PHP include make the whole page appear? [closed] - php

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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?

Answer 1

"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.

Answer 2

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.

Answer 3

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';
?>

Answer 4

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?

Answer 5

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;

Answer 6

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;