小编Cri*_*jon的帖子

我有一个DataGridView,我需要添加自定义对象.请考虑以下代码:

DataGridView grid = new DataGridView();
grid.DataSource = objects;

使用此代码,我得到一个DataGridView对象,其所有属性都是列.就我而言,我不想展示所有这些信息; 我想只展示两三列.我知道我可以设置

AutoGenerateColumns = false.

但我不知道如何继续进行.一种选择是隐藏所有我不感兴趣的列,但我认为最好以相反的方式进行.我怎样才能做到这一点？

我有一个架构public和数据库schema_A.我需要创建一个schema_b结构相同的新模式schema_a.我发现下面的函数,问题是它不复制外键约束.

CREATE OR REPLACE FUNCTION clone_schema(source_schema text, dest_schema text)
  RETURNS void AS
$BODY$
DECLARE
  object text;
  buffer text;
  default_ text;
  column_ text;
BEGIN
  EXECUTE 'CREATE SCHEMA ' || dest_schema ;

  -- TODO: Find a way to make this sequence's owner is the correct table.
  FOR object IN
    SELECT sequence_name::text FROM information_schema.SEQUENCES WHERE sequence_schema = source_schema
  LOOP
    EXECUTE 'CREATE SEQUENCE ' || dest_schema || '.' || object;
  END LOOP;

  FOR object IN
    SELECT …

我有以下方法在DataGridView上加载产品

private void LoadProducts(List<Product> products)
{
    Source.DataSource = products;  // Source is BindingSource
    ProductsDataGrid.DataSource = Source;
}

现在我试图让我回来保存它们如下图所示.

private void SaveAll()
{
   Repository repository = Repository.Instance;
   List<object> products = (List<object>)Source.DataSource; 
   Console.WriteLine("Este es el número {0}", products.Count);
   repository.SaveAll<Product>(products);
   notificacionLbl.Visible = false;
}

但是我得到了InvalidCastException这条线:

List<object> products = (List<object>)Source.DataSource;

那么如何将DataSource转换为List呢？

我读过collection_check_boxes但我不明白如何设置检查值.我有以下型号:

class Objective < ActiveRecord::Base

  has_many :indicators
  has_many :objective_children, class_name: "Objective", foreign_key: "parent_id"

  def objective_ids
    objective_children.collect{|o| o.id}
  end

  def objective_ids= objectives_ids
    objectives_ids.each do |id|
      objective_children << Objective.find(id)
    end
  end
end

编辑视图:

<%= form_for(@objective) do |f| %>
  <%= f.collection_check_boxes :objective_ids, Objective.all, :id, :name %>
  <%= f.submit %>
<% end %>

html复选框没问题,但我不知道如何设置值objective.我被尝试定义objective_ids= objectives_ids但没有任何反应.

在控制器中:

class ObjectivesController < ApplicationController
    def objective_params
      params.require(:objective).permit(:name, :code, :description, :objective_ids)
    end
end

编辑 日志文件说Unpermitted parameters: perspective_id, objective_ids

我正在开发一篇关于这篇文章的多租户应用程序.问题是我第一次运行所有迁移.在schema.rb文件中只有公共模式的表,但其他模式发生了什么？如何创建具有不同结构的其他模式,public我不想使用宝石.

请参阅下面的示例

要为公共模式创建的表

class CreatePerspectives < ActiveRecord::Migration
  include MultiSchema
  def up
      with_in_schemas :only => :public do
         # Create table perspectives
      end
  end


  def down
    with_in_schemas :only => :public do
      drop_table :prespectives
    end
  end
end

要为私有模式创建的表

class CreateObjectives < ActiveRecord::Migration

  include MultiSchema

  def change
    with_in_schemas :except => :public do
        # Create objectives table
    end
  end
end

schema.rb

ActiveRecord::Schema.define(version: 20130810172443) do

  create_table "perspectives", force: true do |t|
    t.string   "name"
    t.datetime "created_at"
    t.datetime "updated_at"
  end
end

