HttpManager.PostBody

方法:

public static SendWebRequest PostBody(string url, Dictionary<string, object> param, Dictionary<string, string> headers);

说明:

发起Post请求，参数以Body形式提交

参数

返回值

类型 : SendWebRequest

字段介绍 :

代码示例:

using UnityEngine;

public class TestHttpManager : MonoBehaviour
{
   private IEnumerator Start() 
   {
       string url = "http://127.0.0.1/test";

       Dictionary<string,object> param = new Dictionary<string, object>();

       SendWebRequest request = HttpManager.PostBody(url, param, null); 
       yield return request;
       if (string.IsNullOrEmpty(request.error))
       {
           // 接口请求成功
           Debug.LogFormat("接口返回内容:{0}",request.Text);
       }
       else { 
           Debug.LogErrorFormat("接口请求失败:{0}",request.error);
       }
   } 
}