Question

我有一个基于结果的AJAX调用 - 我发送另一个电话。

uploadDocument = function (doc1, doc2) {
    $.ajax({
        type: "POST",
        url: "/API/UploadDocs/addDocument",
        data: doc1,
        contentType: "application/json"
    }).then(function (result) {
        console.log(result);
        doc2.id=result;
        return $.ajax({
            type: "POST",
            url: "/API/UploadDocs/addDocument",
            data: doc2,
            contentType: "application/json"
        }).then(function (result) {
        });
    });
}

但是我收到Illegal invocation错误，我做错了什么？

Answer 1

当通过AJAX传递的数据出现错误时，会出现非法调用错误

检查doc1和doc2的类型。也尝试将def keypress(self, event): print(time.time(), "KEYPRESS {char!r} keysym='{keysym}' keycode={keycode} keysym_num={keysym_num} state={state}".format(**vars(event))) def keyrelease(self, event): print(time.time(), "KEYPRESS {char!r} keysym='{keysym}' keycode={keycode} keysym_num={keysym_num} state={state}".format(**vars(event)))传递给ajax。

Answer 2

你正在承诺错误的链接！当您返回承诺时，您必须继续调用您正在解决的承诺的then。

阅读链接部分：https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Promise/then

uploadDocument = function (doc1, doc2) {
    $.ajax({
        type: "POST",
        url: "/API/UploadDocs/addDocument",
        data: doc1,
        contentType: "application/json"
    }).then(function (result) {
        console.log(result);
        doc2.id=result;
        return $.ajax({
            type: "POST",
            url: "/API/UploadDocs/addDocument",
            data: doc2,
            contentType: "application/json"
        });
    }).then(function (result) {
      //Continue here
    });
}

嵌套的JavaScript承诺

2 个答案: