Python SQLAlchemy 等效于 SQL "LIKE" 语句
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SQLAlchemy equivalent to SQL "LIKE" statement
提问by Gary Oldfaber
A tags column has values like "apple banana orange" and "strawberry banana lemon". I want to find the SQLAlchemy equivalent statement to
标签列具有“苹果香蕉橙”和“草莓香蕉柠檬”等值。我想找到 SQLAlchemy 等效语句
SELECT * FROM table WHERE tags LIKE "%banana%";
What should I pass to Class.query.filter()to do this?
我应该通过什么Class.query.filter()来做到这一点？
采纳答案by Daniel Kluev
Each column has like()method, which can be used in query.filter(). Given a search string, add a %character on either side to search as a substring in both directions.
每列都有like()方法，可以在query.filter(). 给定一个搜索字符串，%在任一侧添加一个字符以作为两个方向的子字符串进行搜索。
tag = request.form["tag"]
search = "%{}%".format(tag)
posts = Post.query.filter(Post.tags.like(search)).all()
回答by igsm
Adding to the above answer, whoever looks for a solution, you can also try 'match' operator instead of 'like'. Do not want to be biased but it perfectly worked for me in Postgresql.
添加到上面的答案中，无论谁寻找解决方案，您也可以尝试使用“匹配”运算符而不是“喜欢”。不想有偏见，但它在 Postgresql 中对我来说非常有效。
Note.query.filter(Note.message.match("%somestr%")).all()
It inherits database functions such as CONTAINSand MATCH. However, it is not available in SQLite.
它继承了CONTAINS和MATCH等数据库功能。但是，它在 SQLite 中不可用。
For more info go Common Filter Operators
有关更多信息，请访问Common Filter Operators
回答by waruna k
try this code
试试这个代码
output = dbsession.query(<model_class>).filter(<model_calss>.email.ilike('%' + < email > + '%'))
回答by konstantin
Using PostgreSQL like(see accepted answer above) somehow didn't work for me although cases matched, but ilike(case insensisitive like) does.
使用 PostgreSQL like（见上面接受的答案）不知何故对我不起作用，尽管案例匹配，但ilike（案例我nsensisitive like）确实如此。
