ServiceStack JOIN 3 个表答案

【问题标题】：ServiceStack JOIN 3 tablesServiceStack JOIN 3 个表
【发布时间】：2014-10-25 17:11:56
【问题描述】：

是否可以使用 ServiceStack 加入 2 个以上的表？我尝试了类似的方法，但我需要链接 evrything：

// Join PatientDetails and BedDetails
SqlExpression<PatientDetails.PatientDetails> q = _dbConnection.From<PatientDetails.PatientDetails>();
q.Join<PatientDetails.PatientDetails, BedDetails.BedDetails>((patient, bed) => patient.ByBedId == bed.BedDetailsId);
List<PatientDetails.PatientDetails> PatientBedJOIN = _dbConnection.Select(q);

// Join PatientSession and PatientDetails
SqlExpression<PatientSession> q1 = _dbConnection.From<PatientSession>();
q1.Join<PatientSession, PatientDetails.PatientDetails>((session, patientd) => session.ByPatientId == patientd.PatientDetailsId);
List<PatientSession> SessionPatientJOIN = _dbConnection.Select(q1);

我正在使用 OrmLite 和 Sqlite，谢谢

【问题讨论】：

标签： c# sqlite servicestack ormlite-servicestack

【解决方案1】：

OrmLite 的 Advanced Join Example in 展示了如何继续向同一查询添加连接，例如：

List<FullCustomerInfo> rows = db.Select<FullCustomerInfo>(  // Map results to FullCustomerInfo POCO
  db.From<Customer>()                                       // Create typed Customer SqlExpression
    .LeftJoin<CustomerAddress>()                            // Implicit left join with base table
    .Join<Customer, Order>((c,o) => c.Id == o.CustomerId)   // Explicit join and condition
    .Where(c => c.Name == "Customer 1")                     // Implicit condition on base table
    .And<Order>(o => o.Cost < 2)                            // Explicit condition on joined Table
    .Or<Customer,Order>((c,o) => c.Name == o.LineItem));    // Explicit condition with joined Tables

在您的示例中如下所示：

var q = _dbConnection.From<PatientDetails.PatientDetails>()
    .Join<PatientDetails.PatientDetails, BedDetails.BedDetails>(
        (patient, bed) => patient.ByBedId == bed.BedDetailsId)
    .Join<PatientDetails.PatientDetails, PatientSession>(
        (patientd, session) => session.ByPatientId == patientd.PatientDetailsId);

var SessionPatientJOIN = _dbConnection.Select(q);

【讨论】：