java 用于搜索部分字符串的休眠查询
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Hibernate query for searching part of string
提问by Mahin
I'm trying to write a hibernate query to search if table Room contains roomname which contains part of string.The string value is in a variable. I wrote a query to get exact room name from the table.
我正在尝试编写一个休眠查询来搜索表 Room 是否包含包含部分字符串的房间名。字符串值在一个变量中。我写了一个查询来从表中获取确切的房间名称。
findRoom(String name) {
Query query = em.createQuery("SELECT a FROM Room a WHERE a.roomname=?1");
query.setParameter(1, name);
List rooms = query.getResultList();
return rooms;
}
In sql the query is something like this:
在 sql 中，查询是这样的：
mysql_query("
SELECT *
FROM `table`
WHERE `column` LIKE '%"name"%' or '%"name"' or '"name"%'
");
I want to know the hql query for searching the table that matches my query. I can not use string directly, so the search query has to be veriable based and I need all three types in a query, if it's begin with name, or contains name or ends name.
我想知道用于搜索与我的查询匹配的表的 hql 查询。我不能直接使用字符串，所以搜索查询必须是基于可验证的，我需要查询中的所有三种类型，如果它以名称开头，或者包含名称或结尾名称。
回答by Maurice Perry
I would do something like that:
我会做这样的事情：
findRoom(String name) {
Query query = em.createQuery("SELECT a FROM Room a"
+ "WHERE a.roomname LIKE CONCAT('%',?1,'%')");
query.setParameter(1, name);
List rooms = query.getResultList();
return rooms;
}
回答by Jens
Use likeinstead of =:
使用like代替=：
Query query = em.createQuery("SELECT a FROM Room a WHERE a.roomname like ?1");
query.setParameter(1, "%"+name+"%");
