13

我有一个具有以下属性的 Customer 类：

public int Id { get; set; }
public string Name { get; set; }
public int AddressId { get; set; }
public Address Address { get; set; }

我的目标是编写一个 Dapper 查询，该查询将使用内部联接来填充返回的每个 Customer 中的整个 Address 属性。

这是我所拥有的，它正在工作，但我想知道这是否是最干净/最简单的方法：

StringBuilder sql = new StringBuilder();
using (var conn = GetOpenConnection())
{
    sql.AppendLine("SELECT c.Id, c.Name, c.AddressId, a.Address1, a.Address2, a.City, a.State, a.ZipCode ");
    sql.AppendLine("FROM Customer c ");
    sql.AppendLine("INNER JOIN Address a ON c.AddressId = a.Id ");

    return conn.Query<Customer, Address, Customer>(
        sql.ToString(),
        (customer, address) => {
            customer.Address= address;
            return userRole;
        },
        splitOn: "AddressId"
    ).ToList();
}

我对添加另一个属性有些担心，例如：

public Contact Contact { get; set; }

我不确定如何切换上面的语法来填充地址和联系人。

4

2 回答 2

13

我已经使用 Dapper 1.40 版进行了编码，并且我已经编写了如下方式的查询，我没有遇到任何问题来填充多个对象，但是我遇到了可以在查询中映射的 8 个不同类的限制。

public class Customer {
    public int Id { get; set; }
    public string Name { get; set; }
    public int AddressId { get; set; }  
    public int ContactId { get; set; }
    public Address Address { get; set; }
    public Contact Contact { get; set; }
}

public class Address {
    public int Id { get; set; }
    public string Address1 {get;set;}
    public string Address2 {get;set;}
    public string City {get;set;}
    public string State {get;set;}
    public int ZipCode {get;set;}
    public IEnumerable<Customer> Customer {get;set;}
}

public class Contact {
    public int Id { get; set; }
    public string Name { get; set; }
    public IEnumerable<Customer> Customer {get;set;}
}

using (var conn = GetOpenConnection())
{
    var query = _contextDapper
        .Query<Customer, Address, Contact, Customer>($@"
            SELECT c.Id, c.Name, 
                c.AddressId, a.Id, a.Address1, a.Address2, a.City, a.State, a.ZipCode,
                c.ContactId, ct.Id, ct.Name
            FROM Customer c
            INNER JOIN Address a ON a.Id = c.AddressId
            INNER JOIN Contact ct ON ct.Id = c.ContactId", 
            (c, a, ct) =>
            {
                c.LogType = a;
                c.Contact = ct;
                return c; 
            }, splitOn: "AddressId, ContactId")
        .AsQueryable();

    return query.ToList();          
}
于 2017-07-08T00:39:32.123 回答
7

看一下我的示例，其中包含一个大查询，请注意，每个查询行都是不同的对象。

public List<Appointment> GetList(int id)
{
    List<Appointment> ret;
    using (var db = new SqlConnection(connstring))
    {
        const string sql = @"SELECT AP.[Id], AP.Diagnostics, AP.Sintomns, AP.Prescription, AP.DoctorReport, AP.AddressId,
        AD.Id, AD.Street, AD.City, AD.State, AD.Country, AD.ZIP, Ad.Complement,
        D.Id, D.Bio, d.CRMNumber, D.CRMNumber, D.CRMState,
        P.Id,
        S.Id, S.Name,
        MR.Id, MR.Alergies, MR.BloodType, MR.DtRegister, Mr.HealthyProblems, MR.HealthyProblems, MR.Height, MR.MedicalInsuranceNumber, MR.MedicalInsuranceUserName, MR.Medications, MR.Weight,
        MI.Id, MI.Name
        from Appointment AP
        inner join [Address] AD on AD.Id = AP.AddressId
        inner join Doctor D on D.Id = AP.DoctorId
        inner join Patient P on P.Id = AP.PatientId
        left join Speciality S on S.Id = D.IDEspeciality
        left join MedicalRecord MR on MR.Id = P.MedicalRecordId
        left join MedicalInsurance MI on MI.Id = MR.MedicalInsuranceId
        where AP.Id = @Id
        order by AP.Id desc";

        ret = db.Query<Appointment, Address, Doctor, Patient, Speciality, MedicalRecord, MedicalInsurance, Appointment>(sql,
            (appointment, address, doctor, patient, speciality, medicalrecord, medicalinsurance) =>
            {
                appointment.Address = address;
                appointment.Doctor = doctor;
                appointment.Patient = patient;
                appointment.Doctor.Speciality = speciality;
                appointment.Patient.MedicalRecord = medicalrecord;
                appointment.Patient.MedicalRecord.MedicalInsurance = medicalinsurance;
                return appointment;
            }, new { Id = id }, splitOn: "Id, Id, Id, Id, Id, Id").ToList();
    }
    return ret;
}
于 2017-10-23T14:04:35.930 回答