如何重写我的SQL查询,以便它以特定的格式返回值(使用mysql或php)？

Question

我有一张名为number:

number_id  |  number_name
   1            aaaa
   2            bbbb
   3            cccc
   4            dddd
   5            eeee
   6            ffff

(它只包含6个条目)

我还有一张名为texts:

text_id  |  start_time  |  number_id  |  text_content
   1        some date1  |     2       |    blabla1
   2        some date2  |     1       |    blabla2
   3        some date3  |     2       |    blabla3
   4        some date4  |     3       |    blabla4
   5        some date5  |     4       |    blabla5
   6        some date6  |     6       |    blabla6
  etc.

(它包含很多条目).

现在,当我正在进行这样的查询时:

SELECT
   * 
FROM
   (SELECT
      DATEDIFF(now(),
      start_time) AS days_ago,
      number_id,
      COUNT(text_id) AS num_texts 
   FROM
      TEXTS 
   WHERE
      start_time BETWEEN  DATE_SUB(NOW(), INTERVAL 100 DAY) AND NOW() 
   GROUP BY
      DATE(start_time),
      number_id) AS temp

它使用number_id返回每个地方过去100天每天的文本数量.结果如下:

当我将它编码为json时,我得到:

[{"days_ago":"19","number_id":"1","num_texts":"179"},
{"days_ago":"19","number_id":"5","num_texts":"1"},
{"days_ago":"18","number_id":"1","num_texts":"61"},
{"days_ago":"18","number_id":"2","num_texts":"1"},
{"days_ago":"18","number_id":"5","num_texts":"1"},
{"days_ago":"18","number_id":"6","num_texts":"3"},
{"days_ago":"17","number_id":"3","num_texts":"1"},
{"days_ago":"8","number_id":"1","num_texts":"2"},
{"days_ago":"8","number_id":"2","num_texts":"2"},
{"days_ago":"7","number_id":"4","num_texts":"1"},

我想写一个不同的查询,或更改现有的查询.有没有办法编写一个查询,将以下列方式返回我的数据:

[{"days_ago": "7", "number_id" : "1", "num_texts" : "18", "number_id" : "2", "num_texts" : "12", "number_id" : "3", "num_texts" : "12" , ... , "number_id" : "6", "num_texts" : "1"},
{"days_ago": "6", "number_id" : "1", "num_texts" : "18", "number_id" : "2", "num_texts" : "12", "number_id" : "3", "num_texts" : "12" , ... , "number_id" : "6", "num_texts" : "1"},
{"days_ago": "5", "number_id" : "1", "num_texts" : "18", "number_id" : "2", "num_texts" : "12", "number_id" : "3", "num_texts" : "12" , ... , "number_id" : "6", "num_texts" : "1"},
{"days_ago": "4", "number_id" : "1", "num_texts" : "18", "number_id" : "2", "num_texts" : "12", "number_id" : "3", "num_texts" : "12" , ... , "number_id" : "6", "num_texts" : "1"},
{"days_ago": "3", "number_id" : "1", "num_texts" : "18", "number_id" : "2", "num_texts" : "12", "number_id" : "3", "num_texts" : "12" , ... , "number_id" : "6", "num_texts" : "1"},
{"days_ago": "2", "number_id" : "1", "num_texts" : "18", "number_id" : "2", "num_texts" : "12", "number_id" : "3", "num_texts" : "12" , ... , "number_id" : "6", "num_texts" : "1"},
{"days_ago": "1", "number_id" : "1", "num_texts" : "18", "number_id" : "2", "num_texts" : "12", "number_id" : "3", "num_texts" : "12" , ... , "number_id" : "6", "num_texts" : "1"}] 

基本上每行应该有6个中每个的信息days_ago和值.我需要限制此查询仅过去7天,所以我知道我可以在这里修改它:num_textsnumber_id

BETWEEN DATE_SUB(NOW(), INTERVAL 100 DAY) AND NOW()

并将100改为7,但有可能以某种方式得到这种特定格式的其他信息吗？

哦还有一件事 - 我正在从php调用这个查询,所以如果它更容易在php中以某种方式解析它并返回作为json,这也是一个选项.

非常感谢任何提示家伙!

=========编辑:

正如@AlexBlex指出的那样 - 我提出的json结构可能无效,因为有多次使用相同的键名.为了帮助您更好地理解我的问题 - http://thevectorlab.net/flatlab/morris.html这里是morris js图表的例子.我希望将该图表生成Quarterly Apple iOS device unit sales为该示例中调用的图表.在水平线上我想要date(days ago),在垂直线上我需要num_texts.我希望有6行,每行代表不同的number_id.

Answer 1

假设您number有6个固定值永远不会改变,可以通过链接texts表的6个副本(或创建6个视图).

但是如果你通过跟踪days_ago更改和添加新的JSON记录来在php中执行它可能会更加清晰.您还必须确保number_id跳过no 并num_texts为任何跳过的值添加零值.

另外,示例伪代码:

$days_ago=0;
while ([read_records]) {
    if (record['days_ago']!=$days_ago {
        $days_ago!=0 [Close line]
        $days_ago=record['days_ago'];
        [Add new line]
        [Add days_ago field]  }
    [Add number_id and num_texts fields]
}
[Close line]

第二个编辑,视图解决方案:

如果view_1是我在下面描述view_2的视图和视图number_id 2,你可以像这样链接它们:

SELECT view_1.days_ago, view_1.num_texts AS num_texts1, view_2.num_texts AS num_texts2
FROM view_1 INNER JOIN view_2 ON view_1.days_ago=view_2.days_ago

如果你number_id为每个人都有条目,那么这个willo只能正常工作days_ago,否则你将不得不创建一个只有的视图,days_ago并使用LEFT/RIGHT连接来加入number_id视图.