由网友(花不自羞)分享简介：我想做的事情是这样的：I want to do something like:SELECT * FROM USER WHERE NAME LIKE '%Smith%';我在阿雷尔的尝试：My attempt in Arel:# params[:query] = 'Smith'User.where("name...

我想做的事情是这样的：

I want to do something like:

SELECT * FROM USER WHERE NAME LIKE '%Smith%';

我在阿雷尔的尝试：

My attempt in Arel:

# params[:query] = 'Smith'
User.where("name like '%?%'", params[:query]).to_sql

然而，这变为：

However, this becomes:

SELECT * FROM USER WHERE NAME LIKE '%'Smith'%';

阿雷尔包裹的查询字符串'Smith的正确，而是因为这是一个LIKE语句它不工作。

Arel wraps the query string 'Smith' correctly, but because this is a LIKE statement it doesnt work.

怎样才能做一个LIKE查询阿雷尔？

How does one do a LIKE query in Arel?

P.S。奖金 - 我其实是想扫描在桌子上两个字段，这两个名称和描述，来看看是否有任何匹配的查询。如何将这项工作？

P.S. Bonus--I am actually trying to scan two fields on the table, both name and description, to see if there are any matches to the query. How would that work?

推荐答案

这是你如何执行AREL类似的查询：

This is how you perform a like query in arel:

users = User.arel_table
User.where(users[:name].matches("%#{user_name}%"))

PS：

users = User.arel_table
query_string = "%#{params[query]}%"
User.where(users[:name].matches(query_string)
                       .or(users[:description].matches(query_string)))