列表没有在C＃中添加所有值

我尝试下面的代码创建json代码，代码工作正常。

我是从db加载值，但最后一个值只有我得到一个输出。未添加的值。

DataTable dt = new DataTable(); var objectToSerialize = new RootObject(); SqlConnection connection = new SqlConnection(ConfigurationManager.ConnectionStrings["dbconnection"].ToString()); SqlCommand command = new SqlCommand("select * from IFlicksMenu", connection); SqlDataAdapter da = new SqlDataAdapter(command); da.Fill(dt); foreach (DataRow dr in dt.Rows) { foreach (DataColumn col in dt.Columns) { objectToSerialize.contacts = new List { new Cont { sno = dr["Cinema"].ToString(), name = dr["Gallery"].ToString(), address = dr["star"].ToString(), gender = dr["video"].ToString(), em = dr["em"].ToString(), phone=new Phone { mobile=dr["phonone"].ToString() } }, }; } JavaScriptSerializer serializer = new JavaScriptSerializer(); Response.Write(serializer.Serialize(new { item = objectToSerialize.contacts })); } 

输出是

 { "item":[ { "phone":{ "mobile":"99528748474" }, "sno":"ACTRESS", "name":"ACTRESS", "em":"scd", "address":"ACTRESS", "gender":"TRAILER" } ] } 

Answer 1

当您想要在循环开头创建并继续添加到列表时，可以为每个项重新创建contacts ：

 var items = new List(); foreach (DataRow dr in dt.Rows) { items.Add(new Cont{ sno = dr["Cinema"].ToString(), name = dr["Gallery"].ToString(), address = dr["star"].ToString(), gender = dr["video"].ToString(), em = dr["em"].ToString(), phone = new Phone{ mobile=dr["phonone"].ToString() } }); } objectToSerialize.contacts = items; 

你还需要删除内部循环

 foreach (DataColumn col in dt.Columns) 

因为它会有效地将行数乘以列数

Answer 2

您在每次迭代中实例化foreach列表，因此它需要最后的迭代。您可以在foreach之外定义List对象，在foreach定义值

 List ContCollection = new List(); foreach (DataRow dr in dt.Rows) { foreach (DataColumn col in dt.Columns) { ContCollection.Add(new Cont(){ sno = dr["Cinema"].ToString(), name = dr["Gallery"].ToString(), address = dr["star"].ToString(), gender = dr["video"].ToString(), em = dr["em"].ToString(), phone=new Phone { mobile=dr["phonone"].ToString() } }); } objectToSerialize.contacts = ContCollection; JavaScriptSerializer serializer = new JavaScriptSerializer(); Response.Write(serializer.Serialize(new { item = objectToSerialize.contacts })); } 

Answer 3

尝试使用

 objectToSerialize.contacts.Add( new List { new Cont{ sno = dr["Cinema"].ToString(), name = dr["Gallery"].ToString(), address = dr["star"].ToString(), gender = dr["video"].ToString(), em = dr["em"].ToString(),phone=new Phone{mobile=dr["phonone"].ToString()} }, }); 

代替

 objectToSerialize.contacts = new List { new Cont{ sno = dr["Cinema"].ToString(), name = dr["Gallery"].ToString(), address = dr["star"].ToString(), gender = dr["video"].ToString(), em = dr["em"].ToString(),phone=new Phone{mobile=dr["phonone"].ToString()} }, };