
...
<?php return; //End of file
I found the selected answer from this thread very helpful.
Solution 1
ob_start();
include dirname( __FILE__ ) . '/my-file.php';
$my_file = ob_get_clean();
You might also find this thread about the ob_start() function very insightful.
Solution 2
Add a return statement in the file that is being included.
E.g my-file.php
<?php
echo "<p>Foo.</p>";
return;
Drawn from the answer provided by #gtamborero.
To help you understand it the way I do now, just take this oversimplification:
There should always be a return statement otherwise include will return 1 on success.
Happy coding!
M5
Use return null in foo.php and bar.php
echo substr(include("foo.php"),1,-1);
I have a file, lets say it's index.php where the very beginning of the file has an include for "include.php". In include.php I set a variable like this:
<?php $variable = "the value"; ?>
then further down the in index.php I have another include, say "include2.php" that is included like this:
<?php include(get_template_directory_uri() . '/include2.php'); ?>
How can I call the "$variable" that I set in the first include, in "include2.php"?
The exact code that I am using is as follows:
The very first line of the index.php I have this line
<?php include('switcher.php'); ?>
Inside switcher.php I have this
<?php $GLOBALS["demo_color"] = "#fffffe"; ?>
If I use this in index.php, it works
<?php echo $GLOBALS["demo_color"]; ?>
However, If I use the following code to include another php file
<?php include(get_template_directory_uri() . '/demo_color.php'); ?>
then inside demo_color.php I have this code:
<?php echo "demo color:" . $GLOBALS["demo_color"]; ?>
The only thing it outputs is "demo color:"
edited for code-formatting
It simply can be used in include2.php, unless the inclusion of include.php happens inside of a different scope (i.e. inside a function call). see here.
If you want to be completely explicit about the intention of using the variable across the app, use the $GLOBALS["variable"] version of it's name, which will always point to the variable called variable in the global scope.
EDIT: I conducted a test against php 5.3.10 to reconstruct this:
// index.php
<?php
include("define.php");
include("use.php");
// define.php
$foo = "bar";
// use.php
var_dump($foo);
This works exactly as expected, outputting string(3) "bar".
<?PHP
//index.php
$txt='hello world';
include('include.php');
<?PHP
//include.php
echo $txt; //will output hello world
So it does work. Though there seems to be a bigger issue since this is likely to be difficult to maintain in the future. Just putting a variable into global namespace and using it in different files is not a best practice.
To make the code more maintainable it might be an idea to use classes so you can attach the variables you need explicit instead of just using them. Because the code around is not showed it is not clear what is your exact need further but it will be likely the code can be put in classes, functions etc. If it is a template you could think about an explicit set() function to send the variable data to the templates and extract() it there.
edit:
In addition based on the information first set your error_reporting to E_ALL and set the display_errors to 1. So you get all errors since the information you placed in your updated question gives indications that a missing variable is used as a constant which should raise errors all over the place.
I'm trying to use PHP variables into echoed file, and couldn't get where is a trouble at first using that script:
$head = new mod_head("head.php");
$id="ASDSSgdfsfsdfS";
echo $head;
mod_head class:
class mod_head
{
private $out="";
function __construct($arg)
{
$this->out=$this->parts($arg);
}
public function __toString()
{
return $this->out;
}
private function parts($file)
{
return fread(#fopen(PATH . "parts/".$file, 'r'), filesize(PATH . "parts/".$file));
}
}
and the file is "head.php"
<h1><center style="background:orange; border-radius:15px;">LOGO</center></h1>
<br><?php print_r($id)?>
<div>BANNER <div>$id <?php echo $id ;?></div></div>
i dont want to create global vars, why it doest echo $id var?
First, you're going to need to parse the file, not just read it. The second problem you'll have is a scope issue. $id is outside of the scope of the parts() function. In order to return the contents of the required file instead of just printing it I've used the output control functions
Try changing your parts function to this:
private function parts($file)
{
ob_start();
require(PATH . "parts/".$file);
$output = ob_get_contents();
ob_end_clean();
return $output;
}
To fix the scope issue try changing $id="ASDSSgdfsfsdfS"; to $head->id = "ASDSSgdfsfsdfS";, then change head.php to be the following:
<h1><center style="background:orange; border-radius:15px;">LOGO</center></h1>
<br><?php print_r($this->id)?>
<div>BANNER <div>$id <?php echo $this->id ;?></div></div>
Simply reading a file with fread will not parse any PHP contained inside. Perhaps you are looking for something like:
http://php.net/manual/en/function.require.php
Using require() is basically like copying and pasting the required file directly where your require() statement is. This means that the required file would only be able to use variables that are within the scope of where the require() statement is.
Because you are reading the contents of head.php and echoing them verbatim; nowhere do you make PHP compile and run that file as code. You could do that using include('head.php'), but that would not work blindly because you also have to make sure that $id is in scope at the point you do the include.
However that's not as easy as it sounds because it is not possible to automatically "pack the whole local scope" for later use because automatically implies code inside a function, and the very act of calling that function causes the scope to change.
My thought is that you are probably using mod_head is passing in a php file so it can be used as a view, right?
if thats the case, usually there is also a way to include a variable so that way it exists in the other file, head.php.. since there isn't one, you'd need to create a method for that. and then make that fread an include or require instead.
If not, and you're simply loading the file, the other comments about fread not parsing is totally correct and you will not be able to simply access $id from the other file.
i code the following
<?php
if ($id = mysql_real_escape_string(#$_GET['pid'])
&& $uid = mysql_real_escape_string(#$_GET['file']))
echo include "foo.php";
else
echo include "bar.php";
?>
When I use the include function in conjunction with a function that's designed to output to the page (e.g., or echo include 'foo.php'), it returns the include but with a "1" after the content that has been included.
echo include "foo.php"
should be
include 'foo.php';
Note that this can also happen when using include with shorthand echo:
<?= include 'foo.php'; ?>
This will also print out the return value of 1 when used inside a script. To get rid of this you need to use the regular PHP opening tag like so:
<?php include 'foo.php'; ?>
PHP will now include the contents of the file without printing the return value.
Okey so the answers here are actually not entirely correct; in some sense even misleading.
include takes the contents of the file and places them in context. One of the more common uses is to pass variable scope around, ie. passing scoped variables in your view by including them in the handler and using include on the view. Common, but there are also other uses; you can also return inside a included file.
Say you have a file like this:
<?php return array
(
'some',
'php'
'based'
'configuration',
'file'
); # config
Doing $config = include 'example-config-above.php'; is perfectly fine and you will get the array above in the $config variable.
If you try to include a file that doesn't have a return statement then you will get 1.
Gotcha Time
You might think that include 'example-config-above.php'; is actually searching for the file in the directory where the file calling the include is located, well it is, but it's also searching for the file in various other paths and those other paths have precedence over the local path!
So if you know you had a file like the above with a return inside it, but are getting 1 and potentially something like weird PEAR errors or such, then you've likely done something like this:
// on a lot of server setups this will load a random pear class
include 'system.php'
Since it's loading a file with out a return you will get 1 instead of (in the case of our example) the configuration array we would be expecting.
Easy fix is of course:
include __DIR__.'/system.php'
That is because the include function returns 1 on success. You get, as you say, 'my name is earl1' because the code inside the included file runs first, printing 'my name is earl' and then you local echo runs printing the return value of include() which is 1.
Let the file.txt contain xxx
echo include("file.txt");
This returns,
xxx1
The '1' is the return value of the include function, denoting the success that the file is accessed. Otherwise it returns nothing, in this case parse error is thrown.
echo print "hello";
This too returns,
hello1
Same as above; '1' denoting the success,it's printed.
echo echo "hello";
print echo "hello";
Both the above cases produces an error.
Since the echo function has no return value, hence undefined.
echo echo "hello"; print echo "hello";
(1st) (2nd)
Now the second 'echo' in the both cases produces an undefined.The first 'echo' or 'print' can't take in the hello output with the undefined (produced by the second echo).
A verification:
if((print "hello")==1)
echo "hey!";
output: hellohey! ('echo' in the 2nd line can be a print, it doesn't matter)
Similarly,
if((include ("file.txt"))==1)
echo "hey!";
output: xxxhey!
Other hand,
if((echo "hello")==1)
echo "hey!";
output: an error
In the first two cases the functions (print and include) returned 1, in the third case 'echo' produces no return value (undefined) hence the third case produces an error.
Well... I am using Codeigniter(php frame work). And I encountered the same problem. What I concluded is that when we try to print/echo the include method then it prints 1 on screen and when we just simply write the include command(example given below) it will only do what it is supposed to do.
<?php include('file/path'); ?> // this works fine for me
<?= include('file/path'); ?> // this works fine but prints "1" on screen
Hope my explaination will be helful to someone
= is assigning operator
== is for checking equal to
check for php operators
I have solved it returning nothing at the end of the included file:
$data = include "data-row.php";
return $data;
Inside data-row.php:
<div>etc</div>
...
<?php return; //End of file
I found the selected answer from this thread very helpful.
Solution 1
ob_start();
include dirname( __FILE__ ) . '/my-file.php';
$my_file = ob_get_clean();
You might also find this thread about the ob_start() function very insightful.
Solution 2
Add a return statement in the file that is being included.
E.g my-file.php
<?php
echo "<p>Foo.</p>";
return;
Drawn from the answer provided by #gtamborero.
To help you understand it the way I do now, just take this oversimplification:
There should always be a return statement otherwise include will return 1 on success.
Happy coding!
M5
Use return null in foo.php and bar.php
echo substr(include("foo.php"),1,-1);