Question

Mthod: Post Link: www.link.com Headers: 1: appKey="ABC" 2: Content-Type="text/csv" 如何编写C＃脚本以通过发布发送数据？

Answer 1

您可以执行以下操作：

void Start()
{
    StartCoroutine(PostCrt());
}

IEnumerator PostCrt()
{
    WWWForm form = new WWWForm();
    form.AddField("appKey", "ABC");
    form.AddField("Content-Type", "text/csv");

    using (UnityWebRequest www = UnityWebRequest.Post("www.link.com", form))
    {
        yield return www.SendWebRequest();

        if (www.isNetworkError || www.isHttpError)
        {
            Debug.Log(www.error);
        }
        else
        {
            Debug.Log("Post Request Complete!");
        }
    }
}

别忘了调用“使用UnityEngine.Networking;”。才能使用UnityWebRequest。

编码愉快！

Answer 2

您可以使用System.Net.HTTP命名空间：）

HttpClient client = new client client.PostAsync("url",data)

在https://msdn.microsoft.com/en-us/library/hh138190(v=vs.118).aspx上引用

Answer 3

public void Request()
{
    try
    {
        string url = "www.link.com";

        var request = UnityWebRequest.Post(url, "");
        request.SetRequestHeader("Content-Type", "application/json");
        request.SetRequestHeader("Accept", "text/csv");
        request.SetRequestHeader("appKey", "ABC");
        StartCoroutine(onResponse(request));
    }
    catch (Exception e) { Debug.Log("ERROR : " + e.Message); }
}
private IEnumerator onResponse(UnityWebRequest req)
{

    yield return req.SendWebRequest();
    if (req.isError)
      Debug.Log("Network error has occured: " + req.GetResponseHeader(""));
    else
        Debug.Log("Success "+req.downloadHandler.text );
        byte[] results = req.downloadHandler.data;
    Debug.Log("Second Success");
    // Some code after success

}

此代码对我有用...

如何使用UnityWebRequest在Unity中编写HTTP发布请求？

3 个答案: