标签: orm

我有两个实体：项目和投标，每个项目都有很多投标，因此投标是项目的集合属性。

在显示项目的页面中，我只想显示有关此“项目”的出价的前10条记录。

所以，我这样做查询：

from item
left join fetch item.bids
where item.id=3
...

但这将获取有关某个项目的所有出价，因此，如何限制该项目的出价？

我们有一个示例模型:

#models.py
class Author(models.Model):
    name = models.CharField(max_length=255)

class Book(models.Model):
    name = models.CharField(max_length=255)
    author = models.ManyToManyField(Author, blank=True)

    def get_authors(self):
        return self.authors.all().order_by('id').values_list('name')

#views.py
class BooksView(ListView):
    model = Book

    def get_queryset(self):
        q = Book.select_related('authors').all()


#template
{% for book in books %}

    {{ book.name }} ({%for author in book.get_authors %} {{ author }} {% endfor %}

{% endfor %}

当我尝试使用get_authors函数从模板获取数据时,我看到多个SQL查询会大大降低性能(SQL工作大约5秒).是否可以减少查询？现在我在循环中看到每个作者的SQL查询.

from sqlalchemy.orm import subqueryload, joinedload, eagerload
from sqlalchemy import Column, DateTime, String, Integer, ForeignKey, func,Float, sql
from sqlalchemy.orm import relation
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker
from sqlalchemy import create_engine

engine = create_engine('sqlite:///testdb.sqlite')
engine.echo = True
Base = declarative_base()
session = sessionmaker()
session.configure(bind=engine)
Base.metadata.create_all(engine)
s= session()

class Stock(Base):
__tablename__ = 'stock'
stock_id = Column(Integer, primary_key=True)
name = Column(String)
prices = relation("StockPrice")

class StockPrice(Base):
__tablename__ = 'stock_price'
stock_id = Column(Integer, ForeignKey('stock.stock_id'), primary_key=True)
date = Column(String, primary_key=True)
price = Column(Float) …

我正在我的项目中处理某种"复杂"的表单,其中实体在每个步骤上都是持久的,因为各个表单都是分开的.然后我迈出了第一步(让我们称之为步骤1),我坚持一个实体并将其存储在会话中,请参阅下面的代码:

$productoSolicitudEntity = new Entity\ProductoSolicitud();
$productoSolicitudForm = $this->createForm(new Form\ProductoSolicitudForm(), $productoSolicitudEntity);
$productoSolicitudForm->handleRequest($request);

if ($productoSolicitudForm->isValid()) {
    $productoSolicitudRequest = $request->get('productoSolicitud');

    try {
        $producto = $em->getRepository("AppBundle:Producto")->find($productoSolicitudRequest['producto']['nombre']);
        $productoSolicitudEntity->setProducto($producto);

        $condicionProducto = $em->getRepository("AppBundle:CondicionProducto")->find($productoSolicitudRequest['condicion_producto']);
        $productoSolicitudEntity->setCondicionProducto($condicionProducto);

        $finalidadProducto = $em->getRepository("AppBundle:FinalidadProducto")->find($productoSolicitudRequest['finalidad_producto']);
        $productoSolicitudEntity->setFinalidadProducto($finalidadProducto);

        $procedenciaProducto = $em->getRepository("AppBundle:ProcedenciaProducto")->find($productoSolicitudRequest['procedencia_producto']);
        $productoSolicitudEntity->setProcedenciaProducto($procedenciaProducto);

        $productoSolicitudEntity->setLote($productoSolicitudRequest['lote']);
        $solicitudUsuario = $em->getRepository("AppBundle:SolicitudUsuario")->find($session->get('solicitudUsuarioEntity')->getId());
        $productoSolicitudEntity->setSolicitudUsuario($solicitudUsuario);

        $em->persist($productoSolicitudEntity);
        $em->flush();

        $session->set('productoSolicitudEntity', $productoSolicitudEntity);
        $response['success'] = true;
    } catch (Exception $ex) {
        $status = 400;
        $response['error'] = $ex->getMessage();
    }
} else {
    $status = 400;
    $response['error'] = $this->get('translator')->trans('formularioNoValido');
    $response['formError'] = $this->getFormErrors($productoSolicitudForm);
}

然后在四个步骤(让我们称之为步骤4)我需要将该实体附加到一个新实体,因为它们是相关的,这是涉及的代码:

$productoSolicitud = $session->get('productoSolicitudEntity');

if (! $productoSolicitud) { …

我很好奇地看到如何从n到kth获得记录.我不知道我是否错过了,但是通过文档并没有对我有利.

