Send data to php file using ajax

Question

I have a trouble receiving data sent using ajax in a php file. It seems that ajax sends data, cause I get a success message, but php page shows me an error "Notice: Undefined index: ab".

Here is my jquery code for sending data using AJAX:

$('#download-btn').click(function(){
    var ab="abc";
    $.ajax({
        url:'generate_url.php',
        data:{'ab':ab},
        type:'post',
        success:function(data) {alert("sent");},
        error: function(request, status, error) {alert(request,status);}
    }); 
}

And that's how I extract data in a generate_url.php:

<?php
    $a = $_POST['ab'];
    echo $a;
?>

Thanks in advance for your help!

It looks like it should work. What does var_dump($_POST); show? — Barmar
– Barmar, Commented Dec 8, 2016 at 20:30
How are you seeing what's in the PHP page? The success function doesn't display data. — Barmar
– Barmar, Commented Dec 8, 2016 at 20:32
success:function(data) { alert("sent...next see the data..."); alert(data); }, is one way, but you need to find your javascript console. what browser are you using? — WEBjuju
– WEBjuju, Commented Dec 8, 2016 at 20:34
@WEBjuju I know that's how he could see it, but I wonder what he's actually doing. My suspicion is that he's looking at a different page, and expecting the variable to stick around. — Barmar
– Barmar, Commented Dec 8, 2016 at 20:38

Junius L · Accepted Answer · 2016-12-09 06:03:22Z

1

You are missing the ); from your code. For simple POST I'll advice you to use $.post (there's nothing special about $.post, it's a shorthand way of using $.ajax for POST requests.

$('#download-btn').on('click', function(){
    var ab="abc";
    $.post('generate_url.php', {'ab':ab}, function(data){
        console.log(data);
    }); 
});

edited Dec 9, 2016 at 6:03

answered Dec 9, 2016 at 5:57

Junius L

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No name · Accepted Answer · 2016-12-09 05:37:13Z

1

You are missing ); at the end of the ajax code.

answered Dec 9, 2016 at 5:37

No name

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Ram Pukar · Accepted Answer · 2016-12-09 06:02:48Z

0

<script type="text/javascript">
        $(document).ready(function($) {
            $('#download-btn').click(function(){
                var ab='abc';
                $.ajax({
                    url: 'generate_url.php',
                    type: 'POST',
                    data: {'ab':ab}
                })
                .done(function(respose) {
                    //todo
                })
            });
        })
    </script>

edited Dec 9, 2016 at 6:02

answered Dec 9, 2016 at 5:54

Ram Pukar

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1 Comment

Junius L Over a year ago

why document instead of button click?

Junius L · Accepted Answer · 2017-05-06 17:44:13Z

0

You were missing ')' in your javascript code

This will work:

$('#download-btn').click(function () {
    var ab = "abc";
    $.ajax({
        url: 'generate_url.php',
        data: {'ab': ab},
        type: 'post',
        success: function (data) {
            alert("sent");
        },
        error: function (request, status, error) {
            alert(request, status);
        }
    });
});

edited May 6, 2017 at 17:44

Junius L

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answered Dec 9, 2016 at 6:03

Dani Akash

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