I have 1 html, 1 javascript file and 3 php files, purposed as below:
script.js : used to pass user-entered variables from html to calculate.php without refreshing the html page. then get the result from calculate.php and pass it to certain div in my html.
calculate.php : include parameters.php and functions.php, after that, do the calculation.
<?php
include 'parameters.php';
include 'functions.php';
call_user_func("do_something()");
?>
parameters.php: where I define variables.
<?php
$var = $_POST['data_from_js']
?>
functions.php: where I define functions and their "if" conditions, and these "if" conditions uses variables defined in parameters.php
<?php
function do_something()
{
if ($var == 100)
{
echo 200;
}
}
?>
everything works fine, except that when calculate.php called the user function "do_something", the "if" condition doesn't work, because the $var is always 0
and if I move the "if" condition from functions.php to calculate.php, like below:
functions.php:
<?php
function do_something()
{
echo 200;
}
?>
Calculate.php:
<?php
include 'parameters.php';
include 'functions.php';
if($var == 100)
{
call_user_func("do_something()");
}
?>
Then it works, the value of $var is just exactly what user entered in the html, and it returns 200 into somewhere in the html if the user enters 100 in the field.
so, variables defined in parameters.php,
can be read and get the value (by $_post method) inside calculate.php, where my javascript sends data to.
But it cannot be read inside the function which I defined in functions.php,
when included and called by calculate.php.
It confuses me a little bit, I guess it's something about synchronization, and how php works on its sequence...
my question is:
Is there a simple way I can make it work without moving the "if" condition from functions.php to calculate.php, or make my javascript send data directly into functions.php?
The reason I'd like to do this is to keep the code organized, categorized and explicit so I can know what to modify if needed. The real calculation.php is actually a little bit complicated that I want to separate functions from it.
It's my first post here so if there's something confusing please let me know, thank you guys !
This is an issue with the variable scope. Check this article from the official PHP site: http://php.net/manual/en/language.variables.scope.php
Try using global $val; at the beginning of the do_something() function, so the function knows that $var is a global version of the variable, instead of the local one to the function:
<?php
function do_something()
{
global $var;
if ($var == 100)
{
echo 200;
}
}
?>
Alternatively, and as suggested by other user, you may want to pass the value as a parameter to the function, and avoid the use of global variables (and their potential issues).
you have to pass the variable
$var = $_POST['data_from_js'];
do_something($var);
function do_something($var){
...
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 have two files that are included in my page. Like this:
// mypage.php
include 'script1.php';
include 'script2.php';
I need both of them in first, and then I need to remove one of them, something like this:
if ($var == 'one') {
// inactive script1.php file
} else {
// inactive script2.php file
}
That's hard to explain why I need to inactive one of them, I Just want to know, how can I do that? Is it possible to do unlike include?
The simple answer is no, you can't.
The expanded answer is that when you run PHP, it makes two passes. The first pass compiles the PHP into machine code. This is where includes are added. The second pass is to execute the code from the first pass. Since the compilation pass is where the include was done, there is no way to remove it at runtime.
Since you're having a function collision, here's how to get around that using objects(classes)
class Bob {
public static function samename($args) {
}
}
class Fred {
public static function samename($args) {
}
}
Note that both classes have the samename() function but they live within a different class so there's no collision. Because they are static you can call them like so
Bob::samename($somearghere);
Fred::samename($somearghere);
If you need just the output of either file you could do this
ob_start();
include('file1.php');
$file1 = ob_get_contents();
ob_start();
include('file2.php');
$file2 = ob_get_contents();
Then later if you need to call them
if ($var == 'one') {
echo $file2;
} else {
echo $file1;
}
Your only option is something like this:
if ($var == 'one') {
include('script2.php');
} else {
include('script1.php');
}
You can't 'remove' code, you can only choose to not include/execute it in the first place.
As by your comments you said its because of duplicated function names, i'm assuming you use both files elsewhere separately, but what your trying to achieve now is to merge these files together for a different reason (both files have functions/variables, etc that you need)
If your first file had a function like so my_function:
my_function() {
// code here
}
and your second file also had the same named function you can wrap an if statement around it to exclude it:
if (!function_exists('my_function')) {
my_function() {
// code here
}
}
This way the second file's function wont be available when merging the two files together but using file separately both functions will be available.
For the sake of providing options for others that come here, some solutions I've used myself on occasion...
If the files you're including, you're including for some function with a return value or some execution that doesn't need to be displayed on the page (such as mailing stuff out), and let's say you can't alter either of the target files (let's say they're somebody else's code or part of some highly integrated other piece of software that you really don't want to untangle).
A solution is to create a quick and dirty restful interface for both files to pull from them what you need from them, and then call that interface with your program, effectively bypassing the need to include them.
A worse method but if your situation is truly desperate, and is truly the route of last resort, and will only work in some cases (for example, will break on namespacing)...
$bad_idea = file_get_contents("foo.php");
eval($bad_idea);
unset($bad_idea);
Again, note, this is an option of last resort.
How do I set a piece of code to a function in WordPress and then call that function...this was my first guess, but of course this doesn't work...
Could someone suggest how I can get this to work so I can avoid redundant code. Also if I define code in a function can I call it in different .php files, or can I only call it within that file?
<?php
// This code will never change.
function $test {
echo('test this will be a long string and repeated many times.')
}
?>
<?php
echo $test;
?>
PHP function names (unlike variables) are declared without dollar signs:
function test (
To call a function, use this format:
test();
Your code example will look like this:
<?php
// This code will never change.
function test {
echo('test this will be a long string and repeated many times.')
}
?>
<?php
test();
?>
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.
I include a PHP file to the HEADER of my WordPress site, which searches through a CSV file and returns me the variable I need. I am trying to have it included in the header because it's a variable I will need throughout the site later on. If I test it out and try to echo this variable from the included script, it works fine. However, in the rest of the site, if I try to call that variable it doesn't return anything.
I know that the variable is being created because I try to echo it and it works. But when the I try to use that variable from a different script, it doesn't work. Is there some kind of code I need to pass the variable over to the rest of the site?
Variables default to function level only, you have to pass them or globalize them if you want to use them elsewhere. Depending on how your script is laid out, you might make it an object property, in which case it will be available anywhere your object is available and in all methods of that object - another option is to use global $var, but that's a bad idea and bad coding practice - another is to put it into a session using $_SESSION['myVar'] = $var and then call it the same way - yet another way is to pass it through arguments such as $database->getRows($var) and then on the other side "public function getRows ($var)", now you have $var in that function by passing it.
Make sure you global $variable the variable everytime you want to use it in a new function, or even within a new script. This will make sure that the variable is available everywhere that you need it.
3 files:
a.php:
<?php
include("c.php");
var_dump("c is ".$c . " after include()");
function incit(){
include("b.php");
var_dump("b is ".$b . " inside incit()");
}
incit();
var_dump("b is ".$b . " after incit()");
?>
b.php:
<?php
$b="bear";
?>
c.php:
<?php
$c="car";
?>
output looks like this:
string(24) "c is car after include()"
string(24) "b is bear inside incit()"
string(19) "b is after incit()"
so $b is only defined INSIDE the scope of the function while $c on the other hand is "globally" definde.
So you have to watch in what scope you are using the include.