To make an Ajax request using jQuery you can do this by the following code.
HTML:
<form id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
<!-- The result of the search will be rendered inside this div -->
<div id="result"></div>
JavaScript:
Method 1
/* Get from elements values */
var values = $(this).serialize();
$.ajax({
url: "test.php",
type: "post",
data: values ,
success: function (response) {
// You will get response from your PHP page (what you echo or print)
},
error: function(jqXHR, textStatus, errorThrown) {
console.log(textStatus, errorThrown);
}
});
Method 2
/* Attach a submit handler to the form */
$("#foo").submit(function(event) {
var ajaxRequest;
/* Stop form from submitting normally */
event.preventDefault();
/* Clear result div*/
$("#result").html('');
/* Get from elements values */
var values = $(this).serialize();
/* Send the data using post and put the results in a div. */
/* I am not aborting the previous request, because it's an
asynchronous request, meaning once it's sent it's out
there. But in case you want to abort it you can do it
by abort(). jQuery Ajax methods return an XMLHttpRequest
object, so you can just use abort(). */
ajaxRequest= $.ajax({
url: "test.php",
type: "post",
data: values
});
/* Request can be aborted by ajaxRequest.abort() */
ajaxRequest.done(function (response, textStatus, jqXHR){
// Show successfully for submit message
$("#result").html('Submitted successfully');
});
/* On failure of request this function will be called */
ajaxRequest.fail(function (){
// Show error
$("#result").html('There is error while submit');
});
The .success()
, .error()
, and .complete()
callbacks are deprecated as of jQuery 1.8. To prepare your code for their eventual removal, use .done()
, .fail()
, and .always()
instead.
MDN: abort()
. If the request has been sent already, this method will abort the request.
So we have successfully send an Ajax request, and now it's time to grab data to server.
PHP
As we make a POST request in an Ajax call (type: "post"
), we can now grab data using either $_REQUEST
or $_POST
:
$bar = $_POST['bar']
You can also see what you get in the POST request by simply either. BTW, make sure that $_POST
is set. Otherwise you will get an error.
var_dump($_POST);
// Or
print_r($_POST);
And you are inserting a value into the database. Make sure you are sensitizing or escaping All requests (whether you made a GET or POST) properly before making the query. The best would be using prepared statements.
And if you want to return any data back to the page, you can do it by just echoing that data like below.
// 1. Without JSON
echo "Hello, this is one"
// 2. By JSON. Then here is where I want to send a value back to the success of the Ajax below
echo json_encode(array('returned_val' => 'yoho'));
And then you can get it like:
ajaxRequest.done(function (response){
alert(response);
});
There are a couple of shorthand methods. You can use the below code. It does the same work.
var ajaxRequest= $.post("test.php", values, function(data) {
alert(data);
})
.fail(function() {
alert("error");
})
.always(function() {
alert("finished");
});