CODEWITHSUNDEEP
javascriptphphtml
jvnna
jvnna

Reputation: 61

Pass HTML form data in a PHP file to ONE OF 3 other PHP files

I'm building a website with a simple login interface. The user can log in as one of 3 different users types, A B and C using 1 form. The code inside the form for handling this is:

                    <input type="radio" name="user_type" value="a" id="a">
                    <label for="a">A</label>

                    <input type="radio" name="user_type" value="b" id="b">
                    <label for="b">B</label>

                    <input type="radio" name="user_type" value="c" id="c">
                    <label for="c">C</label>

Depending on the user type, the login page should redirect the user to one of three PHP files on click of the form submit button. I know in PHP you can use

<form method="post" action="otherFile.php">

In the login page for leading to one other file. And use 

<?php
if(isset($_POST['submit']))
{
    $firstname = $_POST['firstname'];
    $lastname = $_POST['lastname'];
    echo $firstname . "<br/>";
    echo $lastname;
}
?>

In otherFile.php to retrieve form data. But I want to do this for one of three PHP files, depending on the user's selection of user_type, and 

<form method="post" action="otherFile.php|otherFile2.php|otherFile3.php">

clearly doesn't work.

Upvotes: 1

Views: 996

Answers (3)

GROVER.
GROVER.

Reputation: 4378

You can use $_SESSION to store data for use in other areas of your site.

To simplify it, you could do something like:

session_start();
$_SESSION['firstname'] = $_POST['firstname']; // store firstname, lastname as sessions
$_SESSION['lastname'] = $_POST['lastname'];

And then, inside of another page, you can call these sessions to output:

$firstname = $_SESSION['firstname'];
print("$firstname"); // will print the first name.

I hope this helps :)

EDIT: Before you create a $_SESSION and try to print it into another page, create a core.php page and include it into both page1.php & page2.php

Inside of your core.php script, you need to start up a session, like so:

session_start();

Once you have included the core.php file into both of your pages, try printing out the desired $_SESSION variable again.

Upvotes: 1

Alisson Linneker
Alisson Linneker

Reputation: 292

Using javascript:

<script src="https://code.jquery.com/jquery-1.11.3.js"></script>
<script type="text/javascript">
jQuery("#myForm").submit(function( event ) {
    var radio = jQuery('input[name=user_type]:checked', this).val();
    if(radio == 'a')
        this.action = 'otherFile.php';
    else if(radio == 'b')
        this.action = 'otherFile2.php';
    else if(radio == 'c')
        this.action = 'otherFile3.php';
    else
        return false;
});
</script>

You form:

<form method="post" action="" id="myForm">
    <input type="radio" name="user_type" value="a" id="a">
    <label for="a">A</label>

    <input type="radio" name="user_type" value="b" id="b">
    <label for="b">B</label>

    <input type="radio" name="user_type" value="c" id="c">
    <label for="c">C</label>

    <input type="submit" value="Submit">
</form>

Upvotes: 0

sloaxleak
sloaxleak

Reputation: 106

I'm not sure if this is you want, but you can create a single php, and if is a value, require it using include or require functions.

For example, processor.php

    <? 
$user_type = $_POST["user_type"];
switch($user_type) {
    case "a":
        require_once 'otherFile.php';
        break;
    case "b":
        require_once 'otherFile2.php';
        break;
    case "c":
        require_once 'otherFile3.php';
        break;
    default:
        die("Not allowed");
        break;
}
?>

Upvotes: 0

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