Question

我有一个MaraiDB（10.2.14）数据库，其中包含一个表，其中存储了一列JSON数据。

我很困惑如何从这个专栏中提取数据。

示例数据

SELECT 1 AS ID
     , '[{"name":"x","score":2},{"name":"y", "score":8},{"name":"z","score":3}]' AS REPLY
UNION ALL
SELECT 2 AS ID
     , '[{"name":"x","score":5},{"name":"y", "score":4},{"name":"z","score":3}]' AS REPLY
UNION ALL
SELECT 3 AS ID
     , '[{"name":"x","score":2},{"name":"y", "score":2},{"name":"z","score":6}]' AS REPLY
UNION ALL
SELECT 4 AS ID
     , '[{"name":"x","score":5},{"name":"y", "score":8},{"name":"z","score":6}]' AS REPLY

那么如何找到所有具有“名称”的条目：“x”和“得分”：5。另外，我需要得到“名称”的“得分”值：该条目的“y”。

我目前的肮脏方法是

WITH JT1 AS (
        SELECT 1 AS ID
             , '[{"name":"x","score":2},{"name":"y", "score":8},{"name":"z","score":3}]' AS REPLY
        UNION ALL
        SELECT 2 AS ID
             , '[{"name":"x","score":5},{"name":"y", "score":4},{"name":"z","score":3}]' AS REPLY
        UNION ALL
        SELECT 3 AS ID
             , '[{"name":"x","score":2},{"name":"y", "score":2},{"name":"z","score":6}]' AS REPLY
        UNION ALL
        SELECT 4 AS ID
             , '[{"name":"x","score":5},{"name":"y", "score":8},{"name":"z","score":6}]' AS REPLY
)
SELECT ID
     , REGEXP_REPLACE(
         REGEXP_REPLACE( EXTRACTED, '^.*"y",\\s', '')
       , '[,\\]].*$', '') AS Y
     , EXTRACTED
FROM (
        SELECT ID
             , JSON_EXTRACT(REPLY, '$[*].name','$[*].score') EXTRACTED
        FROM JT1
) JT2
WHERE EXTRACTED RLIKE '"x", 5\\b'
;

所以我首先提取“名称”和“得分”，它给出了像["x", 5, "y", 4, "z", 3]这样的列数据。有了这个我做了一些讨厌的REGEXP搜索＆amp;取代

我觉得必须有更好的方法。

我尝试使用COLUMN_CREATE，但“COLUMN_CREATE”似乎无法接受来自JSON_EXTRACT的结果作为输入。现在我想一想，这似乎是合乎逻辑的，因为这里的“名称”和“得分”是正确排序的，但我能确定它总是那个序列吗？

任何人都可以给我一个如何做得更好的提示吗？

Answer 1

如果我了解您的需求，则如下查询可能会有用：

WITH `JT1` AS (
  SELECT 1 AS `ID`
  , '[{"name":"x","score":2},{"name":"y", "score":8},{"name":"z","score":3}]' AS `REPLY`
  UNION ALL
  SELECT 2 AS `ID`
  , '[{"name":"x","score":5},{"name":"y", "score":4},{"name":"z","score":3}]' AS `REPLY`
  UNION ALL
  SELECT 3 AS `ID`
  , '[{"name":"x","score":2},{"name":"y", "score":2},{"name":"z","score":6}]' AS `REPLY`
  UNION ALL
  SELECT 4 AS `ID`
  , '[{"name":"x","score":5},{"name":"y", "score":8},{"name":"z","score":6}]' AS `REPLY`
)
SELECT
  `ID`,
  `REPLY`,
  JSON_VALUE(
    `REPLY`,
    JSON_UNQUOTE(
      REPLACE(
        JSON_SEARCH(`REPLY`, 'one', 'y', NULL, '$[*].name'),
        'name',
        'score'
      )
    )
  ) `"name":"y"`
FROM
  `JT1`
WHERE
  JSON_VALUE(
    `REPLY`,
    JSON_UNQUOTE(
      REPLACE(
        JSON_SEARCH(`REPLY`, 'one', 'x', NULL, '$[*].name'),
        'name',
        'score'
      )
    )
  ) = 5;

请参见dbfiddle。

Mariadb查询并提取JSON数据

1 个答案: