1

誰かが私のSQLクエリを解決するのを手伝ってください私はすでにこれに2日を費やしました....

以下に示すMYSQLクエリがあります

(SELECT
  c.cl_list as cl_list,
  c.name as name,
  pc.value as value,
  count( pc.value) as total
FROM
  projs p 
  LEFT JOIN classify_proj_new pc 
    ON p.proj_id = pc.proj_id_fk
  LEFT JOIN classify_list c 
    ON c.cl_list = pc.class_id_fk
WHERE
  MATCH ( p.title ) AGAINST ( 'jerm'  IN BOOLEAN MODE )
GROUP BY 
  c.cl_list,
  pc.value)
UNION ALL
(SELECT
  c.cl_list as cl_list,
  c.name as name,
  pc.value as value,
  count( pc.value) as total
FROM
  jerm p 
  LEFT JOIN classify_jerm_new pc
    ON p.jerm_id = pc.jerm_id_fk
  LEFT JOIN classify_list c
    ON c.cl_list = pc.class_id_fk
WHERE
  MATCH ( p.jermname ) AGAINST ( 'jerm'  IN BOOLEAN MODE )
GROUP BY
  c.cl_list,
  pc.value)

これは（以下）の結果をもたらします：

  cl_list      name              value        total
------------------------------------------------------------------------------------
    1       department         jewller          2
    3       price                 50            2
    6       color                blue           1
    6       color                Red            2
    1       department         jewller          1
    6       color                Red            1

しかし、繰り返し値の合計を追加し、繰り返し値を回避できる結果を取得しようとしています....次のようなもの（以下）：

  cl_list      name              value        total
------------------------------------------------------------------------------------
    1       department         jewller          3
    3       price                 50            2
    6       color                blue           1
    6       color                Red            3

誰かが私を助けてください私は私の出力について非常に悲しいです...

事前にどうもありがとうございました...

4

3 に答える 3

3

クエリから選択し、cl_list、名前、値でグループ化します。

SELECT
  cl_list,
  name,
  value,
  sum(total) as total
FROM (
  -- your current query here ...
) data
GROUP BY
  cl_list,
  name,
  value
于 2013-01-29T10:05:18.030 に答える
1

以下のコードを試してください。一意のレコードを表示し、重複を回避します。

GROUP BY c.name
于 2013-01-29T10:03:19.207 に答える
0

group名前と値の結果 

(SELECT c.cl_list as cl_list, c.name as name, pc.value as value, count( pc.value) as total
FROM projs p 
LEFT JOIN classify_proj_new pc ON p.proj_id = pc.proj_id_fk
LEFT JOIN classify_list c ON c.cl_list = pc.class_id_fk
WHERE MATCH ( p.title ) AGAINST ( 'jerm'  IN BOOLEAN MODE )
GROUP BY c.name, pc.value)
UNION ALL
(SELECT  c.cl_list as cl_list, c.name as name, pc.value as value, count( pc.value) as total
 FROM jerm p 
 LEFT JOIN classify_jerm_new pc ON p.jerm_id = pc.jerm_id_fk
 LEFT JOIN classify_list c ON c.cl_list = pc.class_id_fk
 WHERE MATCH ( p.jermname ) AGAINST ( 'jerm'  IN BOOLEAN MODE )
 GROUP BY c.name, pc.value)
于 2013-01-29T10:05:10.033 に答える