简体   繁体   中英

How can I convert regexp to like?

I came across this question and apparently regexp are much slower than like. MySQL regexp much slow than like

How can I convert the following regexp to like?

WHERE category REGEXP  '[[:<:]](5|50|51|52|53|54|55|56|57|58|59|60|61|62|63|64|65|66|67|68|69|70|71|72|73|74|75|76|77|78|79|80|81|82|83|84|85|86|87|88|89|90|91|92|93|94|95|96|97|98|99|100|101|102|103|104|105|106|107|108|109|110|111|112|113|114|115|116|117|118|119|120|121|122|123|124|125|126|127|128|129|130|131|132|133|134|135|136|137|138|139|140|141|142|143|144|145|146|147|148|149|150|151|152|153|154|155|156|157|158|159|160|161|162|163|164|165|166|167|168|169|170|171|172|173|174|175|176|177|178|179|180|181|182|183|184|185|186|187|188|189|190|191|192|193|194|195|196|197|198|199|200|201|202|203|204|205|206|207|208|209|210|211|212|213|214|215|216|217|218|219|220|221|222|223|224|225|226|227|228|229|230|231|232|233|234|235|236|237)[[:>:]]'

categories stored the following way per post:

15,19,66,12

This is not a direct answer to your question. Instead, it is a brief explanation of how data structures should be stored in SQL.

You are storing a comma delimited list of categories on each row. You then want to see if one of a large set of categories are in the list. The appropriate data structure is an association table between the two. It would be defined as something like:

create table ItemCategory (
    ItemCategoryId int auto_increment not null,
    ItemId int not null,
    CategoryId int not null
);

Then, when you want a comma delimited list, you would do:

select ic.ItemId, group_concat(ic.CategoryId) as categories
from ItemCategory ic
group by ic.ItemId;

When you wanted to find items that have one of a large set of categories:

select distinct ItemId
from ItemCategory ic
where ic.CategoryId in (<long list here>);

Then if you have an index on CategoryId , this will use the index and perform well.

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM