mysql匹配字符串与表中字符串的开头 [英] mysql match string with start of string in table

问题描述

我意识到，如果在创建表时可以对其进行修改，会容易得多，但是假设我做不到，那么我的表将如下所示:

I realise that it would be a lot easier if I could modify the table when it was created, but assuming I can't, I have a table that is such as:

abcd
abde
abdf
abff
bbsdf
bcggs
... snip large amount
zza

表中的值不是固定长度. 我有一个要匹配的字符串，例如abffagpokejfkjs. 如果相反，我可以做

The values in the table are not fixed length. I have a string to match such as abffagpokejfkjs . If it was the other way round, I could do

SELECT * from table where value like 'abff%'

但是我需要选择与所提供的字符串开头匹配的值.

but I need to select the value that matches the start of a string that is provided.

是否有一种快速的方法，还是需要在桌子上烦躁才能找到匹配项?

Is there a quick way of doing that, or does it need an itteration through the table to find a match?

推荐答案

尝试一下:

SELECT col1, col2 -- etc...
FROM your_table
WHERE 'abffagpokejfkjs' LIKE CONCAT(value, '%')

请注意，这将无法有效地使用索引，因此如果您有很多记录，它将变慢.

Note that this will not use an index effectively so it will be slow if you have a lot of records.

还要注意，value中的某些字符(例如%)可能被LIKE解释为具有特殊含义，这可能是不希望的.

Also note that some characters in value (e.g. %) may be interpreted by LIKE as having a special meaning, which may undesirable.

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