NETWORK_ERR：Android同步模式下的XMLHttpRequest异常101

Question

我需要以同步模式向服务器发送请求，所以我设置了async = false，但我总是收到NETWORK_ERR：XMLHttpRequest例外101.这是我的代码，有人知道如何改进它来解决这个问题？

var request = new XMLHttpRequest();
request.open("GET", 'ip-address ', false);
request.onreadystatechange = function () {
    if (request.readyState == 4) {
        if (request.status == 200 || request.status == 0) {
            var data = JSON.parse(request.responseText);
            var object = "";
            var string = "";
            var find = false;
            window.localStorage.setItem(data.id, JSON.stringify(data));
            $.each(data.img, function (i, item) {
                if (data.img[i].principal) {
                    string = data.img[i].path;

                }

            });
            if (string == "") string = data.img[0].path;
            var splitt = string.substring(2);
            window.localStorage.setItem("img" + data.id, splitt);
            if ($("#img" + (l + 1)).attr("src") === 'ip-address' + splitt) {
                find = true;
            } else {



                object='<img id="img'+(l+1)+'" src="ip-address'+splitt+'" name="'+data.id+'"></img>';                if (!trovata) {
                    if (flag) {
                        $("#" + (l + 1)).empty();
                        $("#" + (l + 1)).html(object);
                        if (l == 0) {
                            $("#" + (l + 1)).flip({
                                direction: 'tb'
                            });
                        } else {
                            switch (l % 2) {
                            case (1):
                                $("#" + (l + 1)).flip({
                                    direction: 'lr'
                                });
                                break;
                            case (0):
                                $("# " + (l + 1)).flip({
                                    direction: 'rl'
                                });
                                break;
                            }

                        }
                    }
                }
            }
        }
    }
    request.send();