SQLServer高效解析JSON格式数据的实例过程
1. 背景
最近碰到个需求,源数据存在posgtreSQL中,且为jsON格式。那如果在SQLServer中则 无法直接使用,需要先解析成表格行列结构化存储,再复用。
样例数据如下
‘[{“key”:“2019-01-01”,“value”:“4500.0”},{“key”:“2019-01-02”,“value”:“4500.0”},{“key”:“2019-01-03”,“value”:“4500.0”},{“key”:“2019-01-04”,“value”:“4500.0”},{“key”:“2019-01-05”,“value”:“4500.0”},{“key”:“2019-01-06”,“value”:“4500.0”},{“key”:“2019-01-07”,“value”:“4500.0”},{“key”:“2019-01-08”,“value”:“4500.0”},{“key”:“2019-01-09”,“value”:“4500.0”},{“key”:“2019-01-10”,“value”:“4500.0”},{“key”:“2019-01-11”,“value”:“4500.0”},{“key”:“2019-01-12”,“value”:“4500.0”},{“key”:“2019-01-13”,“value”:“4500.0”},{“key”:“2019-01-14”,“value”:“4500.0”},{“key”:“2019-01-15”,“value”:“4500.0”},{“key”:“2019-01-16”,“value”:“4500.0”},{“key”:“2019-01-17”,“value”:“4500.0”},{“key”:“2019-01-18”,“value”:“4500.0”},{“key”:“2019-01-19”,“value”:“4500.0”},{“key”:“2019-01-20”,“value”:“4500.0”},{“key”:“2019-01-21”,“value”:“4500.0”},{“key”:“2019-01-22”,“value”:“4500.0”},{“key”:“2019-01-23”,“value”:“4500.0”},{“key”:“2019-01-24”,&ldq编程客栈uo;value”:“4500.0”},{“key”:“2019-01-25”,“value”:“4500.0”},{“key”:“2019-01-26”,“value”:“4500.0”},{“key”:“2019-01-27”,“value”:“4500.0”},{“key”:“2019-01-28”,“value”:“4js500.0”},{“key”:“2019-01-29”,“value”:“4500.0”},{“key”:“2019-01-30”,“value”:“4500.0”},{“key”:“2019-01-31”,“value”:“4500.0”}]’
研究了下方法,可以先将 JSON串 拆成独立的 key-value对,再来对key-value子串做截取,获取两列数据值。
2. 拆串-拆分JSON串至key-value子串
这里主要利用行号和分隔符来组合完成拆分的功能。
参考如下样例。主要利用连续数值作为索引(起始值为1),从源字符串每个位置截取长度为1(分隔符的长度)的字符,如果为分隔符,则为有效的、待处理的记录。有点类似于生物DNA检测中的鸟枪法,先广撒网,再根据标记识别、追踪。/* * Date : 2020-07-01 * Author : 飞虹 * Sample : 拆分 指定分割符的字符串为单列多值 * Input : 字符串'jun,cong,haha' * Output : 列,值为 'jun', 'cong', 'haha' */ declare @s nvarchar(500) = 'jun,cong,haha' ,@sep nvarchar(5) = ','; with cte_Num as ( select 1 as n union all select n+1 n from cte_Num where n<100 ) select d.s, a.n ,n-len(replace(left(s, n), @sep, '')) + 1 as pos, CHARINDEX(@sep, s+@sep, n), substring(s, n, CHARINDEX(@sep, s+@sep, n)-n) as element from (select @s as s) as d join cte_Num a on n<=len编程客栈(s) and substring(@sep+s, n, 1) = @sep
3. 取值-创建函数截取key-value串的值
基于第2步的结果,可以将JSON长串拆分为 key-value字符串,如 “2020-01-01”:“98.99”。到这一步,就好办了。既可以自己写表值函数来返回结果,也可以直接通过substring来截取。这里开发一个表值函数,来进行封装。
/* ******************************************************************************* * Date : 2020-07-0编程客栈1 * Author : 飞虹 * Note : 利用patindex正则匹配字符,在while中对字符进行逐个匹配、替换为空。 * Function : getDateAmt * Input : key-value字符串,如 "2020-01-01":"98.99" * Output : Table类型(日期列,数值列)。值为 2020-01-01, 98.99 ******************************************************************************* */ CREATE FUNCTION dbo.getDateAmt(@S VARCHAR(100)) RETURNS @tb_rs table(dt date, amt decimal(28,14)) AS BEGIN WHILE PATINDEX('%[^0-9,-.]%',@S) > 0 BEGIN -- 匹配:去除非数字 、顿号、横线 的字符 set @s=stuff(@s,patindex('%[^0-9,-.]%',@s),1,'') END insert into @tb_rs select SUBSTRING(@s,1,charindex(',',@s)-1) , substring(@s,charindex(',',@s)+1, len(@s) ) return END GO --测试 select * from DBO.getDateAmt('{"key":"2019-01-01","value":"4500.0"')
4. 完整样例
附上完整脚本样例,全程CTE,直接查询,预览效果。
;with cte_t1 as ( select * from ( values('jun','[{"key":"2019-01-01","value":"4500.0"},{"key":"2019-01-02","value":"4500.0"},{"key":"2019-01-03","value":"4500.0"},{"key":"2019-01-04","value":"4500.0"},{"key":"2019-01-05","value":"4500.0"},{"key":"2019-01-06","value":"4500.0"},{"key":"2019-01-07","value":"4500.0"},{"key":"2019-01-08","value":"4500.0"},{"key":"2019-01-09","value":"4500.0"},{"key":"2019-01-10","value":"4500.0"},{"key":"2019-01-11","value":"4500.0"},{"key":"2019-01-12","value":"4500.0"},{"key":"2019-01-13","value":"4500.0"},{"keyjs":"2019-01-14","value":"4500.0"},{"key":"2019-01-15","value":"4500.0"},{"key":"2019-01-16","value":"4500.0"},{"key":"2019-01-17","value":"4500.0"},{"key":"2019-01-18","value":"4500.0"},{"key":"2019-01-19","value":"4500.0"},{"key":"2019-01-20","value":"4500.0"},{"key":"2019-01-21","value":"4500.0"},{"key":"2019-01-22","value":"4500.0"},{"key":"2019-01-23","value":"4500.0"},{"key":"2019-01-24","value":"4500.0"},{"key":"2019-01-25","value":"4500.0"},{"key":"2019-01-26","value":"4500.0"},{"key":"2019-01-27","value":"4500.0"},{"key":"2019-01-28","value":"4500.0"},{"key":"2019-01-29","value":"4500.0"},{"key":"2019-01-30","value":"4500.0"},{"key":"2019-01-31","value":"4500.0"}]') ,('congc','[{"key":"2019-01-01","value":"347.82608695652175"},{"key":"2019-01-02","value":"347.82608695652175"},{"key":"2019-01-03","value":"347.82608695652175"},{"key":"2019-01-04","value":"347.82608695652175"},{"key":"2019-01-07","value":"347.82608695652175"},{"key":"2019-01-08","value":"347.82608695652175"},{"key":"2019-01-09","value":"347.82608开发者_数据库教程695652175"},{"key":"2019-01-10","value":"347.82608695652175"},{"key":"2019-01-11","value":"347.82608695652175"},{"key":"2019-01-14","value":"347.82608695652175"},{"key":"2019-01-15","value":"347.82608695652175"},{"key":"2019-01-16","value":"347.82608695652175"},{"key":"2019-01-17","value":"347.82608695652175"},{"key":"2019-01-18","value":"347.82608695652175"},{"key":"2019-01-21","value":"347.82608695652175"},{"key":"2019-01-22","value":"347.82608695652175"},{"key":"2019-01-23","value":"347.82608695652175"},{"key":"2019-01-24","value":"347.82608695652175"},{"key":"2019-01-25","value":"347.82608695652175"},{"key":"2019-01-28","value":"347.82608695652175"},{"key":"2019-01-29","value":"347.82608695652175"},{"key":"2019-01-30","value":"347.82608695652175"},{"key":"2019-01-31","value":"347.82608695652175"}]') ) as t(name, jsonStr) ) , cte_rn as ( select 1 as rn union all select rn+1 from cte_rn where rn < 1000 ) , cte_splitJson as ( SELECT a.name ,replace(replace(a.jsonStr,'[',''),']','') as jsonStr ,substring(replace(replace(a.jsonStr,'[',''),']','') , b1.rn , charindex('},', replace(replace(a.jsonStr,'[',''),']','')+'},', b1.rn)-b1.rn ) as value_json from cte_t1 a cross join cte_rn b1 where substring('},'+replace(replace(a.jsonStr,'[',''),']',''), rn, 2) = '},' ) select * from cte_splitJson a cross apply dbo.getDateAmt(a.value_json) as t1 -- 注意这里生成行号时, 需要设置默认递归次数 option(maxrecursion 0)
5. 问题
经过在个人普通配置PC实测,性能有点堪忧,耗时:数据量 约为15mins:50W ,不太能接受。有兴趣或者经历过的伙伴,出手来协助, 怎么提高效率,或者来个新方案?
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