MySQL:JSON 到结果集

Question

我有一个存档表，其中有一列包含未索引的 json 记录。示例:

[{
    "attr1": "val1",
    "attr2": "val2",
    "attr3": "val3",
    .
    .
    "attrN": "valN"
}, {
    "attr1": "val1.2",
    "attr2": "val2.2",
    "attr3": "val3.2",
    .
    .
    "attrN": "valN.2"
},...]

我需要创建一个将 json 作为结果集返回的存储过程或函数:

  attr1   |   attr2    |   attr3  | ... |  attrN
  _______________________________________________
  val1    |    val2    |   val3   | ... |  valN
  val1.2  |    val2.2  |   val2.2 | ... |  valN.2

我需要它作为记录集返回，因为我将它用于其他存储过程中的其他查询。

我能够通过以下方式实现这一目标:Read JSON array in MYSQL

但是我想知道除了创建临时表还有别的方法吗？我在考虑性能和效率。比如，如果 20 或 50 个用户触发了这个怎么办？有更好的方法吗？

注意:json 的平均大小约为 1MB

Answer 1

此存储过程创建一个动态查询，输出您需要的结果，应使用真实数据进行测试以检查性能是否足够好:

CREATE PROCEDURE jsonToCols(IN json JSON, IN target VARCHAR(50))
BEGIN
    SET @j = 0;
    SET @fields = JSON_KEYS(json, "$[0]");
    SET @f_length = JSON_LENGTH(@fields);
    SET @select = "";
    WHILE @j < JSON_LENGTH(json) DO
      SET @i = 0;
      SET @select = CONCAT(@select, "SELECT ");
      WHILE @i < @f_length DO
          SET @attr_name = REPLACE(JSON_EXTRACT(@fields, CONCAT("$[", @i, "]")), '\"', '');
          SET @key = CONCAT('$.', @attr_name);
          SET @select = CONCAT(@select, "JSON_EXTRACT(json, '$[", @j , "].", @attr_name, "') as ", @attr_name, ", ");
          SET @i = @i + 1;
      END WHILE;

      SET @select = CONCAT(LEFT(@select, length(@select)-2), " FROM ", target ," UNION ALL ");
      SET @j = @j + 1;
    END WHILE;

    SET @select = LEFT(@select, length(@select)-11);

    PREPARE stmt FROM @select;
    EXECUTE stmt;
    DEALLOCATE PREPARE stmt;
END;

动态创建的查询如下所示:

SELECT JSON_EXTRACT(json, '$[0].attr1') as attr1, JSON_EXTRACT(json, '$[0].attr2') as attr2, JSON_EXTRACT(json, '$[0].attr3') as attr3, JSON_EXTRACT(json, '$[0].attrN') as attrN 
FROM test 
UNION ALL 
SELECT JSON_EXTRACT(json, '$[1].attr1') as attr1, JSON_EXTRACT(json, '$[1].attr2') as attr2, JSON_EXTRACT(json, '$[1].attr3') as attr3, JSON_EXTRACT(json, '$[1].attrN') as attrN 
FROM test

FIDDLE