今天幫群友整理Dapper基礎教程的時候手腳快了點,然后遇到了一個小問題,Dapper QueryMultiple 返回數據的問題
多個返回值用QueryMultiple ,這個大家都知道,如果不清楚的看下下面的文檔:
這個是官方文檔:
Multiple Results
Dapper allows you to process multiple result grids in a single query.
Example:
var sql =
@" select * from Customers where CustomerId = @id select * from Orders where CustomerId = @id select * from Returns where CustomerId = @id"; using (var multi = connection.QueryMultiple(sql, new {id=selectedId})) { var customer = multi.Read<Customer>().Single(); var orders = multi.Read<Order>().ToList(); var returns = multi.Read<Return>().ToList(); ... }
按照文檔來,為啥沒數據呢,就ID有值?難道多表只能傳一個參數,而且必須有關系???NONONO,如果這么多限制還叫Dapper嗎??
給你3s找錯誤。。。。。
其實就是順序弄顛倒了,園友可以當個經驗==》Dapper QueryMultiple並不會幫我們識別多個返回值的順序
Read獲取的時候必須是按照上面返回表的順序 (article,qqmodel,seotkd)
var articleList = multi.Read<Temp>();//類不見得一定得和表名相同
var QQModelList = multi.Read<QQModel>();
var SeoTKDList = multi.Read<SeoTKD>();
官方文檔是這樣寫的,那我們能不能玩點其他的?就一定得定義一個類來獲取對應的強類型嗎?多返回值就不能動態獲取嗎???NONONO
直接
if (!multi.IsConsumed)
{
var articleList = multi.Read();
var QQModelList = multi.Read();
var SeoTKDList = multi.Read();
}
一樣的效果
周日會有一篇文章詳細說下Dapper的,現在得出省了。。。。立刻,馬上。。。
附錄:
using (SqlConnection conn = new SqlConnection(connStr)) { string sqlStr = @"select Id,Title,Author from Article where Id = @Id select * from QQModel where Name = @Name select * from SeoTKD where Status = @Status"; conn.Open(); using (var multi = conn.QueryMultiple(sqlStr, new { Id = 11, Name = "打代碼", Status = 99 })) { //multi.IsConsumed reader的狀態 ,true 是已經釋放 if (!multi.IsConsumed) { ////強類型 ////注意一個東西,Read獲取的時候必須是按照上面返回表的順序 (article,qqmodel,seotkd) //var articleList = multi.Read<Temp>();//類不見得一定得和表名相同 //var QQModelList = multi.Read<QQModel>(); //var SeoTKDList = multi.Read<SeoTKD>(); ////動態類型 var articleList = multi.Read(); var QQModelList = multi.Read(); var SeoTKDList = multi.Read(); #region 輸出 foreach (var item in QQModelList) { Console.WriteLine(item.Id + " " + item.Name + " " + item.Count); } foreach (var item in SeoTKDList) { Console.WriteLine(item.Id + " | " + item.SeoKeywords); } foreach (var item in articleList) { Console.WriteLine(item.Author); } #endregion } } }