MYSQL AJAX submit form without leave page

Question

So here is my scripts:

my Ajax:

$('#ether-wallet').submit(function (e) {
    e.preventDefault();
    var name = $('#etherwallet').val();
    var dataString = 'name=' + name;
    $.ajax({
        type: 'POST',
        data: dataString,
        url: 'Database/wallet.inc.php',
        success: function (data) {
            alert(data);
        }
    });
});

and I am trying to submit the form without leaving the page, Here is my PHP:

    <form id="ether-wallet" class="wallet-form" method="POST">
        <span class="span">Ether Wallet:  </span>
        <input type="text" name="etherwallet" placeholder="Press to lock" id="etherwallet" class="inputapi" value=""/>
        <button id="etherlock" class="lockbutton" type="submit" name="etherlock">Lock</button>
    </form>

and here's where I want to send the data

all of this is inside the php

session_start();

if(isset($_POST['ether-wallet'])) {

include_once 'dhb.inc.php';

$etherwallet = mysqli_real_escape_string($conn, $_POST['name']);
$idbridge1 = mysqli_real_escape_string($conn, $_SESSION['u_id']);
if (empty($etherwallet)) {
    header("Location: ../signup.php?signup=success1");
    exit();
} else {
    if (strlen($etherwallet) != 42 ) {
        header("Location: ../signup.php?signup=success2");
        exit();
    } else {
        $sql = "UPDATE users SET eth_wallet = '$etherwallet' WHERE user_id = '$idbridge1'";
        mysqli_query($conn, $sql);
        header("Location: ../signup.php?signup=success3");
        exit();
    }
}

}

Thanks for the help boys!!

Answer 1

Use input type="button" onClick="function Name"

Or button type="button"

Dont Use input type="submit"...

May help...

Regards, Vishal Sharma

Answer 2

If you are using the button type "Submit", you may also bind the "submit" event in the form (instead of bind "click" event in the button). You also have to return "false" value in your function after AJAX call statement.

Answer 3

Just change the ajax code as below.

$('#eether-wallet').submit(function (e) {
    e.preventDefault();
    var name = $('#etherwallet').val();
   //var dataString = 'name=' + name;
    $.ajax({
        type: 'POST',
        data: new FormData(this),
        url: 'Database/wallet.inc.php',
        success: function (data) {
            alert(data);
        },
      error: function(jqXHR, textStatus, errorThrown)
    {
      alert(textStatus);
    }
    });
});

Now we are calling the ajax inside form submit event. so that e.preventDefault() will prevent the postback of form.

Answer 4

There are couple of issues in your code, such as:

• You're accessing user submitted data using $_POST['etherlock'], which is wrong. See this statement in your AJAX request, var dataString = 'name=' + name;, you should be accessing user submitted data using $_POST['name'].

• Your UPDATE query is wrong, it should be like this:

  $sql = "UPDATE users SET eth_wallet = '$etherwallet' WHERE user_id = '$idbridge1'";
  

---

Sidenote: Learn about prepared statement because right now your query is susceptible to SQL injection attack. And also see how you can prevent SQL injection in PHP.