将RestSharp Reqest AddParameter转换为JavaScript数据以获取发布请求

Question

任何人都知道restsharp的等同：

request.AddParameter("text/plain", body, ParameterType.RequestBody);

在JavaScript中？那是我缺少的唯一部分。

这是C＃中的工作代码：

        public static void UploadFile(string bearer)
    {
        var client = new RestClient("...");
        var request = new RestRequest(Method.POST);
        request.AddHeader("Authorization", "Bearer " + bearer);
        request.AddHeader("cache-control", "no-cache");
        var imageLenngth = new FileInfo(@"...").Length;
        request.AddHeader("Content-Type", "image/jpeg");
        request.AddHeader("Content-Length", imageLenngth.ToString());
        byte[] body = File.ReadAllBytes(@"...");
        request.AddParameter("text/plain", body, ParameterType.RequestBody);
        IRestResponse response = client.Execute(request);
    }

且未在Ajax JavaScript中进行完全工作的代码（它上传到云服务器，但错误地上传图像。将损坏的映像插入了损坏的图像，它应该应该使用太多的字节）：

    function uploadingImage(binaryImage) {
        
        var settings = {
            "async": true,
            "crossDomain": true,
            "url": "...",
            "method": "POST",
            "headers": {
                "Content-Type": "image/jpeg",
                "Authorization": "Bearer ...",
                "cache-control": "no-cache",
            },
            "data": binaryImage
        }
        $.ajax(settings).done(function (response) {
            console.log(response);
        });
    }

Answer 1

我找到了问题。提德勒帮助调查了正在发送的内容。默认情况下，传递的数据正在序列化。为了防止这种情况，我们需要明确说明不要序列化。" processData"：false做到了。这是工作代码：

    function uploadingImage(binaryImage) {
        
        var settings = {
            "async": true,
            "crossDomain": true,
            "url": "...",
            "method": "POST",
            "headers": {
                "Content-Type": "image/jpeg",
                "Authorization": "Bearer ...",
                "cache-control": "no-cache",
            },
            "processData":false,
            "data": binaryImage
        }
        $.ajax(settings).done(function (response) {
            console.log(response);
        });
    }