Question

我正在尝试进行API调用以创建警报，但我不确定如何正确执行。它对我来说没有多大意义所以我提供了下面的代码，但我无法理解为什么它会返回Bad Request错误。

可能是我的请求格式不正确或者其他什么，但我无法分辨，因为我之前从未做过任何与CURL有关的事。

它应该发布一个类似于：

的卷曲请求

curl -XPOST 'https://api.opsgenie.com/v1/json/alert' -d '
{
     "apiKey": "eb243592-faa2-4ba2-a551q-1afdf565c889",
      "message" : "WebServer3 is down",
      "teams" : ["operations", "developers"]
 }'

但它不起作用。

致电功能：

OpsGenie.createAlert("1b9ccd31-966a-47be-92f2-ea589afbca8e", "Testing", null, null, new string[] { "ops_team" }, null, null);

最后，它返回一个错误的请求。我不知道我是如何输入数据或其他任何东西的，这样任何帮助都会受到高度赞赏。

public static void createAlert(string api, string message, string description, string entity, string[] teams, string user, string[] tags)
    {
        var request = WebRequest.Create(new Uri("https://api.opsgenie.com/v1/json/alert"));
        string json = "{";
        if (api != null)
            json = json + "'apiKey': '" + api + "'";
        if (message != null)
            json = json + ", 'message': '" + message + "'";
        if (description != null)
            json = json + ", 'description': '" + description + "'";
        if (entity != null)
            json = json + ", 'entity': '" + entity + "'";
        if (teams != null)
        {
            json = json + ", 'teams': '['" + string.Join(",", teams) + "']'";
        }
        if (user != null)
            json = json + ", 'user': '" + user + "'";
        if (tags != null)
            json = json + ", 'tags': '" + tags.ToString() + "'";
        json = json + "}";
        Console.WriteLine(json);
        request.Method = "POST";
        try
        {
            using (var streamWriter = new StreamWriter(request.GetRequestStream()))
            {
                streamWriter.Write(json);
                streamWriter.Flush();
                streamWriter.Close();
            }


            var httpResponse = (HttpWebResponse)request.GetResponse();
            using (var streamReader = new StreamReader(stream: httpResponse.GetResponseStream()))
            {
                var result = streamReader.ReadToEnd();
                dynamic obj = JsonConvert.DeserializeObject(result);
                var messageFromServer = obj.error.message;
                Console.WriteLine(messageFromServer);

            }
        }
        catch (WebException e)
        {
            if (e.Status == WebExceptionStatus.ProtocolError)
            {
                Console.WriteLine("Status Code : {0}", ((HttpWebResponse)e.Response).StatusCode);
                Console.WriteLine("Status Description : {0}", ((HttpWebResponse)e.Response).StatusDescription);
            }
            else
            {
                Console.WriteLine(e.Message);
            }

        }
    }

Answer 1

将WebClient用于此类任务可能更容易。它有许多辅助方法，可以很容易地下载和上传字符串等内容。

此外，不要尝试通过连接字符串来构建JSON有效内容，只需使用Newtonsoft.Json。

我重构了您使用WebClient和JsonConvert的方法。现在它变得更简单了！我将调试保留在原位，但您可以在测试后删除控制台记录行：

public static void CreateAlert(string api, string message, string description, string entity, string[] teams,
    string user, string[] tags)
{
    // Serialize the data to JSON
    var postData = new
    {
        apiKey = api,
        message,
        teams
    };
    var json = JsonConvert.SerializeObject(postData);

    // Set up a client
    var client = new WebClient();
    client.Headers.Add("Content-Type", "application/json");

    try
    {
        var response = client.UploadString("https://api.opsgenie.com/v1/json/alert", json);
        Console.WriteLine("Success!");
        Console.WriteLine(response);
    }
    catch (WebException wex)
    {
        using (var stream = wex.Response.GetResponseStream())
        using (var reader = new StreamReader(stream))
        {
            // OpsGenie returns JSON responses for errors
            var deserializedResponse = JsonConvert.DeserializeObject<IDictionary<string, object>>(reader.ReadToEnd());
            Console.WriteLine(deserializedResponse["error"]);
        }
    }
}

Answer 2

由于API v1现已弃用，因此上述解决方案应如下所示：

public static void CreateAlert(string api, string message,
string description, string entity, string[] teams,string user, string[] tags)
{
 // Serialize the data to JSON
    var postData = new
    {
        apiKey = api,
        message,
        teams
    };
    var json = JsonConvert.SerializeObject(postData);

    // Set up a client
    var client = new WebClient();
    client.Headers.Add("Content-Type", "application/json");
    client.Headers.Add("Authorization", $"GenieKey {api}");
    try
    {
        var response = client.UploadString("https://api.opsgenie.com/v2/alerts", json);
        Console.WriteLine("Success!");
        Console.WriteLine(response);
    }
    catch (WebException wex)
    {
        using (var stream = wex.Response.GetResponseStream())
        using (var reader = new StreamReader(stream))
        {
            // OpsGenie returns JSON responses for errors
            var deserializedResponse = JsonConvert.DeserializeObject<IDictionary<string, object>>(reader.ReadToEnd());
            Console.WriteLine(deserializedResponse["error"]);
        }
    }
}

如何向OpsGenie发出POST请求？

2 个答案: