Denys Denys - 1 year ago 62
PHP Question

Sending data via Ajax from form

I need to update row in database by id using Ajax.

My Form:

<form id="update_ajax">
<input type="text" name="name" class="test_hide">
<input type="hidden" name="id" value="<?php echo $row['id']; ?>">
<input type="submit" value="<?php echo $row['id']; ?>" class="pure-button">

My Ajax function:

$("#update_ajax").on("submit", function () {
var id = $(this).attr('id');
var name = $(this).attr('name');
var dataall={'id' : id, 'name' : name};
type: "post",
url: "update.php",
data: dataall,
success: function (data) {

And my php file is:

if ((isset($_GET['id'])) && (isset($_GET['name']))) {
$id = $_GET['id'];
$name = $_GET['name'];
die("Not set");
$pdo = new PDO("mysql:host=localhost;dbname=test", "root", "8169x5it");
$query = $pdo->prepare("UPDATE test SET name=:name WHERE id=:id");
$query->bindParam(':name', $name);
$query->bindParam(':id', $id);

And I have the error that my
value is not set in
. And I can't understand why it's not set. Because I send it in data.

Answer Source

This is wrong:

var id = $(this).attr('id');
var name = $(this).attr('name');

$(this) is the form, so $(this).attr('id') is "update_ajax", not the value of the hidden input. And $(this).attr('name') is undefined because the form doesn't have a name=something attribute. What you really want is:

var id = $(this).find("input[name=id]").val();
var name = $(this).find("input[name=name]").val();

But you can simplify it all to:

var dataall = $(this).serialize();

serialize() will find all the inputs in the form and return a URL-encoded string containing all their values.

Finally, you either have to change the jQuery to use type: 'GET', or change the PHP to use $_POST instead of $_GET.

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