How to call php file inside a form (html)

Question

I have 2 PHP files, called:

user_one.php -> returns a JSON with the info of One user (given by POST parameter)
user_uptade.php -> modify all fieds of One user (given by POST parameter) and returns a JSON with the modified data.

Then I got 1 more file called, update_user.html where I want to:

Call user_one.php
take the JSON and put the info in the form of the update_user
When User modifies his data, when he click accept button, user_update.php has to be called.

What I want to know is how can I fill this form calling the user_one.php given the $_POST['user'].

*The user_one.php has to return a "echo json" because this file is called ina mobile phone request*

user_one.php (where the json is coded and returned)

    <?php

/*
 * Following code will get single product details
 * A product is identified by product id (uid)
 */

// array for JSON response
$response = array();

// include db connect class
require_once __DIR__ . '../../db_connect.php';

// connecting to db
$db = new DB_CONNECT();
// check for post data
if (isset($_POST["user"])) {
    $uid = $_POST['user'];
    // get a product from products table
    $result = mysql_query("SELECT *FROM Usuarios WHERE Usuario = '$uid'");

    if (!empty($result)) {
        // check for empty result
        if (mysql_num_rows($result) > 0) {

            $result = mysql_fetch_array($result);

            $users = array();
            $users["name"] = $result["Nombre"];
            $users["lastname"] = $result["Apellidos"];
            $users["user"] = $result["Usuario"];
            $users["password"] = $result["Contrasena"];
            $users["birthday"] = $result["Fnacimiento"];
            $users["direction"] = $result["Direccion"];
            $users["email"] = $result["Email"];
            $users["phone"] = $result["TelfFijo"];
            $users["mobile_phone"] = $result["TelfMov"];
            $users["user_type"] = $result["TipoUsuario"]; 
            // success
            $response["success"] = 1;
            // user node
            $response["user"] = array();
            // push single users into final response array
            array_push($response["user"], $users); 
            // echoing JSON response
            echo json_encode($response);
        } else {
            // no user found
            $response["success"] = 0;
            $response["message"] = "No user found";

            // echo no users JSON
            echo json_encode($response);
        }
    } else {
        // no user found
        $response["success"] = 0;
        $response["message"] = "No user found";

        // echo no users JSON
        echo json_encode($response);
    }
} else {
    // required field is missing
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";

    // echoing JSON response
    echo json_encode($response);
}
?>

user_uptade.html (Where has to be filled the form with JSON obtained in user_one.php)

<html>
<body> 

<br>Agregue un nuevo Usuario:

<form method="post" action="user_update.php" >
Nombre:
<br>
<input type="Text" name="name" >
<br>
<br>
Apellidos:
<br>
<input type="Text" name="lastname" >
<br>
<br>
Usuario:
<br>
<input type="Text" name="user" >
<br>
<br>
Fecha nacimiento:
<br>
<input type="Text" name="bday" >
<br>
<br>
Direccion:
<br>
<input type="Text" name="direccion" >
<br>
<br>
TelFijo:
<br>
<input type="Text" name="tfijo" >
<br>
<br>
TelMov:
<br>
<input type="Text" name="tmov" >
<br>
<br>
Email:
<br>
<input type="Text" name="email" >
<br>
<br>
Password:<br>
<input type="password" name="password" ><br>

<input type="Submit" name="enviar" value="Agregar">

</form>

</body>
</html>

**Danger**: You are using [an **obsolete** database API](http://stackoverflow.com/q/12859942/19068) and should use a [modern replacement](http://php.net/manual/en/mysqlinfo.api.choosing.php). You are also **vulnerable to [SQL injection attacks](http://bobby-tables.com/)** that a modern API would make it easier to [defend](http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php) yourself from. — Quentin, Aug 30 '13 at 12:04
first of all change user_uptade.html to user_uptade.PHP.. You can use PHP inside HTML if your file is saved with .php extension — , Aug 30 '13 at 12:11
Consider reading about **Ajax** which allows you to send a request with `javascript` to a `php` script. — Ofir Baruch, Aug 30 '13 at 13:54

score 0 · Answer 1 · answered Aug 30 '13 at 12:11

You cannot 'call' PHP files from HTML. You can only do that with PHP files after using include or require to load the other PHP file you want to use. The only thing, that PHP can actually do is post a form to server, which can be done like this:

<form type="post" action="http://yoururl.com/your_desired_php_file.php">
...
</form>

UPDATE: you can easily update your HTML file to PHP by changing the extension from .html to .php After that you can use require("your_filename.php"); to load the file you want to use. And then you can call any function or access any global variables from that file.

How to call php file inside a form (html)

1 Answers1