[英]The Java Persistence Query Language - LIKE
我有一个基于Spring Web model-view-controller(MVC)框架的项目。 Spring Web模型-视图-控制器(MVC)框架的版本为3.2.8。
我的DAO中有这种方法
@Override
public List<Application> findByQuickSearch(String searchString) {
final StringBuilder queryString = new StringBuilder(" select app from Application app where upper (ticket_id) like :searchString or upper (id) like :searchString " );
queryString.append(" and app.status != " + Status.DRAFT.ordinal());
queryString.append(" order by app.submissionTime desc ");
try {
final Query query = getEntityManager().createQuery(queryString.toString());
searchString = searchString.replace("!", "!!")
.replace("%", "!%")
.replace("_", "!_")
.replace("[", "![")
.trim()
.toUpperCase();
System.out.println ("searchString -----> " + searchString);
query.setParameter ("searchString", searchString);
return query.getResultList();
} catch (RuntimeException re) {
log.error("findByCompetentBodyAndStatus failed", re);
throw re;
}
}
但是我意识到查询不执行LIKE而是等于
我要寻找的弹簧是"iOS/032/027"
,对于"iOS/032/027"
,但对于"iOS/"
或"027"
不是
尝试将有效的通配符附加到like操作:
str_param like concat(:searchString,'%')
Jpql文档状态
LIKE评估两个字符串匹配'%'和'_'是否为有效的通配符,而ESCAPE字符是否为可选
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