本文介绍了如何使用linq将所选值分配到列表的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
var result = (from r in db.Routes join tin in db.TripIns on r.RouteId equals tin.RouteId join tout in db.TripOuts on r.RouteId equals tout.RouteId select new { RouteId = r.RouteId, RouteName = r.RouteName, RouteDesc = r.RouteDesc, tripins = new TripIn{ TripInId = tin.TripInId, TripInName = tin.TripInName }, tripouts = new TripOut { TripOutId = tout.TripOutId, TripOutName = tout.TripOutName } }).ToList() .Select(x => new Route() { RouteId = x.RouteId, RouteName = x.RouteName, RouteDesc = x.RouteDesc, TripIns = x.tripins, TripOuts = x.tripouts });
我的尝试:
What I have tried:
var result = (from r in db.Routes join tin in db.TripIns on r.RouteId equals tin.RouteId join tout in db.TripOuts on r.RouteId equals tout.RouteId select new { RouteId = r.RouteId, RouteName = r.RouteName, RouteDesc = r.RouteDesc, tripins = new TripIn{ TripInId = tin.TripInId, TripInName = tin.TripInName }, tripouts = new TripOut { TripOutId = tout.TripOutId, TripOutName = tout.TripOutName } }).ToList() .Select(x => new Route() { RouteId = x.RouteId, RouteName = x.RouteName, RouteDesc = x.RouteDesc, TripIns = x.tripins, TripOuts = x.tripouts });推荐答案
尝试将ToList放在命令的末尾。更改: Try putting the ToList at the end of the command. Change: }).ToList() .Select(x => new Route() { RouteId = x.RouteId, RouteName = x.RouteName, RouteDesc = x.RouteDesc, TripIns = x.tripins, TripOuts = x.tripouts });
收件人:
To:
}) .Select(x => new Route() { RouteId = x.RouteId, RouteName = x.RouteName, RouteDesc = x.RouteDesc, TripIns = x.tripins, TripOuts = x.tripouts }).ToList();
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如何使用linq将所选值分配到列表
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