在 XMLHttpRequest 之后打开页面

Question

我向带有参数的页面发出了 XMLHttpRequest post，在本例中为 {file.nodeRef}。现在，我想打开相同的 URL 但使用 post 参数以便能够访问它们。如何打开页面？

我的 XMLHttpRequest 的代码如下：

    var csrf_token = Alfresco.util.CSRFPolicy.getToken();
    var http = new XMLHttpRequest();
    var url = "hdp/ws/my-new-page";

    var params = "file={"+file.nodeRef+"}";
    http.open("POST", url, true);
    //Send the proper header information along with the request
    http.setRequestHeader("Alfresco-CSRFToken", csrf_token);
    http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    http.setRequestHeader("Content-length", params.length);
    http.setRequestHeader("Connection", "close");

    http.onreadystatechange = function() {//Call a function when the state changes.
        if(http.readyState == 4 && http.status == 200) {
            alert(http.responseText);
        }
    }
    http.send(params);

Answer 1

你为什么不尝试使用 ajaxjquery？

$.ajax({
    type: "POST",
    url: "hdp/ws/my-new-page",
    beforeSend: function (request)
            {
                request.setRequestHeader("Alfresco-CSRFToken", csrf_token);
                request.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
                request.setRequestHeader("Content-length", params.length);
                request.setRequestHeader("Connection", "close");
            },
    data: "file={"+file.nodeRef+"}",
    dataType: "json",
    success: function(data) {
       window.location.href = data.redirect;       
    }
});