[英]JPA select query with where clause
I want to write a select statement but can't figure out how to write the where clause... 我想编写一个select语句,但无法弄清楚如何编写where子句...
My code: 我的代码:
CriteriaQuery query = entityManager.getCriteriaBuilder().createQuery();
query.select(query.from(SecureMessage.class)).where();
This is within a method that I am passing a string to. 这是我传递字符串的方法。 I want to fetch only the rows that match the value of the string that Im passing to the method.
我想只获取与Im传递给方法的字符串值匹配的行。
In Criteria this is something like: 在Criteria中,这类似于:
CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<SecureMessage> query = cb.createQuery(SecureMessage.class);
Root<SecureMessage> sm = query.from(SecureMessage.class);
query.where(cb.equal(sm.get("someField"), "value"));
In JPQL: 在JPQL中:
Query query = entityManager.createQuery("Select sm from SecureMessage sm where sm.someField=:arg1");
query.setParameter("arg1", arg1);
See, http://en.wikibooks.org/wiki/Java_Persistence/Querying#Criteria_API_.28JPA_2.0.29 见, http://en.wikibooks.org/wiki/Java_Persistence/Querying#Criteria_API_.28JPA_2.0.29
As I understand, the method parameter should be the parameter of the query. 据我了解,方法参数应该是查询的参数。
So, should looks like: 所以,应该看起来像:
Query query = entityManager.getCriteriaBuilder().createQuery("from SecureMessage sm where sm.someField=:arg1");
query.setParameter("arg1", arg1);
where arg1
- your method String parameter 其中
arg1
- 您的方法String参数
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