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我正在用 C# 编写一个简单的控制台应用程序来与 Slack.com 进行通信。我正在通过他们的 WebApi 进行此操作。目前我知道如何发布消息（带有附件、彩色、链接、用户等）并将文件发送到服务器。

如果您以正常方式发送文件（键入文本框左侧的“上传文件”），该文件将显示在主对话窗口中。但是你怎么能通过 WebAPI 实现同样的事情呢？目前发送文件后，我可以在右侧面板中看到它，其中仅列出了所有文件。

还有第二个问题：是否可以更改消息中文本的颜色（附件中没有“行”）？

这是通过https://slack.com/api/files.upload发送文件后的响应

{
    "ok": true,
    "file": {
        "id": "F04EX4***",
        "created": 1429279966,
        "timestamp": 1429279966,
        "name": "Testing.txt",
        "title": "Testing",
        "mimetype": "text\/plain",
        "filetype": "text",
        "pretty_type": "Plain Text",
        "user": "U*********",
        "editable": true,
        "size": 28,
        "mode": "snippet",
        "is_external": false,
        "external_type": "",
        "is_public": false,
        "public_url_shared": false,
        "url": "https:\/\/slack-files.com\/files-pub\/T*******\
/testing.txt",
        "url_download": "https:\/\/slack-files.com\/files-pub\/T************\
/download\/testing.txt",
        "url_private": "https:\/\/files.slack.com\/files-pri\/T*******\
/testing.txt",
        "url_private_download": "https:\/\/files.slack.com\/files-pri\/T**********\
 /download\/testing.txt",
        "permalink": "https:\/\/******.slack.com\/files\/******\
/F0******\/testing.txt",
        "permalink_public": "https:\/\/slack-files.com\/********",
        "edit_link": "https:\/\/******.slack.com\/files\/****\/F******\/testing.txt\/edit",
        "preview": "This is a test file number2.",
        "preview_highlight": "<div class=\
"sssh-code\"><div class=\"sssh-line\"><pre>This is a test file number2.<\/pre><\/div>\n
<\/div>",
        "lines": 1,
        "lines_more": 0,
        "channels": [],
        "groups": [],
        "ims": [],
        "comments_count": 0
    }
}

抱歉，我不知道如何很好地格式化它。

“is_external”和“is_public”都是假的，也许这就是原因 - 但我怎样才能将它们设置为“真”？

->> 感谢您的编辑！:) 这是我正在使用的整个功能：

public static void SlackSendFile()
    {
        FileStream str = File.OpenRead(@"C:\Users\Eru\Desktop\Testing.txt");
        byte[] fBytes = new byte[str.Length];
        str.Read(fBytes, 0, fBytes.Length);
        str.Close();

        var webClient = new WebClient();
        string boundary = "------------------------" + DateTime.Now.Ticks.ToString("x");
        webClient.Headers.Add("Content-Type", "multipart/form-data; boundary=" + boundary);
        var fileData = webClient.Encoding.GetString(fBytes);
        var package = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"file\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n{3}\r\n--{0}--\r\n", boundary, "Testing.txt", "multipart/form-data", fileData);

        var nfile = webClient.Encoding.GetBytes(package);
        string url = "https://slack.com/api/files.upload?token=" + Token + "&content=" + nfile + "&channels=[" + Channel + "]";

        byte[] resp = webClient.UploadData(url, "POST", nfile);

        var k = System.Text.Encoding.Default.GetString(resp);
        Console.WriteLine(k);
        Console.ReadKey();
    }

EDIT1：在这一行：

 byte[] resp = webClient.UploadData(url, "POST", nfile);

网址是：

 https://slack.com/api/files.upload?token=*********&content=System.Byte[]&channels=[%23*****]

然后我传递字节数组。

编辑：

我已经解决了这个问题。问题是频道应该是频道的ID，而不是频道的名称......愚蠢的错误:(

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3 回答 3

2

如果在 channel 属性之后添加 pretty=1 ，效果会更好。

例子：

string url = "https://slack.com/api/files.upload?token=" + token + "&content=" + nfile + "&channels=[" + Channel + "]&pretty=1"
于 2017-05-17T20:26:22.083 回答
2

嗨，这里是 RestSharp 的一个干净的例子

public void UploadFile(string token, string filePath, string channel)
{
    var client = new RestClient("https://slack.com");

    var request = new RestRequest("api/files.upload", Method.POST);
    request.AddParameter("token", token);
    request.AddParameter("channels", channel);

    var fileInfo = new FileInfo(filePath);
    request.AddFile("file", File.ReadAllBytes(filePath), fileInfo.Name, contentType:"multipart/form-data");

    //Execute the request
    var response = client.Execute(request);
    var content = response.Content;
}
于 2017-09-15T09:31:44.707 回答
1

我不得不稍微更改您的代码以使其适用于仍在寻找此功能的任何人。我需要将文件数组转换为 ASCII 编码字符串，然后将详细信息提取到参数中。

    public static void SlackSendFile(string token, string channelId, string filepath)
    {
        FileStream str = File.OpenRead(filepath);
        byte[] fBytes = new byte[str.Length];
        str.Read(fBytes, 0, fBytes.Length);
        str.Close();

        var webClient = new WebClient();
        string boundary = "------------------------" + DateTime.Now.Ticks.ToString("x");
        webClient.Headers.Add("Content-Type", "multipart/form-data; boundary=" + boundary);
        var fileData = webClient.Encoding.GetString(fBytes);
        var package = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"file\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n{3}\r\n--{0}--\r\n", boundary, "Testing.txt", "multipart/form-data", fileData);

        var nfile = webClient.Encoding.GetBytes(package);
        var encodedfile = System.Text.Encoding.ASCII.GetString(nfile);
        string url = "https://slack.com/api/files.upload?token=" + token + "&content=" + encodedfile + "&channels=" + channelId + "";

        byte[] resp = webClient.UploadData(url, "POST", nfile);

        var k = System.Text.Encoding.Default.GetString(resp);
        Console.WriteLine(k);
    }
于 2017-10-31T20:17:39.920 回答