java - JPA where 子句

Question

在 JPA 中，查询是:

Query q = entityManager.createQuery("select o from Product o WHERE o.category = :value");
q.setParameter("category", category);

如何在 JPA 中将类别设置为任何类别？因此，如果通过了空类别，我会简单地忽略类别参数，选择所有产品。

Answer 1

How can I set category to any category in JPA? So if the null category passed, I simple ignore the category parameter, select all products.

您必须在此处动态构建查询。使用 HQL(这是一个简化的示例):

Map<String, Object> params = new HashMap<String, Object>();
StringBuffer hql = new StringBuffer("from Product p");
boolean first = true;

if (category != null) {
    hql.append(first ? " where " : " and ");
    hql.append("p.category = :category");
    params.put("category", category);
}

// And so on...

Query query = session.createQuery(hql.toString());

Iterator<String> iter = params.keySet().iterator();
while (iter.hasNext()) {
    String name = iter.next();
    Object value = params.get(name);
    query.setParameter(name, value);
}

List results = query.list()

但实际上，我的建议是在此处使用 Criteria API:

Criteria criteria = session.createCriteria(Product.class);
if (category != null) {
    criteria.add(Expression.eq("category", category);
}
// And so on...
List results = criteria.list();

复杂的动态查询要简单得多。