VanillaJS ajax wrapper.

vanillajs-ajax.js

// http://stackoverflow.com/questions/8567114/how-to-make-an-ajax-call-without-jquery/18078705#18078705

var ajax = {};
ajax.xhr = function() {
    try {
        return new ActiveXObject('Msxml2.XMLHTTP')
    } catch (e1) {
        try {
            return new ActiveXObject('Microsoft.XMLHTTP')
        } catch (e2) {
            return new XMLHttpRequest()
        }
    }
};

ajax.send = function(url, callback, method, data, sync) {
    var xhr = ajax.xhr();
    xhr.open(method, url, sync);
    xhr.onreadystatechange = function() {
        if (xhr.readyState == 4) {
            callback(xhr.responseText)
        }
    };
    if (method == 'POST') {
        xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
    }
    xhr.send(data)
};

ajax.get = function(url, data, callback, sync) {
    var query = [];
    for (var key in data) {
        query.push(encodeURIComponent(key) + '=' + encodeURIComponent(data[key]));
    }
    ajax.send(url + '?' + query.join('&'), callback, 'GET', null, sync)
};

ajax.post = function(url, data, callback, sync) {
    var query = [];
    for (var key in data) {
        query.push(encodeURIComponent(key) + '=' + encodeURIComponent(data[key]));
    }
    ajax.send(url, callback, 'POST', query.join('&'), sync)
};