c# Linq Join Into语句编辑器错误

var joinedTables =  from tableRow in filteredTable.AsEnumerable()
                    join contactsRow in contacts.AsEnumerable()
                    on tableRow.Field<double>("Opportunity: Store Number") equals contactsRow.Field<double>("National Store .")
                    into lj
                    from r in lj.DefaultIfEmpty()
                    select resultTable.LoadDataRow(new object[]
                    {
                        tableRow.Field<double>("Opportunity: Store Number"),
                        tableRow.Field<DateTime>("Target Circuit Completion (FOC)"),
                        tableRow.Field<string>("Vendor Name"),
                        contactsRow.Field<string>("Contacts - ACM - Email")
                    }, false);

我正在尝试使用LinQ使用这个答案在两个表上做左外连接。但是，当我尝试将字段从contactsRow添加到. loaddatarow()函数的对象数组参数中时，编辑器说'contactsRow'在当前上下文中不存在。我的代码与链接中的答案有何不同?我一直在努力学习LineQ，以避免疯狂的嵌套循环，但这让我难住了。这里有更多的代码

var joinedTables =  from tableRow in filteredTable.AsEnumerable()
                            join contactsRow in contacts.AsEnumerable()
                            on tableRow.Field<double>("Opportunity: Store Number") equals contactsRow.Field<double>("National Store .")
                            into lj
                            from r in lj.DefaultIfEmpty()
                            select resultTable.LoadDataRow(new object[]
                            {
                                tableRow.Field<double>("Opportunity: Store Number"),
                                tableRow.Field<DateTime>("Target Circuit Completion (FOC)"),
                                tableRow.Field<string>("Vendor Name"),
                                r.Field<string>("Contacts - ACM - Email"),
                                r.Field<string>("OO - Ops Mgr Name"),
                                r.Field<string>("Contacts - Area Sup / BC - Email"),
                                r.Field<string>("Contacts - OTP - Email"),
                                r.Field<string>("OTM Email Address")
                            }, false);
        return resultTable;

按照建议，我将select语句中的resultTable代码更改为r。代码现在编译了，但是resultTable是空的。