SQL Regex - 替换为另一个字段的子字符串

时间:2016-01-25 16:41:02

标签: sql regex oracle regexp-replace regexp-substr

我有一个问卷反馈的数据库表(Oracle 11g),包括多项选择,多个答案问题。 Options列具有用户可以选择的每个值,Answers列具有他们选择的数值。

ID_NO     OPTIONS                               ANSWERS
1001      Apple Pie|Banana-Split|Cream Tea      1|2
1002      Apple Pie|Banana-Split|Cream Tea      2|3
1003      Apple Pie|Banana-Split|Cream Tea      1|2|3

我需要一个能够解码答案的查询,并将答案的文字版本作为单个字符串。

ID_NO     ANSWERS     ANSWER_DECODE
1001      1|2         Apple Pie|Banana-Split
1002      2|3         Banana-Split|Cream Tea
1003      1|2|3       Apple Pie|Banana-Split|Cream Tea

我已尝试使用正则表达式来替换值并获取子字符串,但我无法找到正确合并这两者的方法。

WITH feedback AS (
  SELECT 1001 id_no, 'Apple Pie|Banana-Split|Cream Tea' options, '1|2' answers FROM DUAL UNION
  SELECT 1002 id_no, 'Apple Pie|Banana-Split|Cream Tea' options, '2|3' answers FROM DUAL UNION
  SELECT 1003 id_no, 'Apple Pie|Banana-Split|Cream Tea' options, '1|2|3' answers FROM DUAL )
SELECT 
  id_no,
  options,
  REGEXP_SUBSTR(options||'|', '(.)+?\|', 1, 2) second_option,
  answers,
  REGEXP_REPLACE(answers, '(\d)+', ' \1 ') answer_numbers,
  REGEXP_REPLACE(answers, '(\d)+', REGEXP_SUBSTR(options||'|', '(.)+?\|', 1, To_Number('2'))) "???"
FROM feedback

我不想手动定义或解码SQL中的答案;有许多调查都有不同的问题(以及不同数量的选项),所以我希望有一个能够为所有问题动态运作的解决方案。

我试图通过LEVEL将选项和答案拆分成单独的行,并在代码匹配的地方重新加入它们,但实际数据集的运行速度非常慢(带有600行的5选项问题)响应)。

WITH feedback AS (
  SELECT 1001 id_no, 'Apple Pie|Banana-Split|Cream Tea' options, '1|2' answers FROM DUAL UNION
  SELECT 1002 id_no, 'Apple Pie|Banana-Split|Cream Tea' options, '2|3' answers FROM DUAL UNION
  SELECT 1003 id_no, 'Apple Pie|Banana-Split|Cream Tea' options, '1|2|3' answers FROM DUAL )
SELECT
    answer_rows.id_no,
    ListAgg(option_rows.answer) WITHIN GROUP(ORDER BY option_rows.lvl)
FROM
  (SELECT DISTINCT
    LEVEL lvl,
    REGEXP_SUBSTR(options||'|', '(.)+?\|', 1, LEVEL) answer
  FROM
    (SELECT DISTINCT
      options,
      REGEXP_COUNT(options||'|', '(.)+?\|') num_choices
    FROM
      feedback)
  CONNECT BY LEVEL <= num_choices
  ) option_rows
  LEFT OUTER JOIN
  (SELECT DISTINCT
    id_no,
    to_number(REGEXP_SUBSTR(answers, '(\d)+', 1, LEVEL)) answer
  FROM
    (SELECT DISTINCT
      id_no,
      answers,
      To_Number(REGEXP_SUBSTR(answers, '(\d)+$')) max_answer
    FROM
      feedback)
  WHERE
    to_number(REGEXP_SUBSTR(answers, '(\d)+', 1, LEVEL)) IS NOT NULL
  CONNECT BY LEVEL <= max_answer
  ) answer_rows
    ON option_rows.lvl = answer_rows.answer
GROUP BY
    answer_rows.id_no
ORDER BY
  answer_rows.id_no

如果没有使用正则表达式的解决方案,是否有比LEVEL更有效的方法来拆分值?或者是否有另一种方法可行?

4 个答案:

答案 0 :(得分:1)

它很慢,因为你将每一行扩展太多次;您正在使用的连接条款正在查看所有行,因此您需要输入大量数据才能进行排序 - 这可能是您最终使用DISTINCT的原因在那里。

你可以在连接方面添加两个PRIOR子句,首先保留ID_NO,第二个避免循环 - 任何非确定性函数都会为此做,我&# 39;已选择dbms_random.value,但如果您愿意,可以使用sys_guid或其他内容。你也不需要很多子查询,你可以用两个子查询;或者作为我认为更清楚的CTE:

WITH feedback AS (
  SELECT 1001 id_no, 'Apple Pie|Banana-Split|Cream Tea' options, '1|2' answers FROM DUAL UNION
  SELECT 1002 id_no, 'Apple Pie|Banana-Split|Cream Tea' options, '2|3' answers FROM DUAL UNION
  SELECT 1003 id_no, 'Apple Pie|Banana-Split|Cream Tea' options, '1|2|3' answers FROM DUAL
),
option_rows AS (
  SELECT
    id_no,
    LEVEL answer,
    REGEXP_SUBSTR(options, '[^|]+', 1, LEVEL) answer_text
  FROM feedback
  CONNECT BY LEVEL <= REGEXP_COUNT(options, '[^|]+')
  AND id_no = PRIOR id_no
  AND PRIOR dbms_random.value IS NOT NULL
),
answer_rows AS (
  SELECT
    id_no,
    REGEXP_SUBSTR(answers, '[^|]+', 1, LEVEL) answer
  FROM feedback
  CONNECT BY LEVEL <= REGEXP_COUNT(answers, '[^|]+')
  AND PRIOR id_no = id_no
  AND PRIOR dbms_random.value IS NOT NULL
)
SELECT
  option_rows.id_no,
  LISTAGG(option_rows.answer, '|') WITHIN GROUP (ORDER BY option_rows.answer) AS answers,
  LISTAGG(option_rows.answer_text, '|') WITHIN GROUP (ORDER BY option_rows.answer) AS answer_decode
FROM option_rows
JOIN answer_rows
ON option_rows.id_no = answer_rows.id_no
AND option_rows.answer = answer_rows.answer
GROUP BY option_rows.id_no
ORDER BY option_rows.id_no;

获得:

     ID_NO ANSWERS    ANSWER_DECODE                          
---------- ---------- ----------------------------------------
      1001 1|2        Apple Pie|Banana-Split                  
      1002 2|3        Banana-Split|Cream Tea                  
      1003 1|2|3      Apple Pie|Banana-Split|Cream Tea  

我也改变了您的正则表达式模式,因此您不必附加或删除|

答案 1 :(得分:1)

查看这个紧凑的解决方案:

   with sample_data as
(
  select 'ala|ma|kota' options, '1|2' answers from dual
  union all
  select 'apples|oranges|bacon', '1|2|3' from dual
  union all
  select 'a|b|c|d|e|f|h|i','1|3|4|5|8' from dual
)
select answers, options,
regexp_replace(regexp_replace(options,'([^|]+)\|([^|]+)\|([^|]+)','\' || replace(answers,'|','|\')),'[|]+','|') answer_decode
from sample_data;

输出:

  ANSWERS   OPTIONS              ANSWER_DECODE
--------- -------------------- ---------------------------
1|2       ala|ma|kota          ala|ma
1|2|3     apples|oranges|bacon apples|oranges|bacon
1|3|4|5|8 a|b|c|d|e|f|h|i      a|c|d|f|h|i

答案 2 :(得分:0)

我在MySQL中编写了一个密切的解决方案(现在没有安装Oracle) - 但我已经编写了需要更改的内容,以便查询在Oracle中运行。

此外,我的代码中最丑陋的部分在Oracle中会更好看,因为它具有更好的INSTR功能。

我的想法是用一个数字列表(1到10进行CROSS JOIN,以便每次调查最多支持10个选项),并将OPTIONS字段分解为不同的行...(你这样做使用数字列表和Oracle的INSTR函数,请参阅注释。

从那里过滤掉未选中的行,并将所有内容重新组合在一起。

-- I've used GROUP_CONCAT in MySQL, but in Oracle you'll have to use WM_CONCAT
select ID_NO, ANSWERS, group_concat(broken_down_options,'|') `OPTIONS`
from (
    select your_table.ID_NO, your_table.ANSWERS, 
            -- Luckily, you're using ORACLE so you can use an INSTR function that has the "occurrence" parameter
            -- INSTR(string, substring, [position, [occurrence]])
            -- use the nums.num field as input for the occurrence parameter
            -- and just put '1' under "position"
            case when nums.num = 1 
                then substr(your_table.`OPTIONS`, 1, instr(your_table.`OPTIONS`, '|') - 1)
                when nums.num = 2
                then substr(substr(your_table.`OPTIONS`, instr(your_table.`OPTIONS`, '|') + 1), 1, instr(substr(your_table.`OPTIONS`, instr(your_table.`OPTIONS`, '|') + 1), '|') - 1)
                else substr(your_table.`OPTIONS`,  length(your_table.`OPTIONS`) - instr(reverse(your_table.`OPTIONS`), '|') + 2) end broken_down_options
    from (select 1 num union all
        select 2 num union all
        select 3 num union all
        select 4 num union all
        select 5 num union all
        select 6 num union all
        select 7 num union all
        select 8 num union all
        select 9 num union all
        select 10 num
        ) nums 
        CROSS JOIN
        (select 1001 ID_NO, 'Apple Pie|Banana-Split|Cream Tea' `OPTIONS`, '1|2' ANSWERS union
        select 1002 ID_NO, 'Apple Pie|Banana-Split|Cream Tea' `OPTIONS`, '2|3' ANSWERS union
        select 1003 ID_NO, 'Apple Pie|Banana-Split|Cream Tea' `OPTIONS`, '1|2|3' ANSWERS
        ) your_table
    -- for example: 2|3 matches 2 and 3 but not 1
    where your_table.ANSWERS like concat(concat('%',nums.num),'%')
) some_query
group by ID_NO, ANSWERS

答案 3 :(得分:0)

创建存储的预测并执行以下步骤

  • 声明您的大小数组。
  • 从第一行获取option个数据。使用正则表达式或level在管道之间提取值,然后将它们存储在数组中。注意:这只是一次性的。所以你不需要为每一行重复它。
  • 现在循环,对于每一行,选择answers并使用数组值来指定answers
  • 的值