由Dapper QueryMultiple 返回数据的问题得出==》Dapper QueryMultiple并不会帮我们识别多个返回值的顺序
今天帮群友整理Dapper基础教程的时候手脚快了点,然后遇到了一个小问题,Dapper QueryMultiple 返回数据的问题
多个返回值用QueryMultiple ,这个大家都知道,如果不清楚的看下下面的文档:
这个是官方文档:
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();
...
}
按照文档来,为啥没数据呢,就ID有值?难道多表只能传一个参数,而且必须有关系???NONONO,如果这么多限制还叫Dapper吗??
给你3s找错误。。。。。
其实就是顺序弄颠倒了,园友可以当个经验==》Dapper QueryMultiple并不会帮我们识别多个返回值的顺序
Read获取的时候必须是按照上面返回表的顺序 (article,qqmodel,seotkd)
var articleList = multi.Read<Temp>();//类不见得一定得和表名相同
var QQModelList = multi.Read<QQModel>();
var SeoTKDList = multi.Read<SeoTKD>();
官方文档是这样写的,那我们能不能玩点其他的?就一定得定义一个类来获取对应的强类型吗?多返回值就不能动态获取吗???NONONO
直接
if (!multi.IsConsumed)
{
var articleList = multi.Read();
var QQModelList = multi.Read();
var SeoTKDList = multi.Read();
}
一样的效果
周日会有一篇文章详细说下Dapper的,现在得出省了。。。。立刻,马上。。。
附录:
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 = "打代码", Status = 99 })) { //multi.IsConsumed reader的状态 ,true 是已经释放 if (!multi.IsConsumed) { ////强类型 ////注意一个东西,Read获取的时候必须是按照上面返回表的顺序 (article,qqmodel,seotkd) //var articleList = multi.Read<Temp>();//类不见得一定得和表名相同 //var QQModelList = multi.Read<QQModel>(); //var SeoTKDList = multi.Read<SeoTKD>(); ////动态类型 var articleList = multi.Read(); var QQModelList = multi.Read(); var SeoTKDList = multi.Read(); #region 输出 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 } } }