var requestData = new { lineid = 27, areaid = 22, starttime = "2023-05-07 09:54:22", endtime = "2023-06-07 09:54:22", datatype = 1, pageSize = 5, pageNumber = 1 };请求参数是这些，时间参数类型为string，unity wbgl代码怎么写

您可以使用Unity的WebRequest类来发送POST请求并传递JSON数据。可以按照以下步骤进行编写：

1. 首先，将请求参数转换为JSON格式的字符串：

string json = JsonUtility.ToJson(requestData);

2. 然后，创建一个WebRequest对象并设置请求的URL和方法：

UnityWebRequest request = UnityWebRequest.Post(url, "POST");

3. 接下来，将JSON数据添加到请求体中：

byte[] bodyRaw = Encoding.UTF8.GetBytes(json);
request.uploadHandler = (UploadHandler)new UploadHandlerRaw(bodyRaw);

4. 最后，发送请求并在接收到响应后处理它：

yield return request.SendWebRequest();

if (request.result == UnityWebRequest.Result.Success)
{
    string responseJson = request.downloadHandler.text;
    // 处理响应JSON数据
}
else
{
    Debug.LogError(request.error);
}

完整的代码示例如下：

using UnityEngine;
using UnityEngine.Networking;
using System.Collections;
using System.Text;

public class RequestExample : MonoBehaviour
{
    void Start()
    {
        StartCoroutine(SendRequest());
    }

    IEnumerator SendRequest()
    {
        var requestData = new
        {
            lineid = 27,
            areaid = 22,
            starttime = "2023-05-07 09:54:22",
            endtime = "2023-06-07 09:54:22",
            datatype = 1,
            pageSize = 5,
            pageNumber = 1
        };
        
        string url = "http://example.com/api/data";
        string json = JsonUtility.ToJson(requestData);

        UnityWebRequest request = UnityWebRequest.Post(url, "POST");
        byte[] bodyRaw = Encoding.UTF8.GetBytes(json);
        request.uploadHandler = (UploadHandler)new UploadHandlerRaw(bodyRaw);

        yield return request.SendWebRequest();

        if (request.result == UnityWebRequest.Result.Success)
        {
            string responseJson = request.downloadHandler.text;
            // 处理响应JSON数据
        }
        else
        {
            Debug.LogError(request.error);
        }
    }
}