say i have a main file :
<form id="form1" name="form1" action="" method="post">
<input title="Masukkan kalimat disini" type="text" />
<?php
//main.php
$a = ...;
$b = ....;
?>
and included file
<?php
//included.php
include 'main.php';
?>
certainly included.php file will load all parts of main.php, so how to include some variable, such as include $a only, and not showing the textfield
Thanks.
If you want to include just a variable (or set of variables), they need to be in their own file. Ideally, you would also put them in a different scope (such as a class), so they're not cluttering up the global namespace.
E.g.,
config.php:
<?php
class Foo {
public $a = "I am a variable";
public $b = "I am also a variable";
const $c = "I am an immutable variable!";
}
?>
main.php:
<form id="form1" name="form1" action="" method="post">
<input title="Masukkan kalimat disini" type="text" />
<?php
#require_once "config.php";
?>
included.php:
<?php
#require_once "config.php";
?>
Programming is just driving your thoughts :)
So what i want to say that your question is how you can include just some part of an included file and my answer is that you can achieve that by doing a test each time the main file is included from withing this file to see if the file is included internally or not and you can be more precise in a way that you split your main file into block which are loaded due suitable variable
Take a look for this workaround and hope you will understand what i mean
Supposing we have the main file named main.php contains that contents
<?php
echo 'I am a java programmer';
echo 'I know also PHP very well';
echo 'When the jquery is my preferred toast !';
?>
now i have three external files that will include that file each file is specific for one of this 3 programming language
So i will create my 3 files in this way :
File : java.php
<?php
$iamjavadevelopper = 1;
include_once("main.php");
?>
File : phpfav.php
<?php
$iamphpdevelopper = 1;
include_once("main.php");
?>
File : jquery.php
<?php
$iamjquerydevelopper = 1;
include_once("main.php");
?>
and my main.php will be coded in this way
<?php
if(isset($iamjavadevelopper))
echo 'I am a java programmer';
if(isset($iamphpdevelopper))
echo 'I know also PHP very well';
if(isset($iamjquerydevelopper))
echo 'When the jquery is my preferred toast !';
?>
By this way each one of our three external files will show just a part of the included file :)
Hope that was clear and helpfull :) Any help needed i am just ready :)
Providing you are including the page, you will beable to use your set variables which are set on a different page (if that makes sense)
Global.inc.php
<?php
$Var_1 = "This will work";
$Var_2 = "Another set variable.";
?>
index.php
<?php
include "Global.inc.php";
echo $var_1; // This will output 'This will work'
?>
As you have included a file, all the contents will be usable; You just have to work with the variables that you want. There will be no performance hits for including a 300 line page, to use less than 50 lines of code for example.
Update:
Show within HTML Structures:
Inputs:
<input type='text' name='Name' value='<?php echo $Var_1; ?>'>
Text Box:
<textarea><?php echo $Var_1; ?></textarea>
Related
So I have an HTML file that is a basic form. Say
<form method="POST" action="test1.php">
Name: <input type="text" name="name" id="name" />
<input type="submit" value="Submit" />
</form>
Now I have that execute my test1.php file which goes as follows:
<?php
$name = $_POST['name'];
?>
So all it is doing is getting the value from the HTML form. Now I have a second PHP file test2.php that needs to get the value from the first test1.php file $name and output it via an echo statement.
I'm new to PHP and am fine with using one PHP file to output values from an HTML form, but I don't know how to approach a second one.
I'm aware that you can use the include statement to carry over the variables and their values, but it didn't seem to work in my instance. I'm almost positive the issue is that I don't have test2.php actually being executed. And I don't know how to approach that. Any help is appreciated.
EDIT: This is all I want test2.php to do. $name has to be the same value as retrieved from the HTML form in test1.php
<?php
echo $name;
?>
Just use include your file test2.php inside test1.php:
<?php
$name = $_POST['name'];
include('test2.php');
?>
Whenever you include a file using either include() or require(), the file always gets "executed", so maybe there's something else wrong in you code.
EDIT
test1.php:
<?php
$name = $_POST['name'];
include('test2.php');
?>
test2.php:
<?php
echo $name;
?>
I would suggest starting a session in test1, assign the var $name to the session and then picking the session up on test2.
test1.php
session_start();
$name = filter_var($_POST['name'],FILTER_SANITIZE_STRING);
$_SESSION['test1']['name'] = $name;
test2.php
session_start();
$name = $_SESSION['test1']['name'];
Try that.
