使用LINQ从查询其结果集一起返回的项目数 [英] Using Linq to return the count of items from a query along with its resultset

问题描述

我使用C＃MVC4使用LINQ。

I am using C# MVC4 with Linq.

我已经使用依赖注入我的项目，这导致了我一个单独的存储库项目一起有一个单独的模型的项目（一个用于测试等）。所有这没有问题。

I have used dependency injection for my project which resulted in me having a separate Model's project along with a separate Repository project (and one for testing ect). All this no problem.

我把我的查询出控制器（老款）和到库中（新DI风格），并注入他们。它工作正常。

I moved my queries out of the controllers (old style) and into the repository (new DI style), and injected them. It works fine.

我有一个标准的LINQ查询（挑选任何例子，他们有足够的基础），它返回一组从数据库正常项目。这里没有任何问题。

I have a standard linq query (pick any example, they are basic enough), which returns a set of items from the database as normal. No problems here either.

我的问题是，我想实现分页，我教这将是足够简单到。下面是我的步骤：

My problem is, that I want to implement paging, and I taught it would be simple enough to. Here is my steps:

取（注入控制器）将其存储在一个变种。它看起来是这样的：

Take in the results of the linq query from the repository (injected into the controller) store it in a var. It looks something like:

var results = _someInjectedCode.GetListById(SomeId);

之前，我能够做一些简单的像：

Before, I was able to do something simple like:

results.Count()
results.Skip(SomeNum).Take(SomeOtherNum)

但现在，我想分页，我需要做我跳过拿这样的事情：

But now that I want paging, I need to do my Skip Take something like this:

var results = from xyz in _someInjectedCode.GetListById(SomeId).SomeId).Skip(SomeNum).Take(SomeOtherNum)
select new[] {a,id, a.fName, a.lName .....}

这样做的问题是，我不再有机会获得物品的总数之前，名单被缩短为pre跳过......就拿状态，除非我做两个查询，这意味着击中两次DB。

The problem with this is that I no longer have access to the total count of items before the list was shortened to the Pre Skip...Take state unless I do two queries which means hitting the DB twice.

什么是解决这一问题的最佳途径。

What is the best way to resolve this issue.

推荐答案

我只是做这样的：

var result = (from n in mycollection 
              where n.someprop == "some value" 
              select n).ToList();
var count = result.Count;

有可能是其他的方法，但是这是我所知道的最简单的。

There are probably other ways, but this is the simplest that I know of.

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