我正在使用FosRestBundle,我正在使用手动路由声明一个控制器.

namespace Cboujon\PropertyBundle\Controller;

use FOS\RestBundle\Controller\Annotations\QueryParam;
use FOS\RestBundle\Controller\Annotations\RouteResource;
use FOS\RestBundle\View\View;
use FOS\RestBundle\Controller\Annotations\Get;


/**
 * Property controller.
 * @RouteResource("Property")
 */
class PropertyRESTController extends \FOS\RestBundle\Controller\FOSRestController {


    /**
     * 
     * @return type
     * @Get("/types")
     * 
     */
    public function getTypesAction() {
        ...
        return $this->get('fos_rest.view_handler')->handle($view);
    }

}

使用routing.yml

cboujon_property_property_api:
    resource: "@CboujonPropertyBundle/Controller/PropertyRESTController.php"
    type:     rest
    prefix:   /api

当我提出要求http://localhost:8000/api/properties/types或者http://localhost:8000/api/property/types我得到了

找不到404 - NotFoundHttpException.

但如果http://localhost:8000/api/types它有效!

我需要配置网址http://localhost:8000/api/properties/types.我究竟做错了什么？

更新1

没有Property实体.PropertyController应该执行自定义操作(无CRUD).

更新2

你可以看到git存储库

我正在使用JHipster,当我运行时,sudo mvn liquibase:diff我得到以下错误

[INFO] Settings
----------------------------
[INFO]     driver: org.postgresql.Driver
[INFO]     url: jdbc:postgresql://localhost/gastos8
[INFO]     username: gastos8
[INFO]     password: *****
[INFO]     use empty password: false
[INFO]     properties file: null
[INFO]     properties file will override? false
[INFO]     prompt on non-local database? true
[INFO]     clear checksums? false
[INFO]     changeLogFile: src/main/resources/config/liquibase/master.xml
[INFO]     context(s): null
[INFO]     label(s): null
[INFO]     referenceDriver: null
[INFO]     referenceUrl: hibernate:spring:com.cboujon.domain?dialect=org.hibernate.dialect.PostgreSQL82Dialect
[INFO]     referenceUsername: null
[INFO]     referencePassword: null
[INFO]     referenceDefaultSchema: null
[INFO]     diffChangeLogFile: src/main/resources/config/liquibase/changelog/20150807132702_changelog.xml
[INFO] ------------------------------------------------------------------------
[INFO] ------------------------------------------------------------------------
[INFO] BUILD FAILURE
[INFO] …

我是NHibernate的新手.我正在尝试将此ORM与SQLite一起使用.我有以下内容:

Product.cs

namespace Stock.Models.Classes
{
    class Product
    {
        public virtual string Name {get; set;}
        public virtual int Id { get; set; }
        public virtual decimal Price { get; set; }
        public virtual decimal Quantity { get; set; }
    }
}

Product.hbm.xml

<?xml version="1.0" encoding="utf-8" ?>
<hibernate-mapping xmlns="urn:nhibernate-mapping-2.2"
                  assembly="Stock.Models"
                  namespace="Stock.Models.Classes">
  <class name="Product" table="products">
    <id name="Id">
      <generator class="int" />
    </id>
    <property name="Name" />
    <property name="Price" />
    <property name="Quantity" />
  </class>
</hibernate-mapping>

的hibernate.cfg.xml

<?xml version="1.0" encoding="utf-8" ?>
<hibernate-configuration xmlns="urn:nhibernate-configuration-2.2">
  <session-factory>
    <property name="connection.provider">NHibernate.Connection.DriverConnectionProvider</property>
    <property …

我有一个现有的数据库.让我们说我有一个名为的表"transactions",我想创建相应的实体名称"Transaction".我怎样才能做到这一点？

我有一个多类分类器，我需要同时获取概率和标签。

model.predict_proba(X)返回每个数据点的所有训练类的概率。

model.predict(X)返回每个数据点的标签。

我不想对每个数据点预测两次。