#bloodyKnuckles, Your assignment was left open ended because there are a number of ways you can accomplish what you're looking to do.
include('test2.php');
Is one option. Another would be
header("Location: test2.php");
exit();
Using the header option, $_SESSION variable would work. If test1.php had any HTML output, you could consider an AJAX call or urlencode() your string.
test1.php
echo 'Click Me';
test2.php
$name = $_GET['name'];
I'm guessing your assignment task was designed to demonstrate various ways to pass values from script to script.
Good Luck
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Okay, my code is simple:
<?php include 'formvalidation.php';
echo $name; ?>
I want to make only the $name appear, but the whole 'formvalidation.php' shows up.
How can I fix that?
"I want that after I pressed the button, the function I had written will be excute to check the input, and then if the input is right, it'll redirect to another page, which can display the input."
If you wish to redirect after a form has been submitted and show a name afterwards, you will need to use sessions and a header.
Here is an example, (see comments in code) to be added inside formvalidation.php:
<?php
session_start();
// all other codes you already have
$name = "John"; // this is an example
// replace above with your POST variable
// such as $name = $_POST['name'];
$_SESSION['showname'] = $name;
header("Location: your_other_file.php");
exit; // keep this
N.B.: Make sure there is nothing else above <?php session_start(); such as your form, because this will throw a warning, stating Headers already sent...
your_other_file.php
<?php
session_start();
if(isset($_SESSION['showname'])){
$name = $_SESSION['showname'];
echo $name;
}
$name = "John"; // this is an example
Replace above in code with your POST variable.
Such as $name = $_POST['name']; as an example, since you have not provided additional code as to what your superglobal variable is for the name.
You can use session feature to call the variable instead of including a page, because it will indeed load an entire page. Please be more specific about your goal, perhaps we can help more by knowing more.
The whole HTML content of formvalidation.php will come up on the screen (so everything outside of <?php ?> just like all the outputs (echo etc ...) of the included file will come up on the screen.
You can avoid that by removing the outputs and only putting functions and variable declarations into the included file formvalidation.php.
The content of formvalidation.php could eg. be:
<?php
$name = 'test';
?>
It's going to execute whatever's in formvalidation.php. All the include function does is basically replace that line with the contents of your included file.
So if you don't want to show what's in that file - enclose the entire file in a function, or multiple functions. This is the way your library files (that you include) should be. Nothing in that file should be outside of a function definition.
Example (it's a crappy programming practice, but for example) -
formvalidation.php was ->
<?php
echo "hola";
if ($i = 5) {
echo 15;
}
?>
<form><input type="text" value= "10"/>
<?php
echo "macho, macho man.";
?>
Everything gets shown when I include it.
formvalidation.php now ->
<?php
function validate_form() {
echo "hola";
if ($i = 5) {
echo 15;
}
?>
<form><input type="text" value= "10"/>
<?php
echo "macho, macho man.";
}
?>
Now the way I've changed the file - nothing will get printed until I call validate_form();
Does that make sense?
When you use include 'someFile.php'; it's like you are taking all the contents of that file and pasting it in the code. For example if I had:
someFile.php
echo '<h1>Hello World</h1>';
echo '<p>I like pie</p>';
then:
include 'someFile.php';
echo $name;
Is the same as:
echo '<h1>Hello World</h1>';
echo '<p>I like pie</p>';
echo $name;
As others have said, that's what include does. I assume that you define $name inside the formvalidation.php file and that is why you are including it.
The right solution here would be to seperate the code in formvalidation.php into a function/class/file which does the data processing and another which creates the output. The you include/call only the first in the situation where you don't want the output
However, it is possible to capture the output and then discard it, using output buffering:
<?php
ob_start(); //Start capturing output
include 'formvalidation.php';
ob_end_clean(); // Stop capturing output and discard whatever was captured
echo $name;
?>
That said I would really not recommend this. The recommended way to fix this is to seperate your back-end code from your output generating code and only call what you actually need in any given situation.
Edit
An example of how to seperate the issues properly could be the following.
