Question

如何连接结果Json(new{lstrng,a}, JsonRequestBehavior.AllowGet);。我想一起查看lstrng和a的结果。我的代码是

var a = this.abc();  
JsonResult json =Json(new{lstrng,a}, JsonRequestBehavior.AllowGet);  
json.MaxJsonLength = int.MaxValue;  
return json;

这里abc是另一种方法，而lstrng是当前方法的结果

public JsonResult gFbrRecord()
    {
        DataSet ds = dblayer.gfbrdata();
        //this is for getordert tart
        List<fbrData> lstrng = new List<fbrData>(10000);
        foreach (DataRow dr in ds.Tables[0].Rows)
        {
            lstrng.Add(new fbrData
            {  
               FBRInvoiceNumber = dr["FBR_NUM"] != DBNull.Value ? Convert.ToString(dr["FBR_NUM"]) : Convert.ToString(""),
                POSID = dr["REG_NUM"] != DBNull.Value ? Convert.ToInt32(dr["REG_NUM"]) : Convert.ToInt32(0),
               });               

        }

        var a = this.abc();                 
        JsonResult json =Json(new{lstrng,a}, JsonRequestBehavior.AllowGet);
        json.MaxJsonLength = int.MaxValue;
        return json;
    }
     public JsonResult abc() {
        DataSet ds = dblayer.gfbrdata();
        List<fbr1> lstrng = new List<fbr1>(10000);
        foreach (DataRow dr in ds.Tables[0].Rows)
        {
            lstrng.Add(new fbr1
            {  
                DetailSerialNo = dr["SRL_NUM"] != DBNull.Value ? Convert.ToInt32(dr["SRL_NUM"]) : Convert.ToInt32(0),                   
                ItemCode = dr["ITM_COD"] != DBNull.Value ? Convert.ToString(dr["ITM_COD"]) : Convert.ToString(""),
              });

        }

        JsonResult json = Json(lstrng, JsonRequestBehavior.AllowGet);
        json.MaxJsonLength = int.MaxValue;
        return json;
    }
}

Answer 1

您需要查询两个列表并从第二个列表中选择多个项目，与第一个列表中的单个项目具有相同的ID

public JsonResult gFbrRecord()
    {
        DataSet ds1 = dblayer.gfbrdata();
        List<fbrData> lstrng1 = new List<fbrData>(10000);

        //Your other stuff here

        DataSet ds2 = dblayer.gfbrdata();
        List<fbr1> lstrng2 = new List<fbr1>(10000);

        //Your other stuff here

        var result = (from item1 in lstrng1
                      select new
                      {
                          FBRInvoiceNumber = item1.FBRInvoiceNumber,
                          POSID = item1.POSID,
                          SRL_NUM = item1.SRL_NUM,
                          Data = (from item2 in lstrng2
                                  where item2.SRL_NUM == item1.SRL_NUM
                                  select new
                                  {
                                      DetailSerialNo = item2.DetailSerialNo,
                                      ItemCode = item2.ItemCode
                                  }).ToList()
                      }).ToList();

        return Json(result, "application/json", JsonRequestBehavior.AllowGet);
    }

替代

基本上，我为您提供了一个示例代码，该代码应适用于两个json对象的串联

1）此方法从数据库中获取记录并填充列表，然后从该列表中创建JToken

public static JToken Method1()
{
    DataSet ds = dblayer.gfbrdata();
    List<fbrData> lstrng = new List<fbrData>(10000);

    //Your other stuff here

    JToken jToken = JToken.FromObject(lstrng);
    return jToken;
}

JToken通过上述方法返回的样子

[
  {
    "FBRInvoiceNumber": "ABC",
    "POSID": 110034,
    "SRL_NUM": 123
  },
  {
    "FBRInvoiceNumber": "PQR",
    "POSID": 210035,
    "SRL_NUM": 456
  }
]

2）此方法也与上述方法相同，只是区别在于该方法是从数据库中获取除上述方法之外的记录。

public static JToken Method2()
{
    DataSet ds = dblayer.gfbrdata();
    List<fbr1> lstrng = new List<fbr1>(10000);

    //Your other stuff here

    JToken jToken = JToken.FromObject(lstrng);
    return jToken;
}

JToken通过上述方法返回的样子

[
  {
    "DetailSerialNo": 8,
    "ItemCode": "700275",
    "SRL_NUM": 123
  },
  {
    "DetailSerialNo": 9,
    "ItemCode": "800279",
    "SRL_NUM": 456
  }
]

3）现在，以下函数可以将两个数组中的每个对象连接到同一索引上。

自从您在上述评论中提到

您要通过相同的属性键SRL_NUM值将两个数组中的每个对象串联起来。

public static JToken Concat(JToken jToken1, JToken jToken2)
{
    JArray jArray = new JArray();

    foreach (JObject jObject1 in jToken1.ToObject<JArray>())
    {
        var value1 = jObject1.Properties().Where(x => x.Name == "SRL_NUM").FirstOrDefault().Value;

        foreach (JObject jObject2 in jToken2.ToObject<JArray>())
        {
            var value2 = jObject2.Properties().Where(x => x.Name == "SRL_NUM").FirstOrDefault().Value;

            if (Convert.ToInt32(value1) == Convert.ToInt32(value2))
            {
                jObject1.Merge(jObject2, new JsonMergeSettings { MergeArrayHandling = MergeArrayHandling.Concat });
                jArray.Add(jObject1);
            }
        }
    }

    return jArray;
}

4）然后，您可以将上述方法用作您的操作方法

public ContentResult gFbrRecord()
{
    JToken jToken1 = Method1();
    JToken jToken2 = Method2();

    var result = Concat(jToken1, jToken2);

    return Content(result.ToString(), "application/json");
}

输出：

编辑1：

如果您的数组在第二个数组中包含多个SRL_NUM，则以下方法可以将所有具有相同SRL_NUM的项与具有相同SRL_NUM的第一个数组项并置。

public static JToken Concat(JToken jToken1, JToken jToken2)
{
    JArray jArray = new JArray();

    foreach (JObject jObject1 in jToken1.ToObject<JArray>())
    {
        var value = jObject1.Properties().Where(x => x.Name == "SRL_NUM").FirstOrDefault().Value;

        var result = jToken2.ToObject<JArray>().ToObject<JObject[]>().Properties().Where(x => x.Name == "SRL_NUM" && Convert.ToInt32(x.Value) == Convert.ToInt32(value)).Select(x => x.Parent);

        JArray jAr = new JArray();

        foreach (JObject obj in result)
        {
            jAr.Add(obj);
        }

        JObject jObject = new JObject();

        jObject.Merge(jObject1, new JsonMergeSettings { MergeArrayHandling = MergeArrayHandling.Concat });
        jObject.Add("Data", jAr);

        jArray.Add(jObject);
    }

    return jArray;
}

输出：

Answer 2

我想您只希望list的结果仅合并a和lstrng ？？
a.Data.AddRang(lstrng.Data) 该代码将arr的两个合并，您需要在lstrng模型中定义a.Data的列表。

如何在Asp.Net中连接两个Json对象

2 个答案: