6

我有一个问卷反馈的数据库表(Oracle 11g),包括多项选择、多项回答问题。Options 列包含用户可以选择的每个值,Answers 列包含他们选择的数值。

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

我需要一个查询来解码答案,答案的文本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

我已经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;有许多带有不同问题(以及不同数量的选项)的调查,所以我希望有一个解决方案可以动态地适用于所有这些。

我尝试将选项和答案按 LEVEL 拆分为单独的行,然后在代码匹配的地方重新加入它们,但这对于实际数据集运行速度非常慢(一个包含 600 行响应的 5 选项问题)。

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

If there isn't a solution just using Regex, is there a more efficient way than LEVEL to split the values? Or is there another approach that would work?

4

4 回答 4

1

看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
于 2016-01-25T19:27:26.233 回答
1

这很慢,因为您将每一行扩展太多次;您正在使用的 connect-by 子句正在查看所有行,因此您最终会得到大量数据然后进行排序 - 这大概就是您最终DISTINCT在其中找到的原因。

您可以在 connect-by 中添加两个PRIOR子句,第一个ID_NO是保留,第二个是避免循环 - 我选择了任何非确定性函数都可以做到这一点,dbms_random.valuesys_guid如果您愿意,可以使用,或者其他. 你也不需要很多子查询,你可以用两个来完成;或者作为我认为更清晰的 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;

得到:

     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  

我还更改了您的正则表达式模式,因此您不必附加或删除|.

于 2016-01-25T18:34:23.083 回答
0

创建一个存储的 predure 并执行以下步骤

  • 声明一个你的大小的数组。
  • option从第一行获取数据。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:这将只是一次性的。所以你不需要为每一行重复它。
  • 现在在一个循环中,对于每一行,选择answers并使用数组值来分配answers
于 2016-01-25T18:29:44.977 回答
0

我已经在 MySQL 中编写了一个关闭的解决方案(现在没有安装 Oracle)-但我已经编写了需要更改的内容才能使查询在 Oracle 中工作。

此外,我的代码中最难看的部分在 Oracle 中会更好看,因为它具有更好的 INSTR 函数。

我们的想法是使用数字列表(1 到 10 以支持每个调查最多 10 个选项)进行 CROSS JOIN,并将 OPTIONS 字段分解为不同的行...(您可以同时使用数字列表和 Oracle 的 INSTR 函数,请参阅注释)。

从那里您过滤掉未选择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
于 2016-01-25T18:18:23.413 回答