Lets assume your current formvalidation.php is something like:
<?php
$name = $_REQUEST['name'];
if( '' == $name ){ ?>
<p class="error">You must supply a name</p>
<form action="foo.php">
<input name="name" />
<button type="submit">Submit</button>
</form><?php
} else {
echo '<p>Form valid!</p>';
}
In a new file validator.php you could do:
<?php
class MyValidator{
public $name;
function read_data(){
$this->name = $_REQUEST['name'];
}
function validate(){
if( '' == $name ){ ?>
<p class="error">You must supply a name</p>
<form action="foo.php">
<input name="name" />
<button type="submit">Submit</button>
</form><?php
} else {
echo '<p>Form valid!</p>';
}
}
}
Then you change you formvalidtion.php to
<?php
require_once 'validator.php';
$val = new MyValidator();
$val->read_data();
$val->validate();
While the file you have above becomes
<?php
require_once 'validator.php';
$val = new MyValidator();
$val->read_data();
echo $val->name;
When I post my form data:
<form action="layouts.php" method="post">
<input type="text" name="bgcolor">
<input type="submit" value="Submit" align="middle">
</form>
to a php page, "layouts.php", it returns the string, as expected.:
$bgcolor = $_POST['bgcolor'];
echo $bgcolor; //returns "red"
echo gettype($bgcolor); // returns "string"
But when I include "layouts.php" in another page it returns NULL.
<?php
include("php/layouts.php");
echo $bgcolor; //
echo gettype($bgcolor); //returns "NULL"
?>
How do I pass the variable to another page?
You'll have to use a session to have variables float around in between files like that.
It's quite simple to setup. In the beginning of each PHP file, you add this code to begin a session:
session_start();
You can store variables inside of a session like this:
$_SESSION['foo'] = 'bar';
And you can reference them across pages (make sure you run session_start() on all of the pages which will use sessions).
layouts.php
<?php
session_start();
$bgcolor = $_POST['bgcolor'];
$_SESSION['bgcolor'] = $bgcolor;
?>
new.php
<?php
session_start();
echo $_SESSION['bgcolor'];
?>
Give the form two action="" and see what happens. I just tried that on a script of mine and it worked fine.
A more proper way to solve this might exist, but that is an option.
How to pass variables from a php file to another while it is not html inputs ,just i have a link refer to the other file and i want to pass variables or values to it
Example:
File1.php
<?php
$name='OdO';
echo "<a href='File2.php'>Go To File2</a>";
?>
File2.php
<?php
echo $name;
?>
Use sessions to store any small value that needs to persist over several requests.
File1.php:
session_start();
$_SESSION['var'] = 'foo';
File2.php:
session_start();
$var = $_SESSION['var']; // $var becomes 'foo'
Try to use sessions. Or you can send a GET parameters.
You can use URLs to pass the value too.
like
index.php?id=1&value=certain
and access it later like
$id = $_GET['id'];
$value = $_GET['value'];
However, POST might be much reliable. Sessions/Cookies and database might be used to make the values globally available.
Here's one (bad) solution, using output buffering:
File 1:
<?php
$name = 'OdO';
echo 'Go To File2';
?>
File 2:
<?php
ob_start();
include 'File1.php';
ob_end_clean();
echo $name;
?>
I have to show a page from my php script based on certain conditions. I have an if condition and am doing an "include" if the condition is satisfied.
if(condition here){
include "myFile.php?id='$someVar'";
}
Now the problem is the server has a file "myFile.php" but I want to make a call to this file with an argument (id) and the value of "id" will change with each call.
Can someone please tell me how to achieve this?
Thanks.
Imagine the include as what it is: A copy & paste of the contents of the included PHP file which will then be interpreted. There is no scope change at all, so you can still access $someVar in the included file directly (even though you might consider a class based structure where you pass $someVar as a parameter or refer to a few global variables).
You could do something like this to achieve the effect you are after:
$_GET['id']=$somevar;
include('myFile.php');
However, it sounds like you are using this include like some kind of function call (you mention calling it repeatedly with different arguments).
In this case, why not turn it into a regular function, included once and called multiple times?
An include is just like a code insertion. You get in your included code the exact same variables you have in your base code. So you can do this in your main file :
<?
if ($condition == true)
{
$id = 12345;
include 'myFile.php';
}
?>
And in "myFile.php" :
<?
echo 'My id is : ' . $id . '!';
?>
This will output :
My id is 12345 !
If you are going to write this include manually in the PHP file - the answer of Daff is perfect.
Anyway, if you need to do what was the initial question, here is a small simple function to achieve that:
<?php
// Include php file from string with GET parameters
function include_get($phpinclude)
{
// find ? if available
$pos_incl = strpos($phpinclude, '?');
if ($pos_incl !== FALSE)
{
// divide the string in two part, before ? and after
// after ? - the query string
$qry_string = substr($phpinclude, $pos_incl+1);
// before ? - the real name of the file to be included
$phpinclude = substr($phpinclude, 0, $pos_incl);
// transform to array with & as divisor
$arr_qstr = explode('&',$qry_string);
// in $arr_qstr you should have a result like this:
// ('id=123', 'active=no', ...)
foreach ($arr_qstr as $param_value) {
// for each element in above array, split to variable name and its value
list($qstr_name, $qstr_value) = explode('=', $param_value);
// $qstr_name will hold the name of the variable we need - 'id', 'active', ...
