在同一表中的一列上查找不同值的SQL查询

我有下表，其中

Description

字段有

scope=10

，我插入了

Description

=

Cobra-Ni

的第一行，但几天前我意识到我需要扩展更大的范围，正确的值应该是

Cobra-Nisyor

我需要编写查询，从这个示例表中查找第一行和最后一行

我试着这样做：

SELECT t1.*
FROM table as t1
WHERE t1.Description in
      (SELECT t2.Description
       FROM table as t2
       WHERE t1.Doc_nr = t2.Doc_nr
       AND t1.Description != t2.Description)

但是它不起作用。

我假设“范围”是指列的宽度为10。因此，您需要关联行，一个行的长度为10，另一个行的长度大于10，并且以相同的字符串开始。我们可以使用

LEN（）

函数来获取字符字段的长度，并使用

LEFT（）

来获取子字符串，后者可以用来比较“新”和“旧”

例如：

with oldRows as (
    select *
    from myTable
    where LEN(Description) = 10
), newRows as (
    select *, LEFT(Description, 10) as oldKey
    from myTable
    where LEN(Description) > 10
)
select n.*, o.*
from oldRows o
join newRows n on o.Description = n.oldKey
-- Of course add any other comparisons you need to correlate rows:
--    and o.Column_ref = n.Column_ref
--    and o.[Date] = n.[Date]
--    and o.[Money] = n.[Money]
--    and o.Doc_nr = n.Doc_nr

---

为便于将来参考，您可能不应该在意识到问题后在表中插入额外的新行，而应该使用更新来代替。

要查找您要查找的行，您需要在

文档编号

上执行一次操作，仅包括描述不匹配的行

---

上述语法的功劳。

订购列在哪里？您专门搜索具有不同描述的项目，然后您希望它是相同的？即，当同一日期有多行时，您如何选择第一个/最后一个？您是否在寻找具有相同文档编号但不同描述的行？根据您的要求和提供的数据：

其中文档编号='2000/10'

您的答案肯定是正确的。我在SQLFiddle上检查了它，它正常工作了。我想我的数据有问题，但我知道你的解决方案也是正确的。非常感谢SQL Fiddle，我以前不知道这一点！

with oldRows as (
    select *
    from myTable
    where LEN(Description) = 10
), newRows as (
    select *, LEFT(Description, 10) as oldKey
    from myTable
    where LEN(Description) > 10
)
select n.*, o.*
from oldRows o
join newRows n on o.Description = n.oldKey
-- Of course add any other comparisons you need to correlate rows:
--    and o.Column_ref = n.Column_ref
--    and o.[Date] = n.[Date]
--    and o.[Money] = n.[Money]
--    and o.Doc_nr = n.Doc_nr

CREATE TABLE basic ( column_ref INT, description VARCHAR(30), dateField DATETIME, amount DECIMAL(12,2), doc_nr VARCHAR(30) ); INSERT INTO basic (column_ref, description, dateField, amount, doc_nr) VALUES (123, 'Cobra - Ni', '06/11/2015',505.50,'2000/10'), (123, 'Cobra - Toung', '07/11/2015',505.50,'2000/12'), (123, 'Cobra - Brain', '07/11/2015',505.50,'2000/25'), (123, 'Cobra - Nisyor', '07/11/2015',505.50,'2000/10'); SELECT * FROM basic b JOIN basic q ON b.doc_nr = q.doc_nr WHERE b.description != q.description ╔════════════╦════════════════╦════════════════════════╦════════╦═════════╦════════════╦════════════════╦════════════════════════╦════════╦═════════╗ ║ column_ref ║ description ║ dateField ║ amount ║ doc_nr ║ column_ref ║ description ║ dateField ║ amount ║ doc_nr ║ ╠════════════╬════════════════╬════════════════════════╬════════╬═════════╬════════════╬════════════════╬════════════════════════╬════════╬═════════╣ ║ 123 ║ Cobra - Ni ║ June, 11 2015 00:00:00 ║ 505.5 ║ 2000/10 ║ 123 ║ Cobra - Nisyor ║ July, 11 2015 00:00:00 ║ 505.5 ║ 2000/10 ║ ║ 123 ║ Cobra - Nisyor ║ July, 11 2015 00:00:00 ║ 505.5 ║ 2000/10 ║ 123 ║ Cobra - Ni ║ June, 11 2015 00:00:00 ║ 505.5 ║ 2000/10 ║ ╚════════════╩════════════════╩════════════════════════╩════════╩═════════╩════════════╩════════════════╩════════════════════════╩════════╩═════════╝

DELETE b 
FROM basic b
JOIN basic q ON b.doc_nr = q.doc_nr
WHERE LEN(b.description) < LEN(q.description);