跨站 XMLHttpRequest

If you trust the data source then maybe JSONP would be the better option. JSONP involves dynamically adding new SCRIPT elements to the page with the SRC set to another domain, with a callback set as a parameter in the query string. For example:

function getJSON(URL,success){
    var ud = 'json'+(Math.random()*100).toString().replace(/./g,'');
    window[ud]= function(o){
        success&&success(o);
    };
    document.getElementsByTagName('body')[0].appendChild((function(){
        var s = document.createElement('script');
        s.type = 'text/javascript';
        s.src = URL.replace('callback=?','callback='+ud);
        return s;
    })());
}

getJSON('http://YOUR-DOMAIN.com/script.php?dataName=john&dataAge=99&callback=?',function(data){
    var success = data.flag === 'successful';
    if(success) {
        alert('The POST to abc.com WORKED SUCCESSFULLY');
    }
});

因此，您需要托管自己的脚本，该脚本可以使用 PHP/CURL 发布到 abc.com 域，然后以 JSONP 格式输出响应:

<?php
    /* Grab the variables */
    $postURL = $_GET['posturl'];
    $postData['name'] = $_GET['dataName'];
    $postData['age'] = $_GET['dataAge'];

    /* Here, POST to abc.com */
    /* MORE INFO: http://uk3.php.net/curl & http://www.askapache.com/htaccess/sending-post-form-data-with-php-curl.html */

    /* Fake data (just for this example:) */
    $postResponse = 'blahblahblah';
    $postSuccess = TRUE;

    /* Once you've done that, you can output a JSONP response */
    /* Remember JSON format == 'JavaScript Object Notation' - e.g. {'foo':{'bar':'foo'}} */
    echo $_GET['callback'] . '({';
    echo "'flag':' . $postSuccess . ',";
    echo "'response':' . $postResponse . '})";

?>