我必须在wordpress中通过AJAX将数据发布到另一个网站(http://123abc.com/register) .但我总是出错。这是我的jquery AJAX代码：

<script>
$(document).ready(function() {

    // process the form
    $(\'form\').submit(function(event) {

        // get the form data
        // there are many ways to get this data using jQuery (you can use the class or id also)
        var formData = {
            \'fname\'              : $(\'input[name=fname]\').val(),
            \'lname\'              : $(\'input[name=lname]\').val(),
            \'email\'              : $(\'input[name=email]\').val(),
            \'password\'           : $(\'input[name=password]\').val(),
            \'gender\'             : $(\'input[name=gender]\').val(),
            \'accent\'             : $(\'input[name=accent]\').val(),
            \'type\'               : $(\'input[name=type]\').val(),
        };

        // process the form
        $.ajax({
            type        : \'POST\', // define the type of HTTP verb we want to use (POST for our form)
            url         : \'http://123abc.com/register\', // the url where we want to POST
            data        : formData, // our data object
            dataType    : \'json\', 
            encode          : true,
            success: function(data){                
                if(data.error){
                    //show error message here
                    $(\'#name-group\').append(\'<div class="help-block">\' + data.errors + \'</div>\');
                }else{
                    //handle success part       
                    $(\'#name-group\').append(\'<div class="help-block">\' + data.message + \'</div>\');              
                }               
              },
              error: function(jqXHR, textStatus, errorThrown){
                //request error 
                $(\'#name-group\').html(\'<p>status code: \'+jqXHR.status+\'</p><p>errorThrown: \' + errorThrown + \'</p><p>jqXHR.responseText:</p><div>\'+jqXHR.responseText + \'</div>\');
                console.log(\'jqXHR:\');
                console.log(jqXHR);
                console.log(\'textStatus:\');
                console.log(textStatus);
                console.log(\'errorThrown:\');
                console.log(errorThrown);   
              }             
        });

        // stop the form from submitting the normal way and refreshing the page
        event.preventDefault();
    });

});
</script>