如何用fastcrud編寫sql代碼？

Question

類：

[Table("department_types", Schema = "public")] 
public class DepartmentTypes 


[Table("departments", Schema = "public")] 
    public class Department 

查詢：

private string BuildQuery() 
      { 
       return OrmConfiguration.GetSqlBuilder<DepartmentViewModel>().Format(
        $@" SELECT {nameof(Department.Id):C}, 
           {nameof(Department.Name):C}, 
           {nameof(Department.TypeId):C}, 
           {nameof(DepartmentTypes.Name):C} as DepartmentTypeName 
         FROM {nameof(Department):T} 
         LEFT JOIN {nameof(DepartmentTypes):T} 
         ON {nameof(Department.TypeId):C} = {nameof(DepartmentTypes.Id):C}"); 
      } 

結果類：

public class DepartmentViewModel 
    { 
     public int Id { get; set; } 

     public string Name { get; set; } 

     public int TypeId { get; set; } 

     public string DepartmentTypeName { get; set; } 

    } 

我想和部和DepartmentTypes聯接，並得到一個DepartmentViewModel結果...

現在，而不是{nameof(DepartmentTypes):T}（對於實例mple）系統放入「DepartmentViewModel」，而不是DepartmentTypes的名稱，即：「department_types」

Answer 1

我找到了答案：D。

private string BuildDepartmentListQuery() 
      { 
       var departmentTypeTable = Sql.Table<DepartmentType>(); 
       var departmentTable = Sql.Table<Department>(); 

       var departmentColumnId = Sql.TableAndColumn<Department>(nameof(Department.Id)); 
       var departmentColumnName = Sql.TableAndColumn<Department>(nameof(Department.Name)); 
       var departmentColumnTypeId = Sql.TableAndColumn<Department>(nameof(Department.TypeId)); 
       var departmentColumnManagerId = Sql.TableAndColumn<Department>(nameof(Department.ManagerId)); 

       var departmentTypeColumnId = Sql.TableAndColumn<DepartmentType>(nameof(DepartmentType.Id)); 
       var departmentTypeColumnName = Sql.TableAndColumn<DepartmentType>(nameof(DepartmentType.Name)); 

       string sqlBuilderResult = OrmConfiguration.GetSqlBuilder<Department>() 
        .Format(
         $@"SELECT 
     {departmentColumnId} 
    , {departmentColumnName} 
    , {departmentColumnTypeId} 
    , {departmentColumnManagerId} 
    , {departmentTypeColumnName} as ""DepartmentTypeName"" 
    FROM {departmentTable} 
    LEFT JOIN {departmentTypeTable} 
    ON {departmentColumnTypeId} = {departmentTypeColumnId}"); 
       ; 

       return sqlBuilderResult; 
    }