如何使用linq将所选值分配到列表

编程入门 行业动态 更新时间:2024-10-23 15:24:29
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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