Xamarin Forms 将图像和变量上传到网页答案

【问题标题】：Xamarin Forms Upload image and variable to webpageXamarin Forms 将图像和变量上传到网页
【发布时间】：2020-01-02 06:22:42
【问题描述】：

我有问题。我想将图像和变量上传到我的网页，所以我创建了以下代码：

using (var formContent = new MultipartFormDataContent("NKdKd9Yk"))
{
    formContent.Headers.ContentType.MediaType = "multipart/form-data";

    StringContent UserIdContent = new StringContent(App.User.Id.ToString(), Encoding.UTF8, "application/x-www-form-urlencoded");
    formContent.Add(UserIdContent, "id");

    StringContent CreatedImageContent = new StringContent(CreatedImage, Encoding.UTF8, "binary/octet-streOpenWriteam");
    formContent.Add(CreatedImageContent , "image");

    using (var client = new HttpClient())
    {
        try
        {
            // 4.. Execute the MultipartPostMethod
            var message = await client.PostAsync(url, formContent);
            // 5.a Receive the response
            var result = await message.Content.ReadAsStringAsync();

            if (result == "Success")
            {
                App.Current.MainPage = new SideMenuItems();
            }
        }
        catch (Exception ex)
        {
            // Do what you want if it fails.
            throw ex;
        }
    }
}

现在这两个变量都被服务器接收了，但 image 变量是设备上图像的路径，而不是图像本身。

我做错了什么？

【问题讨论】：

标签： c# xamarin xamarin.forms xamarin.android xamarin.ios

【解决方案1】：

使用路径获取图片流，使用流内容。

using (var formContent = new MultipartFormDataContent("NKdKd9Yk")) {
    formContent.Headers.ContentType.MediaType = "multipart/form-data";

    var id = App.User.Id.ToString();
    StringContent UserIdContent = new StringContent(id, Encoding.UTF8, "application/x-www-form-urlencoded");
    formContent.Add(UserIdContent, "id");

    FileStream fs = System.IO.File.OpenRead(CreatedImage);
    formContent.Add(new StreamContent(fs), "image");

    using (var client = new HttpClient()) {
        try {
            // 4.. Execute the MultipartPostMethod
            var message = await client.PostAsync(url, formContent);
            // 5.a Receive the response
            var result = await message.Content.ReadAsStringAsync();

            if (result == "Success") {
                App.Current.MainPage = new SideMenuItems();
            }
        } catch (Exception ex) {
            // Do what you want if it fails.
            throw ex;
        }
    }
}

【讨论】：