Question

我想将文件发送到带有其他参数的Web服务像这样：

C:\wamp64\www\symfony>php ../composer.phar update
Loading composer repositories with package information
Updating dependencies (including require-dev)
Package operations: 2 installs, 10 updates, 0 removals

我怎么能在C＃中做到这一点我试试这个：

param1=value1&param2=value2&param3=IMAGE_FILE

但那不起作用。
有解决方案吗？

更新：

此外，我尝试使用&param3=Convert.ToBase64String(File.ReadAllBytes(txtlogo.Text))这样的（我不知道它是正确的还是没有）

HttpContent

并且服务器不会收到HttpContent stringContent = new StringContent("param1=param1&param2=param2"); FileStream file = new FileStream(txtlogo.Text, FileMode.Open); HttpContent fileStreamContent = new StreamContent(file); using (var client = new HttpClient()) using (var formData = new MultipartFormDataContent()) { formData.Add(stringContent, "param1", "param1"); formData.Add(fileStreamContent, "file1", "file1"); var response = client.PostAsync("url", formData).Result; if (!response.IsSuccessStatusCode) { string ss = "s"; } string my = response.Content.ReadAsStringAsync().Result; } ...

的任何数据

Answer 1

尝试使用get方法（简单）

    try
{
    using (var wc = new WebClient())
    {
        string url = "https://url.com/index.php?x=1222&y=filebase64-sdfsdfsdfsdfsdfsdfsdfsdfsd";
        // Get account from accessToken
        response = wc.DownloadString(url);
    }
}
catch (Exception ee)
{
    Console.WriteLine("Coś nie tak z pobieraniem kont");
}

使用C＃发布文件

1 个答案: