Input into database using PHP and Javascript

Question

I am new to javascript and am trying to use the following code to input a value into the database using php and javascript.

When the checkbox is checked, all the code works, except for the "load('../check.php?checked=Y');". I want that to input the "Y/N" into the database. I thought "load" would do the trick but it didn't. What other function would I use?

I appreciate the help!

if (checkbox.checked)

{

$('#btn').show();$('#hide').show();$('#count').show();$('#message').show();$('#notify').show();$('#responds').show();

load('../check.php?checked=Y');



}else{
  $('#btn').hide();$('#hide').hide();$('#count').hide();$('#message').hide();$('#notify').hide();$('#responds').hide();

load('../check.php?checked=N');}

}

you want [.ajax](http://api.jquery.com/jQuery.ajax/) or [.post](http://api.jquery.com/jQuery.post/), [.load](http://api.jquery.com/load) just loads in data from that url it doesnt send (except in the GET variables). You can do it with load with the GET url variables but its better to use the proper functions. — Patrick Evans, Jan 27 '14 at 22:21
Just the mere question title suggests it's going to be a bad question. Why do you start learning by mixing 3 technologies you don't know? Just learn one by one! — Tomas, Jan 27 '14 at 22:22

score 2 · Accepted Answer · answered Jan 27 '14 at 22:21

2

You need to look into $.ajax() or the shorthand counterpart $.post()

http://api.jquery.com/jquery.post/

$.ajax({
    type: "POST",
    url: "load.php",
    data: {"checked" : "Y"},
    success: function( d ){
        console.log( d ); // returned by script on server
    }
});

This assuming you have some script in load.php on your server to actually handle the data and insert into the database.

answered Jan 27 '14 at 22:21

Patrick Moore

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1

Thanks! This did the trick. I was not aware that ".load" did not post data. – Sam J. Jan 27 '14 at 22:29

score 1 · Answer 2 · answered Jan 27 '14 at 22:21

You can use $.ajax to communicate between JS and the server (PHP) without refreshing the page:

$.ajax({
            url: 'check.php?checked=N',
            type: 'get',
            success: function(data) {
             alert('great success!')
            }
});

Hope this helps!