Ruby On Rails 和 redmine Aglie 插件修复答案

【问题标题】：Ruby On Rails and redmine Aglie plugin fixingRuby On Rails 和 redmine Aglie 插件修复
【发布时间】：2015-02-12 16:35:05
【问题描述】：

好吧，也许你们知道，在 redmine 基础结构中，有表 issues、projects。顺便说一句，我以前从未用 ruby 编写代码。但是我有任务，所以。这里是issues 表结构：这是project 表结构：问题是我还需要获取所有issues，其中project_id equals @project and project.parent_id equals @project。问题是，首先我不明白当前的@project 内容是什么。其次，我不知道如何在这里访问@project 的 parent_id 字段。 Parent_id 顺便引用了 project.id 字段。

我在这里得到了 ruby 函数。老板说我需要编辑范围变量。所以我尝试像这样编辑它：

def find_no_version_issues
    @backlog_version = @project.versions.open.where("LOWER(#{Version.table_name}.name) LIKE LOWER(?)", "backlog").first ||
        @project.versions.open.where(:effective_date => nil).first ||
        Version.open.where(:project_id => @project).order("effective_date ASC").first

         @current_version = Version.open.
        where(:project_id => @project).
        where("#{Version.table_name}.id <> ?", @backlog_version).
        order("effective_date DEcSC").first

    q = (params[:q] || params[:term]).to_s.strip
    #scope = (params[:scope] == "all" || @project.nil? ? Issue : @project.issues).open.visible.where(:fixed_version_id => nil).sorted_by_rank
    scope = Issue.open.visible.where(:fixed_version_id => nil).where("project_id = ?", ) OR @project.or(:project_id.parent_id => @project)).sorted_by_rank
    #scope = Issue.open.visible.where(:project_id => @current_version) #(:project_id => @project || @project.parent_id == :project_id || @project ).sorted_by_rank
    if q.present?
      if q.match(/^#?(\d+)\z/)
        scope = scope.where("(#{Issue.table_name}.id = ?) OR (LOWER(#{Issue.table_name}.subject) LIKE LOWER(?))", $1.to_i, "%#{q}%")
      else
        scope = scope.where("LOWER(#{Issue.table_name}.subject) LIKE LOWER(?)", "%#{q}%")
      end
    end
    @issue_count = scope.count
    @issue_pages = Redmine::Pagination::Paginator.new @issue_count, 20, params['page']
    @version_issues = scope.offset(@issue_pages.offset).limit(@issue_pages.per_page).all
  end

那么，请问有人可以帮助我吗？我怎么能理解，@project 包含什么？如何将查询变成范围？

这是有效的 SQL 查询（或者至少它有我需要的行）。但我真的不知道如何让这个scope 变量包含这些数据？

SELECT * FROM `issues` WHERE `project_id` IN (SELECT `id` FROM `projects` WHERE `parent_id` IS NOT NULL OR `parent_id` IS NULL ) ORDER BY `id` ASC

【问题讨论】：

标签： mysql ruby-on-rails ruby redmine

【解决方案1】：

其实我是这样修的

 def find_no_version_issues
    q = (params[:q] || params[:term]).to_s.strip
    #scope = (params[:scope] == "all" || @project.nil? ? Issue : @project.issues).open.visible.where(:fixed_version_id => nil).sorted_by_rank
    @subprojects = @project.descendants.visible.all
    @subprojects = @subprojects.to_a.push @project
    scope = Issue.open.visible.where(project_id:@subprojects).where(:fixed_version_id => nil).sorted_by_rank
    if q.present?
      if q.match(/^#?(\d+)\z/)
        scope = scope.where("(#{Issue.table_name}.id = ?) OR (LOWER(#{Issue.table_name}.subject) LIKE LOWER(?))", $1.to_i, "%#{q}%")
      else
        scope = scope.where("LOWER(#{Issue.table_name}.subject) LIKE LOWER(?)", "%#{q}%")
      end
    end
    @issue_count = scope.count
    @issue_pages = Redmine::Pagination::Paginator.new @issue_count, 20, params['page']
    @version_issues = scope.offset(@issue_pages.offset).limit(@issue_pages.per_page).all
  end

【讨论】：