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为树节点使用外部组的lambda表达式

wpf c#

Alan Wayne · 技术社区 · 6 年前

下面的LINQ代码正确地为顶级专业代码的子级提供记录组。如何使用lambda?

最上面的查询从数据库中获取视图咨询记录,并按专业名称对其进行正确分组:

// get the records from the database.
            view_consulting[] v = MyNetwork.Medical.Client.GetConsultingStaff();

            // All records will first be grouped by the speciality name. The specialityGroup will hold all the records for the specific speciality.
            var s = v.GroupBy(p => p.speciality_name)
                .Select((IGrouping<string, view_consulting> specialityGroup) =>
                      new SpecialityNode(specialityGroup.Key, specialityGroup, new speciality { speciality_name = specialityGroup.Key }));
            var ss = s.ToList();

然后,每个特殊代码都有一个构造函数,如下所示:

// The specialityGroup holds all the records for the speciality.
        public SpecialityNode(string name, IGrouping<string, view_consulting> specialityGroup, speciality data) : base(name, null, null, data)
        {
            // Create a subgroup for each office under each SpecialityNode.           
            var offices = new List<OfficeNode>();

            var officeGroups = from t in specialityGroup
                               group t by t.office_name;

            foreach (var officegroup in officeGroups)
            {
                foreach (var _office in officegroup)
                {
                    var _officenode = new OfficeNode(_office.office_name, this, this, officegroup, new office { city = _office.city, office_name = _office.office_name, phone = _office.phone, fax = _office.fax, state = _office.state, street = _office.street });
                    offices.Add(_officenode);
                }
            }
            Children = offices.ToList<ITreeNode>();
        }

这一切都符合我的需要。请注意,最后一个foreach()使用的是对specialinode构造函数中外部foreach()的引用。

所有这些“foreach”都可以组合成一个lambda表达式吗?

事先谢谢。

如果有助于澄清,这里是officenode构造函数。注意:它使用这个办公室所属的小组在每个办公室生成顾问。

class OfficeNode : TreeViewBase
    {
        private office data;

        public class MyDoctor
        {
            public string lastname { get; internal set; }
            public string firstname { get; internal set; }
            public string speciality { get; internal set; }
            public consultant data { get; internal set; }

            public string nodename
            {
                get { return string.Format("{0}  {1}  {2}", lastname, firstname, speciality); }
            }
        }

        // The name for the office node is the office name. The office node parent and root are the same and is the speciality node.
        public OfficeNode(string officename, ITreeNode _parent, ITreeNode _root, IGrouping<string, view_consulting> officeGroup, office _data) : base(officename, _parent, _root, _data)
        {
            var _doctors = officeGroup.Where(q => !string.IsNullOrEmpty(q.lastname))
                 .GroupBy(q => new MyDoctor {
                     lastname = q.lastname,
                     firstname = q.firstname,
                     speciality = q.speciality_name,
                     data = new consultant { lastname = q.lastname, firstname = q.firstname }
                 })
                 .Select(doctorGroup => new DoctorNode(doctorGroup.Key.nodename, this, _root, doctorGroup, doctorGroup.Key.data));

            Children = _doctors.ToList<ITreeNode>();

            City = _data.city;
            Phone = _data.phone;
            Fax = _data.fax; 
          //  Order = order ?? throw new ArgumentNullException(nameof(order));

}

1 回复 | 直到 6 年前

TyCobb 6 年前

是的,您可以使用 SelectMany . 选择许多 允许您将多个集合展平为一个集合。因为它似乎是你的 officeGroups 是可枚举的, 选择许多 将返回所有项的单个可枚举项。

var list = officeGroups.SelectMany(o => o)
                        .Select(o => new OfficeNode(constructorParam1,
                                                    constructorParam2,
                                                    etc.))
                        .ToList();

不过,现在看起来还是很干净。一旦你把所有的参数都转储进去,然后 Office 初始化器,它将比您的前臂更难读取和跟踪。它也将更难调试。

编辑: 因为我错过了仍然需要 officeGroup 我们可以稍微透视一下,先从组中构建集合,然后利用匿名对象。

var list = officeGroups.Select(og => new 
                                     {
                                        Offices = og.Select(o => new OfficeNode(o.office_name,
                                                                                this,
                                                                                this,
                                                                                og,
                                                                                etc.))
                                     }
                        .SelectMany(x => x.Offices)
                        .ToList();