Message:

I'm trying to use an HTML dropdown form in order to let the user choose which table from the database to display. This part works fine, but I want them to be able to add to whichever table they choose. I've created the form for it but once they hit submit on that form it unloads the table they selected and fails to add the data to the table.

This is the code for the main page; body.php:

<!DOCTYPE html>

<html xmlns="http://www.w3.org/1999/xhtml">

<head>

<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />

<title>Contact Form - PHP/MySQL Demo Code</title>

</head>

<style>

table, th, td {

  border:1px solid black;

}

</style>

<body>



<form action="<?php echo htmlspecialchars($_SERVER['PHP_SELF']) ?>" method="post">

    <div>

        <label for="dbselect">Select Table:</label><br>

        <select name="dbselect" id="dbselect">

            <option value=""><-- Choose Table --></option>

            <option value="messages">messages</option>

            <option value="messages2">messages2</option>

        </select>

    </div>

    <div>

        <button type="submit">Select</button>

    </div>   

</form>



<?php

    session_start();

    

    include 'get.php'; 

    $_SESSION["dbselect"] = filter_input(INPUT_POST, 'dbselect', FILTER_SANITIZE_STRING);                               

    if ($_SESSION["dbselect"] == "messages") {

        dbMessages("*");

        include 'Forms/contact_form.php';

    }

    if ($_SESSION["dbselect"] == "messages2") {

        dbMessages2("*");

    }                    

    

?>

</body>

This is the code that actually prints the tables, which I believe works fine; get.php:

<?php

function display_data($data) {

    $output = "<table>";

    foreach($data as $key => $var) {

        //$output .= '<tr>';

        if($key===0) {

            $output .= '<tr>';

            foreach($var as $col => $val) {

                $output .= "<td>" . $col . '</td>';

            }

            $output .= '</tr>';

            foreach($var as $col => $val) {

                $output .= '<td>' . $val . '</td>';

            }

            $output .= '</tr>';

        }

        else {

            $output .= '<tr>';

            foreach($var as $col => $val) {

                $output .= '<td>' . $val . '</td>';

            }

            $output .= '</tr>';

        }

    }

    $output .= '</table>';

    echo $output;

}



function dbMessages($selector) {

    include 'db.php';

    $query = "SELECT $selector FROM messages";

    $res = mysqli_query($conn,$query);

    display_data($res);

}



function dbMessages2($selector) {

    include 'db.php';

    $query = "SELECT $selector FROM messages2";

    $res = mysqli_query($conn,$query);

    display_data($res);

}

And this is the code that is supposed to display the form and upload the data into the table; contact_form.php:

<?php



include 'db.php';

if ((isset($_POST["Submit"])) && (!empty($_POST["txtName"])) && (!empty($_POST["txtPhone"])) && (!empty($_POST["txtEmail"])) && (!empty($_POST["txtMessage"]))) {

    $txtName = $_POST['txtName'];

    $txtEmail = $_POST['txtEmail'];

    $txtPhone = $_POST['txtPhone'];

    $txtMessage = $_POST['txtMessage'];

    $query = "insert into messages(ID, name, email, phone, message) values (NULL, 

                '$txtName', '$txtEmail', '$txtPhone', '$txtMessage')";

    

    $res = mysqli_query($conn , $query);

    

    mysqli_close($conn);

    unset($_POST);

    echo "<meta http-equiv='refresh' content='0'>";

}



echo '

<fieldset>

<legend>Contact Form</legend>

<form name="frmContact" method="post" action="index.php?send=1" target="_self">

<p>

<label for="Name">Name </label>

<input type="text" name="txtName" id="txtName" >

</p>

<p>

<label for="email">Email</label>

<input type="text" name="txtEmail" id="txtEmail">

</p>

<p>

<label for="phone">Phone</label>

<input type="text" name="txtPhone" id="txtPhone">

</p>

<p>

<label for="message">Message</label>

<textarea name="txtMessage" id="txtMessage"></textarea>

</p>

<p>&nbsp;</p>

<p>

<input type="submit" name="Submit" id="Submit" value="Submit">

</p>

</form>

</fieldset>

'

?>

This is the observed process:

I select which table to display using dropdown menu and hit select.

Selected table successfully appears with its form.

I enter data into form and hit submit.

The dropdown menu returns to it's default state and the table unloads.

Entered data fails to be added to table.

This is the expected process:

I select which table to display using dropdown menu and hit select.

Selected table successfully appears with its form.

I enter data into form and hit submit.

Selected table stays on the page and submitted data is successfully added.

It should be noted that the contact_form.php file is only for the first table, messages, and not for messages2.

My Answer:
To achieve the expected process, you need to make a few changes in your code. Here's how you can modify it:

1. Modify the form action in contact_form.php to post the form data back to the same page (body.php) instead of index.php:

php

2. Update the form submission check in contact_form.php to check for the form submission and the selected table:

php

if ((isset($_POST["Submit"])) && (!empty($_POST["txtName"])) && (!empty($_POST["txtPhone"])) && (!empty($_POST["txtEmail"])) && (!empty($_POST["txtMessage"])) && $_SESSION["dbselect"] == "messages") {

3. Remove the line that unsets $_POST in contact_form.php as it is not necessary:

php

unset($_POST);

With these changes, the form data should be submitted to the same page (body.php) and the selected table should remain displayed after the form submission. Make sure to also handle form submissions for the messages2 table if needed.

John Davidson

html - PHP How to choose which table to display and then keep that table displayed after a form submission

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