How to pass data using jQuery/ Ajax to a URL

Question

How to pass the fetched data using $.get() from a URL to a particular URL with query_string that can be retrieved using $_GET() in PHP. I'm using following but not working. I want to pass the fetched data into following parameter http://example.com/data?data=FETECHED_DATA

<?php
$ip_address=$_SERVER['REMOTE_ADDR'];    
echo '<script>
$.get( "http://ipinfo.io/'.$ip_address.'/org", function( data ) {
$.ajax({
  type: "GET",
  url: "http://example.com/data",
  data: data,
  success: success,
  dataType: dataType
});
});
</script>';?>

Well, don't recommended PHP to fetch and send data

You're using $.get and wrapping $.ajax inside. This is not intended to be used this way. You don't need $.ajax, because $.get is a shortcut for this. And please indent your code properly. Thanks. — cezar
– cezar, Commented Jul 5, 2015 at 9:36
data: "data="+data, - should pass your received data as get request but as people said - you need to rewrite it — metamorph_online
– metamorph_online, Commented Jul 5, 2015 at 9:43
Last thing first: please indent the code in your question (add 4 spaces for every block). This makes the code more readable. The chances are higher someone will help you, if the people can read your code with ease. — cezar
– cezar, Commented Jul 5, 2015 at 9:43
@cezar, nothing wrong with doing a $.ajax inside a $.get if that's what is required. That snippet is trying to get the "org" of an IP from ipinfo.io and using it in an ajax call to example.com/data with the end result processed by a function (not shown) called success — Jaromanda X
– Jaromanda X, Commented Jul 5, 2015 at 9:46

Jaromanda X · Accepted Answer · 2015-07-05 10:09:51Z

1

I believe this will do what you want

<?php
$ip_address=$_SERVER['REMOTE_ADDR'];    
echo '<script>
$.get("http://ipinfo.io/'.$ip_address.'/org", function(data) {
    $.ajax({
        type: "GET",
        url: "http://example.com/data",
        data: {data : data},
        success: success,
        dataType: dataType
    });
});
</script>';?>

the only change is data: data changed to data: {data: data}

that's a lot of data :p

see this cut down mock up, http://jsfiddle.net/cm8o5dk8/ if you have a decent browser that can show you the network "usage" you'll see that the mockup fails trying to GET http://example.com/data?data=AS15169+Google+Inc.%0A - the %0A comes back from ipinfo.io

edited Jul 5, 2015 at 10:09

answered Jul 5, 2015 at 9:49

Jaromanda X

1

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9 Comments

Disk01 Over a year ago

It is not passing the fetched info in the URL example.com/data?data=FETECHED_DATA

Jaromanda X Over a year ago

My apologies - jQueery changes it's behaviour every minor release, so I'm probably 1000 behaviours behind since last year

cezar Over a year ago

url: "http://example.com/" + data, ??? Or stands data in the URL for some example parameter?

Jaromanda X Over a year ago

just tested - the code I posted is SOUND and works and DOES try to get http://example.com/data?data=FETCHED_DATA - any failure is due to you not having a success function defined or the dataType defined corectly

Jaromanda X Over a year ago

@cezar - read the $.ajax documentation you linked to, find the section explaining the "data" parameter, you'll see why my code is correct

|

Disk01 · Accepted Answer · 2015-07-05 10:27:46Z

0

Thanks it worked

$.get("http://ipinfo.io/8.8.8.8/org", function(data) {
    $.ajax({
        type: "GET",
        url: "http://example.com/data",
        data: {data : data},
        success: function() {},
        dataType: 'json'
    });
});

answered Jul 5, 2015 at 10:27

Disk01

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