java可选参数JPA2.1
我有这门课
@Entity
public class Message {
@Id @GeneratedValue(strategy=GenerationType.IDENTITY)
private long id;
@Column( nullable = false )
private String mobile;
@Column( nullable = false )
private String message;
@Column( nullable = false )
private Lang lang;
@Column( nullable = false )
private int status;
@Column( nullable = false )
private Calendar creationDate;
...
}
我希望能够使用表单中的可选参数查询表。我正在使用JPA2.1和Hibernate。有没有办法在不使用多个查询语句或不将我的应用程序与hibernate或我正在使用的特定数据库紧密耦合的情况下做到这一点
以下是我正在使用的表单:
<form action="${pageContext.request.contextPath}/search">
<label>From</label>
<input type="datetime" placeholder="dd/mm/yy hh/mm" name="from" required>
<label>To</label>
<input type="datetime" placeholder="dd/mm/yy hh/mm" name="to" required>
<select name="lang">
<option value="english">English</option>
<option value="arabic">Arabic</option>
</select>
<input type="tel" placeholder="Recipient" name="recipient">
<select name="status">
<option value="new">New</option>
<option value="picked">Picked</option>
<option value="sent">Sent</option>
<option value="rejected">Rejected</option>
</select>
<input type="submit" value="Search">
更新::
我试过这个:
public List<Message> getMessages(Calendar fromDate, Calendar toDate, String lang, String recipient, String status) {
/*TypedQuery<Message> query = em.createQuery(
"SELECT e FROM Message e WHERE e.creationDate BETWEEN :fromDate AND :toDate", Message.class);
query.setParameter("fromDate", fromDate);
query.setParameter("toDate", toDate);
List<Message> list = query.getResultList();
return list;*/
List<Predicate> predicates = new ArrayList<>();
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Message> query = cb.createQuery(Message.class);
Root<Message> root = query.from(Message.class);
predicates.add(cb.between(root.get("creationDate"), fromDate, toDate));
if (lang != null && !lang.equalsIgnoreCase("any")) {
predicates.add(cb.equal(root.get("lang"), lang));
}
if (recipient != null && !recipient.equals("")) {
predicates.add(cb.equal(root.get("mobile"), recipient));
}
if (status != null && !status.equalsIgnoreCase("any")) {
predicates.add(cb.equal(root.get("mobile"), recipient));
}
query.where(cb.and(predicates.toArray(new Predicate[predicates.size()])));
List<Message> list = em.createQuery(query).getResultList();
return list;
}
但我得到了以下错误:
java.lang.IllegalArgumentException: Parameter value [0] did not match expected type [entity.Lang (n/a)]
at org.hibernate.jpa.spi.BaseQueryImpl.validateBinding(BaseQueryImpl.java:897)
at org.hibernate.jpa.internal.QueryImpl.access$000(QueryImpl.java:61)
at org.hibernate.jpa.internal.QueryImpl$ParameterRegistrationImpl.bindValue(QueryImpl.java:235)
at org.hibernate.jpa.spi.BaseQueryImpl.setParameter(BaseQueryImpl.java:638)
at org.hibernate.jpa.spi.AbstractQueryImpl.setParameter(AbstractQueryImpl.java:163)
at org.hibernate.jpa.spi.AbstractQueryImpl.setParameter(AbstractQueryImpl.java:32)
at org.hibernate.jpa.criteria.compile.CriteriaCompiler$1$1.bind(CriteriaCompiler.java:109)
at org.hibernate.jpa.criteria.CriteriaQueryImpl$1.buildCompiledQuery(CriteriaQueryImpl.java:369)
at org.hibernate.jpa.criteria.compile.CriteriaCompiler.compile(CriteriaCompiler.java:130)
at org.hibernate.jpa.spi.AbstractEntityManagerImpl.createQuery(AbstractEntityManagerImpl.java:699)
at dao.MessageDAO.getMessages(MessageDAO.java:63)
# 1 楼答案
简短回答:是的。您不需要使用特定于hibernate的类来构建满足您需求的查询
但是,您可能希望在生成查询之前过滤掉可选参数。 如果} 以按照需要的方式构建
tel
是null/empty
,您可能不希望在查询中包含它。所以不要把它添加到查询中。 查看the Java EE persistence tutorial和^{TypedQuery
伪代码(显然未经测试…但我希望您能理解):
也许以下内容也有帮助:Really dynamic JPA CriteriaBuilder
# 2 楼答案
这里有一种更好的方法,可以根据可选参数进行更新