ajax simple js variable to php

Question

I've seen many js var to php question asked but none seem to help me as its either to complicated or doesn't fit my requirements. I'm looking to transfer three variable in js called title, body and author. Body being a very long string therefor I cannot place in URL. I am looking for the simplest Ajax code possible that follows this: js var already defined -> ajax -> php var Here is the code I've got so far:

<script>
var title;
var body;
var author;
function post(){
    title = document.getElementById("title").value;
    body = document.getElementById("body").value;
    author = document.getElementById("author").value;
}
    //Insert ajax code here
<?php


$title;
$body;
$author;
$post = "INSERT INTO post (title, body, author) VALUES ($title,$body,$author)";
$sql = mysqli_query($conn,$post);
?>
</script>

Thank you in advance!

"//Insert ajax code here" creates so many questions, it should not be there.... — Rayon
– Rayon, Commented Nov 24, 2016 at 7:15

Nidhi Barhate · Accepted Answer · 2016-11-24 07:28:05Z

3

You can do this way simple

<script>
var title = document.getElementById("title").value;
var body = document.getElementById("body").value;
var author = document.getElementById("author").value;

$.ajax({
    type: 'POST',
    url: 'insert.php',                
    data: {title:title,body:body,author:author},
 success: function(data) {

 }  
})
</script>

inside, insert.php file

<?php
$title = $_POST['title'];
$body = $_POST['body'];
$author = $_POST['author'];
$post = "INSERT INTO post (title, body, author) VALUES ($title,$body,$author)";
$sql = mysqli_query($conn,$post);
?>

answered Nov 24, 2016 at 7:28

Nidhi Barhate

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5 Comments

Yoan Poulmarc'k Over a year ago

Thanks to this my JS function post() works but the Ajax doesn't seem to work as it doesn't update the database. I've tried to put a JS alert inside insert.php to test but it doesn't activate. Thank you for help!

Nidhi Barhate Over a year ago

you can't put a JS alert inside insert.php, you can print $_POST value, it will get your values. your ajax

Nidhi Barhate Over a year ago

sorry, it is unanswered, you can't put a JS alert inside insert.php,your ajax is not working, I think your script not added, please add it "<script type="text/javascript" src="code.jquery.com/jquery-2.2.4.min.js"></script>" and you can print $_POST value, it will get your values. I hope it help.

Yoan Poulmarc'k Over a year ago

Sorry I explained this incorrectly. I put an alert inside a script in the insert.php and have put an alert in the function(data) in ajaz which activetes so I do not know what is wrong with code> It would probably be connection between both documents. Thank you!

Chetan Panchal Over a year ago

@NidhiBarhate Gud Keep it up

Aron · Accepted Answer · 2016-11-24 07:29:22Z

0

$.post('url', {title: title, body: body, author: author})

Then use $_POST global variable to get the data you wanted.

$title = $_POST['title'];
$body = $_POST['body'];
$author = $_POST['author'];

More info on ajax post shorthand here:
https://api.jquery.com/jquery.post/

answered Nov 24, 2016 at 7:29

Aron

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