Question

好的，所以我在查询我的Job表时遇到了麻烦。我尝试做的是填写日历，其中包含与各自日期相对应的作业。在我的片段中，我使用了宝石撬。根据我当前的逻辑（控制台中的第28行），我的目标是搜索我的工作表，它将在同一天返回所有工作。现在当我渲染calendar时，每个日期都填充了所有5个Job对象。当我撬开你在控制台中看到的东西时，我得到第一个3.前3个都有相同的created_at日期，但我不知道为什么当我在控制台中只要求一个时，它会返回那些＃39;案例。

供参考： @calender_days是一个填充Time对象格式的数组ex＆＃34; 2018-05-16 08:30:00 -0400＆＃34;

如果您需要更多信息并提前感谢您的帮助，请告诉我。

编辑：澄清。我不知道为什么当我使用时间对象搜索日期为16日时，我会在日期16日22日和16日获得3个对象。

＆＃13;

    21:         <% 5.times do %>
    22:           <tr>
    23:             <% 7.times do %>
    24:               <td>
    25:                 <% date = @calender_days[count] %>
 => 26:                 <% binding.pry %>
    27:                 <u><%= date.strftime('%d') %></u><br />
    28:                 <% @jobs.where(:datetime == date.day).each do |job| %>
    29:                   <%= job.name %><br />
    30:                 <% end %>
    31:                 <% count += 1 %>
[1] pry(#<#<Class:0x6fc59a0>>)> @jobs.where(:datetime == Time.new("2018-05-16 08:30:00 -0400"))
  Job Load (0.0ms)  SELECT "jobs".* FROM "jobs"
=> [#<Job:0x74f20b0
  id: 1,
  name: "Sweep Floor",
  datetime: Wed, 16 May 2018 08:30:00 EDT -04:00,
  created_by: 1,
  client_id: 9,
  notes: "Get under couch",
  workers: "3",
  created_at: Thu, 17 May 2018 09:52:49 EDT -04:00,
  updated_at: Thu, 17 May 2018 09:52:49 EDT -04:00>,
 #<Job:0x74f1fd8
  id: 2,
  name: "Wash Windows",
  datetime: Tue, 22 May 2018 11:30:00 EDT -04:00,
  created_by: 2,
  client_id: 10,
  notes: "Remove all spots",
  workers: "4",
  created_at: Thu, 17 May 2018 09:52:49 EDT -04:00,
  updated_at: Thu, 17 May 2018 09:52:49 EDT -04:00>,
 #<Job:0x74f1ee8
  id: 3,
  name: "Clean Kitchen",
  datetime: Wed, 16 May 2018 12:00:00 EDT -04:00,
  created_by: 1,
  client_id: 11,
  notes: "Wipe down cabinets",

＆＃13;

这是我的所有Job对象。

＆＃13;

[0] #<Job:0x65ef2a8> {
            :id => 1,
          :name => "Sweep Floor",
      :datetime => Wed, 16 May 2018 08:30:00 EDT -04:00,
    :created_by => 1,
     :client_id => 9,
         :notes => "Get under couch",
       :workers => "3",
    :created_at => Thu, 17 May 2018 09:52:49 EDT -04:00,
    :updated_at => Thu, 17 May 2018 09:52:49 EDT -04:00
},
[1] #<Job:0x65ec638> {
            :id => 2,
          :name => "Wash Windows",
      :datetime => Tue, 22 May 2018 11:30:00 EDT -04:00,
    :created_by => 2,
     :client_id => 10,
         :notes => "Remove all spots",
       :workers => "4",
    :created_at => Thu, 17 May 2018 09:52:49 EDT -04:00,
    :updated_at => Thu, 17 May 2018 09:52:49 EDT -04:00
},
[2] #<Job:0x65ec578> {
            :id => 3,
          :name => "Clean Kitchen",
      :datetime => Wed, 16 May 2018 12:00:00 EDT -04:00,
    :created_by => 1,
     :client_id => 11,
         :notes => "Wipe down cabinets",
       :workers => "7",
    :created_at => Thu, 17 May 2018 09:52:50 EDT -04:00,
    :updated_at => Thu, 17 May 2018 09:52:50 EDT -04:00
},
[3] #<Job:0x65ec4b8> {
            :id => 4,
          :name => "Dusting",
      :datetime => Wed, 16 May 2018 02:30:00 EDT -04:00,
    :created_by => 1,
     :client_id => 10,
         :notes => "Get the fans.",
       :workers => "8",
    :created_at => Sat, 19 May 2018 08:49:11 EDT -04:00,
    :updated_at => Sat, 19 May 2018 08:49:11 EDT -04:00
},
[4] #<Job:0x65ec3f8> {
            :id => 5,
          :name => "Dusting",
      :datetime => Thu, 03 May 2018 08:30:00 EDT -04:00,
    :created_by => 1,
     :client_id => 11,
         :notes => "TV Stand.",
       :workers => "4",
    :created_at => Sat, 19 May 2018 08:50:33 EDT -04:00,
    :updated_at => Sat, 19 May 2018 08:50:33 EDT -04:00
}

＆＃13;

Answer 1

我假设Jobs.datetime列类型为datetime，但您尝试按日期列出项目，而不是日期时间。我认为以下查询应该有效：

@jobs.where('DATE(datetime) = ?', Date.parse('Wed, 16 May 2018'))

DATE(...)部分将datetime转换为date。

Rails查询时间对象，不返回特定于日期的对象

1 个答案: