Question

我试图查询嵌套链表，其中结果是子值的乘积乘以父列表中的值，然后添加结果。

此代码有效：

var X = (from B in Main.Globals.BookLL
                    from G in B.GreeksLL
                    where B.DealNo == 1 && B.Strategy == "Condor" &&
                    G.BookOrComp == "Book" && G.GreekType == "Delta"
                    select new
                    {
                        Total = G.Data[3] * B.BookPosn * B.FxRate
                    }).Sum(s=>s.Total);

...但我更喜欢使用lambda。下面的代码给出了编译错误，显示为该行末尾的注释。

double Z = Globals.BookLL.Where(B => B.DealNo == 1 && B.Strategy == "Condor").
            SelectMany(G => G.GreeksLL).
            Where(G => G.BookOrComp == "Book" && G.GreekType == "Delta").
            Select(G => new { Total = G.Data[3] * B.BookPosn*B.FxRate }).           // Compile error "B does not exist in the current context"
            Sum();

我不知道如何操作，请查看并更正查询？感谢。

Answer 1

尝试：

double Z = Globals.BookLL.Where(B => B.DealNo == 1 && B.Strategy == "Condor").
           SelectMany(par => par.GreeksLL, (parent, child) => new { G = child, B = parent }).
           Where(both => both.G.BookOrComp == "Book" && both.G.GreekType == "Delta").
           Select(both => new { Total = both.G.Data[3] * both.B.BookPosn*both.B.FxRate }).
           Sum(x => x.Total);

我的命名有点奇怪，但我希望你明白这一点，基本上当你做SelectMany()时你'放弃了'B，这应该就是这样。

未经测试，请告诉我它是否有效。

使用结果选择器功能，查看SelectMany() $this->db->select('title'); // your column $this->db->from('table'); // your table $result = $this->db->get()->result(); // get result $title = $result->first_row()->title; // get ist row using first_row with your field name $graph->title->Set($title); // as you are using for graph。

Answer 2

另一种方法是使用this重载过滤SelectMany中的GreekLL，然后使用Sum扩展名：

double z = Main.Globals.BookLL.Where(book => book.DealNo == 1 && book.Strategy == "Condor")
           .SelectMany(book => book.GreeksLL.Where(greek => greek.BookOrComp == "Book" && greek.GreekType == "Delta")
               ,(book, greek) => new { Greek = greek, Book = book })
           .Sum(greekAndBook => greekAndBook.Book.BookPosn *  greekAndBook.Book.Fxrate * greekAndBook.Greek.Data[3]);

查询嵌套链接列表

2 个答案: