将jQuery转换为等效的JavaScript代码

Question

I know how to use jQuery well, but I don't know so much pure JavaScript.

This is my jQuery code:

$(document).ready(function() {
    $.get('http://jsonip.com/', function(r){
        var ip_address = r.ip;
        my_function(ip_address);
    });
    function my_function(ip_address){
        var url = "Url_to my server hosted on a different domain";
        var data = {number:"1286", ip: ip_address};
        $.ajax({
            url: url, 
            type: "POST", 
            dataType: 'json', 
            crossDomain: true, 
            data: {data: data}, 
            success: function (data) {console.log(JSON.stringify(data));}, 
            error: function (xhr, error) {console.log("There was an error and data was not posted.");}});}
});

What it does: it is pasted in any website, then it picks any visitors ip address and send that as JSON to my server as variable data.

Problem: the code is working perfectly okay in some sites but not all the sites due to jQuery dependency. And I want to remove this and use pure JavaScript.

I am getting great answers but CORS is not working, there failing them. I am using different domains since the site we are sending data to is hosted on another server.

Answer 1

As mentioned in my commment above, you do not need the first ajax request as you can get this information from the request headers (PHP Example below) from your AJAX request.

To make sure that your website(s) have jQuery loaded you can run a check in your script and load it in dynamically. Using some code from this answer . See below for an example:

// Anonymous "self-invoking" function
(function() {
    // Load the script
    var script = document.createElement("SCRIPT");
    script.src = 'https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js';
    script.type = 'text/javascript';
    document.getElementsByTagName("head")[0].appendChild(script);

    // Poll for jQuery to come into existance
    var checkReady = function(callback) {
        if (window.jQuery) {
            callback(jQuery);
        }
        else {
            window.setTimeout(function() { checkReady(callback); }, 100);
        }
    };

    // Start polling...
    checkReady(function($) {
        var url = "Url_to my server hosted on a different domain";
        var data = {number:"1286", ip: ip_address};
        $.ajax({
            url: url, 
            type: "POST", 
            dataType: 'json', 
            crossDomain: true, 
            data: {data: data}, 
            success: function (data) {console.log(JSON.stringify(data));}, 
            error: function (xhr, error) {console.log("There was an error and data was not posted.");
        });
    });
})();

To get the IP Address from your ajax request: (PHP) Source

if (!empty($_SERVER['HTTP_CLIENT_IP'])) {
    $ip = $_SERVER['HTTP_CLIENT_IP'];
} elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) {
    $ip = $_SERVER['HTTP_X_FORWARDED_FOR'];
} else {
    $ip = $_SERVER['REMOTE_ADDR'];
} 

Answer 2

The annoying part is that you need to do a cross-domain POST to send your data. There's a W3C standard for this called Cross-Origin Resource Sharing (CORS). Check out this tutorial for more info.

You'll want to put this at the bottom of the page. Different browsers handle ready state change events differently, so let's just avoid them.

<script>
    // Once the JSONP script loads, it will call this function with its payload.
    function getip(ipJson) {
        var method = 'POST';
        var url = 'URL of your server';

        // The XMLHTTPRequest is the standard way to do AJAX. Try to use CORS.
        var xhr = new XMLHttpRequest();

        if ("withCredentials" in xhr) {
            // XHR for Chrome/Firefox/Opera/Safari.
            xhr.open(method, url, true);
        } else if (typeof XDomainRequest != "undefined") {
            // XDomainRequest for IE.
            xhr = new XDomainRequest();
            xhr.open(method, url);
        }

        // Create your request body. It has to be form encoded. I'm not sure
        // where you get `number` from, so I've hard-coded it.
        xhr.setRequestHeader("Content-type","application/x-www-form-urlencoded");
        xhr.send('number=1286&ip=' + ipJson.ip);
    }
</script>

<!-- Get the IP using JSONP so we can skip implementing that. -->
<script type="application/javascript" src="http://www.telize.com/jsonip?callback=getip"></script>

This probably only works in modern browsers, but it should work in most modern browsers.

Answer 3

Replace $.ready(function...) with document.addEventListener('DOMContentLoaded', function..., false) .

Replace $.ajax with XMLHttpRequest :

var xhr = new XMLHttpRequest();

//var data = {number:"1286", ip: ip_address};
var data = new FormData();
data.append("number", "1286");
data.append("ip", ip_address); // As stated by Scriptable in the question comments, your server should be able to get it from the request headers.

xhr.onload = function() { console.log(JSON.stringify(this.response)); };

xhr.onerror = function() { console.log("There was an error and data was not posted.") };

xhr.open("POST", "URL to your server");
xhr.send(data);