I want to submit a form without reloading the page and without OK message

Question

I want your help with the following issue:

I've tried to submit a form without reloading my page, but without success. I've read other relative questions but i can't do it.

I want the following: When a user clicks on the button, then the data should be sent to server without displaying an OK messsage. Can you help me? Here is my code:

index.html

<html>
    <head>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
    </head>
    <body>
    <p align="center">Example</p>
    <table align="center" width="730" border="0">
    <tr>
    <td align="center">
    <div>
      <table class="blueTable" style="float: left">
        <thead>
    <tr>
     <th colspan="1"><u>Menu</u></th>
    </tr>
    </thead>
    <tbody>
    <tr>
    <td><input type="button" value="New" id="new" onclick="new1();" class="button12"/></td></tr>
    <tr>
    <td><input type="button" value="Load" id="load" onclick="load2()" class="button"/></td></tr>
    <tr>
    <td><form name="SaveGame" id="SaveGame" method="post" action="http://127.0.0.1/PHP/mine2.php" enctype="multipart/form-data">
    <input type="submit" value="Save" id="save" onclick="save3()" class="button"/>
    </form>
    </body>
    </html>

mine2.php

<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("127.0.0.1", "root", "", "mysql3");
// Check connection
if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}

// Attempt insert query execution
$sql = "INSERT INTO components (user_id, book_id, game_id, site_id) VALUES ('6', '6', '6', '6')";
if(mysqli_query($link, $sql)){
    } else{
    }

// Close connection
mysqli_close($link);
?>

You need javascript for that. You already seem to be using javascript in your form as you have some `onclick()` functions. You should show the relevant javascript here as well. — jeroen, Aug 30 '18 at 06:57
Hi, you use the JS function save3(), can you provide this code? — Sfili_81, Aug 30 '18 at 06:58
Possible duplicate of [Submit form without page reloading](https://stackoverflow.com/questions/2866063/submit-form-without-page-reloading) — Ahmed Sunny, Aug 30 '18 at 07:24

Zain Farooq · Accepted Answer · 2018-08-30T07:17:27.633

Change your html file, like this:

<html>
    <head>
    <script
    src="https://code.jquery.com/jquery-3.3.1.min.js"
    integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="
    crossorigin="anonymous"></script>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
    </head>
    <body>
    <p align="center">Example</p>
    <table align="center" width="730" border="0">
    <tr>
    <td align="center">
    <div>
      <table class="blueTable" style="float: left">
        <thead>
    <tr>
     <th colspan="1"><u>Menu</u></th>
    </tr>
    </thead>
    <tbody>
    <tr>
    <td><input type="button" value="New" id="new" onclick="new1();" class="button12"/></td></tr>
    <tr>
    <td><input type="button" value="Load" id="load" onclick="load2()" class="button"/></td></tr>

    <button  id="save" onclick="save3()" class="button">Save</button><!--This button will perform action-->


    <script>



       function save3(){
        var value = 123;//dummy variable to show you its functionality
        $.ajax({
               type: "POST",
               url: "http://127.0.0.1/PHP/mine2.php",
               data: {value: value}, //that value comes from above which I have initialized
               success: function(data)
               {
                   console.log(data); //it will show response from php file in your console
               }



        });
     }
    </script>

    </body>
    </html>

While in php file mine2.php this is how you can access value which is coming from ajax request

<?php
 if(isset($_POST['value']))
  {
    $value = $_POST['value'];
  }
 ?>

thank you so much. You really helped me! – Stelios Aug 30 '18 at 07:16 — Stelios, Aug 30 '18 at 07:16
I am happy that I could help you. Cheers!!! – Zain Farooq Aug 30 '18 at 07:16 — Zain Farooq, Aug 30 '18 at 07:16

score 1 · Answer 2 · answered Aug 30 '18 at 07:00

Change your html file, like this:

<html>
    <head>
    <script
    src="https://code.jquery.com/jquery-3.3.1.min.js"
    integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="
    crossorigin="anonymous"></script>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
    </head>
    <body>
    <p align="center">Example</p>
    <table align="center" width="730" border="0">
    <tr>
    <td align="center">
    <div>
      <table class="blueTable" style="float: left">
        <thead>
    <tr>
     <th colspan="1"><u>Menu</u></th>
    </tr>
    </thead>
    <tbody>
    <tr>
    <td><input type="button" value="New" id="new" onclick="new1();" class="button12"/></td></tr>
    <tr>
    <td><input type="button" value="Load" id="load" onclick="load2()" class="button"/></td></tr>
    <tr>
    <td><form name="SaveGame" id="SaveGame" method="post" action="http://127.0.0.1/PHP/mine2.php" enctype="multipart/form-data">
    <input type="submit" value="Save" id="save" onclick="save3()" class="button"/>
    </form>

    <script>
        $("#SaveGame").submit(function(e) {


    var form = $(this);
    var url = form.attr('action');

        $.ajax({
               type: "POST",
               url: url,
               data: form.serialize(), // serializes the form's elements.
               success: function(data)
               {
                   alert(data); // show response from the php script.
               }
             });

        e.preventDefault(); // avoid to execute the actual submit of the form.
    });
    </script>

    </body>
    </html>

If u add some inputs u can get their values with $_POST, in your php script.

:) Just check for more info, about AJAX :) it's pretty simple :) — MorganFreeFarm, Aug 30 '18 at 07:13

I want to submit a form without reloading the page and without OK message

2 Answers2