// $qstr_value - the corresponding value
// $$qstr_name - this construction creates variable variable
// this means from variable $qstr_name = 'id', adding another $ sign in front you will receive variable $id
// the second iteration will give you variable $active and so on
$$qstr_name = $qstr_value;
}
}
// now it's time to include the real php file
// all necessary variables are already defined and will be in the same scope of included file
include($phpinclude);
}
?>
I'm using this variable variable construction very often.
The simplest way to do this is like this
index.php
<?php $active = 'home'; include 'second.php'; ?>
second.php
<?php echo $active; ?>
You can share variables since you are including 2 files by using "include"
In the file you include, wrap the html in a function.
<?php function($myVar) {?>
<div>
<?php echo $myVar; ?>
</div>
<?php } ?>
In the file where you want it to be included, include the file and then call the function with the parameters you want.
I know this has been a while, however, Iam wondering whether the best way to handle this would be to utilize the be session variable(s)
In your myFile.php you'd have
<?php
$MySomeVAR = $_SESSION['SomeVar'];
?>
And in the calling file
<?php
session_start();
$_SESSION['SomeVar'] = $SomeVAR;
include('myFile.php');
echo $MySomeVAR;
?>
Would this circumvent the "suggested" need to Functionize the whole process?
I have ran into this when doing ajax forms where I include multiple field sets. Taking for example an employment application. I start out with one professional reference set and I have a button that says "Add More". This does an ajax call with a $count parameter to include the input set again (name, contact, phone.. etc) This works fine on first page call as I do something like:
<?php
include('references.php');`
?>
User presses a button that makes an ajax call ajax('references.php?count=1'); Then inside the references.php file I have something like:
<?php
$count = isset($_GET['count']) ? $_GET['count'] : 0;
?>
I also have other dynamic includes like this throughout the site that pass parameters. The problem happens when the user presses submit and there is a form error. So now to not duplicate code to include those extra field sets that where dynamically included, i created a function that will setup the include with the appropriate GET params.
<?php
function include_get_params($file) {
$parts = explode('?', $file);
if (isset($parts[1])) {
parse_str($parts[1], $output);
foreach ($output as $key => $value) {
$_GET[$key] = $value;
}
}
include($parts[0]);
}
?>
The function checks for query params, and automatically adds them to the $_GET variable. This has worked pretty good for my use cases.
Here is an example on the form page when called:
<?php
// We check for a total of 12
for ($i=0; $i<12; $i++) {
if (isset($_POST['references_name_'.$i]) && !empty($_POST['references_name_'.$i])) {
include_get_params(DIR .'references.php?count='. $i);
} else {
break;
}
}
?>
Just another example of including GET params dynamically to accommodate certain use cases. Hope this helps. Please note this code isn't in its complete state but this should be enough to get anyone started pretty good for their use case.
You can use $GLOBALS to solve this issue as well.
$myvar = "Hey";
include ("test.php");
echo $GLOBALS["myvar"];
If anyone else is on this question, when using include('somepath.php'); and that file contains a function, the var must be declared there as well. The inclusion of $var=$var; won't always work. Try running these:
one.php:
<?php
$vars = array('stack','exchange','.com');
include('two.php'); /*----- "paste" contents of two.php */
testFunction(); /*----- execute imported function */
?>
two.php:
<?php
function testFunction(){
global $vars; /*----- vars declared inside func! */
echo $vars[0].$vars[1].$vars[2];
}
?>
Try this also
we can have a function inside the included file then we can call the function with parametrs.
our file for include is test.php
<?php
function testWithParams($param1, $param2, $moreParam = ''){
echo $param1;
}
then we can include the file and call the function with our parameters as a variables or directly
index.php
<?php
include('test.php');
$var1 = 'Hi how are you?';
$var2 = [1,2,3,4,5];
testWithParams($var1, $var2);
Your question is not very clear, but if you want to include the php file (add the source of that page to yours), you just have to do following :
if(condition){
$someVar=someValue;
include "myFile.php";
}
As long as the variable is named $someVar in the myFile.php
I was in the same situation and I needed to include a page by sending some parameters... But in reality what I wanted to do is to redirect the page... if is the case for you, the code is:
<?php
header("Location: http://localhost/planner/layout.php?page=dashboard");
exit();
?>