[英]Sort SQL records based on matched conditions
I have this query: 我有这个问题:
SELECT * FROM table WHERE key LIKE '1,2,3,%' OR key LIKE '1,2,%' OR key LIKE '1,%'
Is it posible to sort records returned from this query based on which conditions was matched first. 是否可以根据首先匹配的条件对从此查询返回的记录进行排序。 I'd like to get all records that match
key LIKE '1,2,3,%'
first, then key LIKE '1,2,%'
and the others after. 我想首先得到所有与
key LIKE '1,2,3,%'
匹配的记录,然后key LIKE '1,2,%'
以及之后的其他记录。
For example, if I have these records: 例如,如果我有这些记录:
key: "1,2,3,4"
key: "1,2,5"
key: "1,4"
key: "1,2,5,6"
key: "1,3"
key: "1,2,3,4,7"
key: "1,2,4"
I would like them to be sorted like so: 我希望他们像这样分类:
key: "1,2,3,4"
key: "1,2,3,4,7"
key: "1,2,4"
key: "1,2,5"
key: "1,2,5,6"
key: "1,3"
key: "1,4"
Is it possible to do? 有可能吗?
Use MATCH ... AGAINST
and order by rank. 使用
MATCH ... AGAINST
并按排名排序。 It exactly does what you want. 它完全符合你的要求。
.... ORDER BY CASE
WHEN key LIKE '1,2,3,%' THEN 1
WHEN key LIKE '1,2,%' THEN 2
ELSE 3
END
Does using "UNION" could do the job? 使用“UNION”可以完成这项工作吗?
SELECT * FROM table WHERE key LIKE '1,2,3,%' UNION SELECT * FROM table WHERE key LIKE '1,2,%' UNION SELECT * FROM table WHERE key LIKE key LIKE '1,%' SELECT * FROM table WHERE key LIKE'1,2,3,%'UNION SELECT * FROM table WHERE key LIKE'1,2,%'UNION SELECT * FROM table WHERE key LIKE key LIKE'1,%'
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