我不是说我想要记录,其中id介于5 - 10之间,但记录介于5 - 10之间.如何进行此操作？

我在使用CFC组件时在ColdFusion中编写了一些代码,当我对使用不同方法创建组件对象感到困惑时.如果有人请让我知道创建对象的哪种方法更好,我将不胜感激. CreateObject(),EntityNew()&& Newkeyword.我阅读了几篇博客并得到了不同的答案,有些人表示,与Create Object相比,Entity New更快.我还发现语法上的差异更好EntityNew().如果我能从任何人那里得到一些想法,我将不胜感激.谢谢.

这是我在DynamicViewsTable.php中的自定义finder方法

public function findAccessibleByUser(Query $query, array $options)
    {
        if (empty($options['User']['id'])) {
            throw new Exception("Current User not set", 1);    
        }

        $query->select(['DynamicViews.id', 'DynamicViews.title', 'UsersAccessDynamicViews.ordinal_ranking'])
              ->contain(['UsersAccessDynamicViews'])
              ->where([
                    'UsersAccessDynamicViews.user_id' => $options['User']['id'],
                ])
              ->order(['UsersAccessDynamicViews.ordinal_ranking' => 'ASC']);
        return $query;
    }

我一直得到的错误是:

Error: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'UsersAccessDynamicViews.ordinal_ranking' in 'field list'

并且错误页面中显示的查询是:

SELECT DynamicViews.id AS `DynamicViews__id`, DynamicViews.title AS `DynamicViews__title`, UsersAccessDynamicViews.ordinal_ranking AS `UsersAccessDynamicViews__ordinal_ranking` FROM dynamic_views DynamicViews WHERE UsersAccessDynamicViews.user_id = :c0 ORDER BY UsersAccessDynamicViews.ordinal_ranking ASC

DynamicViews拥有许多UsersAccessDynamicViews

我试图在Hibernate中使用FetchMode.JOIN来理解但面临某些问题.我有3个班级,员工和员工主类.部门与员工有一个关系.

以下是代码部门类: -

@Entity
@Table(name="DEPARTMENT")
public class Department {

@Id
@GeneratedValue
@Column(name="DEPARTMENT_ID")
private Long departmentId;

@Column(name="DEPT_NAME")
private String departmentName;

@OneToMany(mappedBy="department")
@Fetch(FetchMode.JOIN)
private List<Employee> employees = new ArrayList<>();

public Long getDepartmentId() {
    return departmentId;
}

public void setDepartmentId(Long departmentId) {
    this.departmentId = departmentId;
}

public String getDepartmentName() {
    return departmentName;
}

public void setDepartmentName(String departmentName) {
    this.departmentName = departmentName;
}

public List<Employee> getEmployees() {
    return employees;
}

public void setEmployees(List<Employee> employees) {
    this.employees = employees;
}

}  

员工类: -

@Entity
@Table(name="EMPLOYEE") …

我想知道在Waterline中是否可以定义模型或按照Node-ORM2中的名称获取模型.

定义:

var Person = db.define("person", {
    name      : String,
    surname   : String,
    age       : Number, // FLOAT
    male      : Boolean,
    continent : [ "Europe", "America", "Asia", "Africa", "Australia", "Antartica" ], // ENUM type
    photo     : Buffer, // BLOB/BINARY
    data      : Object // JSON encoded
}, {
    methods: {
        fullName: function () {
            return this.name + ' ' + this.surname;
        }
    },
    validations: {
        age: orm.enforce.ranges.number(18, undefined, "under-age")
    }
});

获得:

var MyPersonModel = db.models["person"];

谢谢!

我在基本字段中有类型datetime,它是time字段.我希望从基础获得所有元素并获得额外的timeu字段 - 它是UNIX_TIMESTAMP(time).我试着通过添加来做到这一点,->select('*','UNIXTIMESTAMP(time) AS timeu')但是Laravel给了我错误.我需要它用于->keyBy().我有下一个错误:

SQLSTATE[42S22]: Column not found: 1054 Unknown column 'UNIX_TIMESTAMP(time)' in 'field list' (SQL: select *, `UNIX_TIMESTAMP(time)` as `timeu` from `values` where UNIX_TIMESTAMP(time) <= 1429135199 and UNIX_TIMESTAMP(time) >= 1428444000 order by `id` asc)

我该如何解决？