如何在mysql中找到序列编号的空白？

Question

我们有一个数据库,其表格的值是从另一个系统导入的.有一个自动增量列,没有重复值,但缺少值.例如,运行此查询:

select count(id) from arrc_vouchers where id between 1 and 100

应该返回100,但它返回87.有没有我可以运行的查询将返回缺失数字的值？例如,id为1-70和83-100的记录可能存在,但没有id为71-82的记录.我想返回71,72,73等

这可能吗？

Answer 1

更新

ConfexianMJS 在性能方面提供了更好的 答案.

(尽可能不快)答案

这里的版本适用于任何大小的表(不仅仅是100行):

SELECT (t1.id + 1) as gap_starts_at, 
       (SELECT MIN(t3.id) -1 FROM arrc_vouchers t3 WHERE t3.id > t1.id) as gap_ends_at
FROM arrc_vouchers t1
WHERE NOT EXISTS (SELECT t2.id FROM arrc_vouchers t2 WHERE t2.id = t1.id + 1)
HAVING gap_ends_at IS NOT NULL

• gap_starts_at - 当前差距中的第一个id
• gap_ends_at - 当前差距中的最后一个ID

Answer 2

这对我来说只是找到一个超过80k行的表中的空白:

SELECT
 CONCAT(z.expected, IF(z.got-1>z.expected, CONCAT(' thru ',z.got-1), '')) AS missing
FROM (
 SELECT
  @rownum:=@rownum+1 AS expected,
  IF(@rownum=YourCol, 0, @rownum:=YourCol) AS got
 FROM
  (SELECT @rownum:=0) AS a
  JOIN YourTable
  ORDER BY YourCol
 ) AS z
WHERE z.got!=0;

结果:

+------------------+
| missing          |
+------------------+
| 1 thru 99        |
| 666 thru 667     |
| 50000            |
| 66419 thru 66456 |
+------------------+
4 rows in set (0.06 sec)

请注意,列的顺序expected和got是至关重要的.

如果您知道YourCol不是从1开始并且无关紧要,则可以替换

(SELECT @rownum:=0) AS a

同

(SELECT @rownum:=(SELECT MIN(YourCol)-1 FROM YourTable)) AS a

新结果:

+------------------+
| missing          |
+------------------+
| 666 thru 667     |
| 50000            |
| 66419 thru 66456 |
+------------------+
3 rows in set (0.06 sec)

如果您需要对缺少的ID执行某种shell脚本任务,您还可以使用此变体来直接生成可以在bash中迭代的表达式.

SELECT GROUP_CONCAT(IF(z.got-1>z.expected, CONCAT('$(',z.expected,' ',z.got-1,')'), z.expected) SEPARATOR " ") AS missing
FROM (  SELECT   @rownum:=@rownum+1 AS expected,   IF(@rownum=height, 0, @rownum:=height) AS got  FROM   (SELECT @rownum:=0) AS a   JOIN block   ORDER BY height  ) AS z WHERE z.got!=0;

FROM(SELECT @rownum:= @ rownum + 1 AS预期,IF(@ rownum = height,0,@ rownum:= height)AS get FROM(SELECT @rownum:= 0)AS JOIN块ORDER BY height)AS z在哪里z.got!= 0;

这会产生类似的输出

$(seq 1 99) $(seq 666 667) 50000 $(seq 66419 66456)

然后,您可以将其复制并粘贴到bash终端中的for循环中,以便为每个ID执行命令

for ID in $(seq 1 99) $(seq 666 667) 50000 $(seq 66419 66456); do
  echo $ID
  # fill the gaps
done

它与上面的内容相同,只是它既可读又可执行.通过更改上面的"CONCAT"命令,可以为其他编程语言生成语法.或者甚至可能是SQL.

Answer 3

快速和肮脏的查询,应该做的伎俩:

SELECT a AS id, b AS next_id, (b - a) -1 AS missing_inbetween
FROM 
 (
SELECT a1.id AS a , MIN(a2.id) AS b 
FROM arrc_vouchers  AS a1
LEFT JOIN arrc_vouchers AS a2 ON a2.id > a1.id
WHERE a1.id <= 100
GROUP BY a1.id
) AS tab

WHERE 
b > a + 1

这将为您提供一个表格,显示其上方缺少ID的ID,以及存在的next_id,以及...之间缺少的ID

 
id  next_id  missing_inbetween
 1        4                  2
68       70                  1
75       87                 11

Answer 4

如果您使用，MariaDB则可以使用序列存储引擎更快（800％）的选择：

SELECT * FROM seq_1_to_50000 WHERE SEQ NOT IN (SELECT COL FROM TABLE);