Question

我正在尝试从数据库中检索两个内容：第一个字段来源等于"imported"（这意味着我们从excel电子表格中导入），第二个来自{{1 （我们从头开始创建）。附上我的代码：

source != imported

我希望在显示之前将两个结果结合起来：

def index
    add_breadcrumb 'Projects', projects_path
    add_breadcrumb @project.name, @project
    add_breadcrumb "List #{@category.display_name} Content",    project_category_contents_path(@project, @category)
    @contents_imported = Content.of_project(@project).with_category(@category).imported.order('contents.created_at asc')
    @contents_not_imported = Content.of_project(@project).with_category(@category).not_imported.order('contents.created_at desc')
    @page = params[:page]
    @contents = @contents_not_imported << @contents_imported    
    @q = @contents.search(params[:q])
    @content = @q.result(distinct: true).page(@page).per(20)
end

@contents_imported = Content.of_project(@project).with_category(@category).imported.order('contents.created_at asc')

@contents_not_imported = Content.of_project(@project).with_category(@category).not_imported.order('contents.created_at desc')

但它不起作用。我怎么能这样做？

Answer 1

如果它们都是数组且具有相同类型的对象，则可以执行Result = Arr1 | Arr1

这也删除了重复项。它就像布尔UNION。在您的情况下@contents = @contents_not_imported | @contents_imported

Answer 2

问题是您想要连接结果，但您还希望继续将组合结果视为ActiveRelation（在其上调用.search）。这是一种更简单的方法，它首先避免了连接的需要。但是，您需要一个更复杂的ORDER BY子句来完成此任务：

@page = params[:page]
@contents = Content.of_project(@project).with_category(@category).
  order('CASE WHEN source <> "imported" THEN contents.created_at END desc, CASE WHEN source = "imported" THEN contents.created_at END asc')
@q = @contents.search(params[:q])

Answer 3

使用加号

连接数组

你得到undefined method search for Array因为，连接会返回一个数组。而且您无法在Array

上调用搜索方法

修改

def index add_breadcrumb 'Projects', projects_path add_breadcrumb @project.name, @project add_breadcrumb "List #{@category.display_name} Content", project_category_contents_path(@project, @category) contents_imported_ids = Content.of_project(@project).with_category(@category).imported.order('contents.created_at asc').map(&:id) contents_not_imported_ids = Content.of_project(@project).with_category(@category).not_imported.order('contents.created_at desc').map(&:id) @page = params[:page] contents_ids = contents_imported_ids + contents_not_imported_ids contents = Content.where(content_ids) @contents = content_ids.collect{|id| contents.detect{|c| c.id == id}} @q = @contents.search(params[:q]) @content = @q.result(distinct: true).page(@page).per(20) end

Answer 4

只需使用导入或未导入的条件创建新的关系，然后订购所有记录（如果订单对@contents和@content很重要）：

def index
  add_breadcrumb 'Projects', projects_path
  add_breadcrumb @project.name, @project
  add_breadcrumb "List #{@category.display_name} Content", project_category_contents_path(@project, @category)
  @contents_imported = Content.of_project(@project).with_category(@category).imported.order('contents.created_at asc')
  @contents_not_imported = Content.of_project(@project).with_category(@category).not_imported.order('contents.created_at desc')
  @page = params[:page]

  imported = @contents_imported.where_values.reduce(:and)
  not_imported = @contents_not_imported.where_values.reduce(:and)

  @contents = Content.where(imported.or(not_ipmorted)).order('CASE contents.imported WHEN true THEN contents.created_at asc ELSE contents.created_at desc END')

  @q = @contents.search(params[:q])
  @content = @q.result(distinct: true).page(@page).per(20)
end

现在您可以在Ransack#search上调用@contents，因为它是一个ActiveRecord :: Relation。我假设导入的范围采用值contents.imported，其值为true。

如果我写的没有错误，这必须有效。

如何在Ruby on Rails中合并两个对象

4 个答案: