本文介绍了如何获得LIKE和COUNT以返回行数少于该行中未包含的值的行数?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
例如:
SELECT COUNT(ID) FROM My_Table
WHERE ID <
(SELECT ID FROM My_Table
WHERE ID LIKE '%4'
ORDER BY ID LIMIT 1)
我的表:
X ID Y
------------------------
| | A1 | |
------------------------
| | B2 | |
------------------------
| | C3 | |
------------------------ -----Page 1
| | D3 | |
------------------------
| | E3 | |
------------------------
| | F5 | |
------------------------ -----Page 2
| | G5 | |
------------------------
| | F6 | |
------------------------
| | G7 | | -----Page 3
没有以4结尾的数据,但仍有5行最后的内容小于 %4
。
There is no data ending in 4 but there still are 5 rows that end in something less than "%4
".
但是,在这种情况下没有匹配项,因此SQLite只返回 0
However, in this case were there is no match, so SQLite only returns 0
我不知道它在那里,但是如何更改此行为以仍然返回它之前的行数,就好像它在那里?
I get it is not there but how do I change this behavior to still return number of rows before it, as if it was there?
有什么建议吗?
谢谢。
推荐答案
SELECT COUNT(ID) FROM My_Table
WHERE ID < (SELECT ID FROM My_Table
WHERE SUBSTRING(ID, 2) >= 4
ORDER BY ID LIMIT 1)
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