标签: criteria-api

有没有人实现过这个,或者知道是否难以实现这个/有任何指针？

public static SpatialRelationCriterion IsWithinDistance(string propertyName, object anotherGeometry, double distance)
{
    // TODO: Implement
    throw new NotImplementedException();
}

来自NHibernate.Spatial.Criterion.SpatialRestrictions

我可以在hql中使用"NHSP.Distance(PROPERTY,:point)".但是想要将此查询与我现有的Criteria查询相结合.

目前我正在创建一个粗糙的多边形,并使用

criteria.Add(SpatialRestrictions.Intersects("PROPERTY", myPolygon));

编辑 通过在SpatialRelationCriterion上重载构造函数来获得原型,添加新的SpatialRelation.Distance

public static SpatialRelationCriterion IsWithinDistance(string propertyName, object anotherGeometry, double distance)
        {
            return new SpatialRelationCriterion(propertyName, SpatialRelation.Distance, anotherGeometry, distance);
        }

为SpatialRelationCriterion添加了一个新字段

private readonly double? distance;

public SpatialRelationCriterion(string propertyName, SpatialRelation relation, object anotherGeometry, double distance)
            : this(propertyName, relation, anotherGeometry)
        {
            this.distance = distance;
        }

编辑ToSqlString

object secondGeometry = Parameter.Placeholder;
                if (!(this.anotherGeometry is IGeometry))
                {
                    secondGeometry = columns2[i];
                }

                if …

我的Java应用程序使用JPA进行对象持久化.业务域非常简单(只有三个类是持久的,每个类有3-5个属性).查询也很简单.问题是我应该使用哪种方法:JPQL或Criteria API？

我想只选择特定的列(例如SELECT a FROM b).我有一个通用的DAO,我想出的是:

public List<T> getAll(boolean idAndVersionOnly) {
    CriteriaBuilder builder = manager.getCriteriaBuilder();
    CriteriaQuery<T> criteria = builder.createQuery(entityClazz);
    Root<T> root = criteria.from(entityClazz);
    if (idAndVersionOnly) {
        criteria.select(root.get("ID").get("VERSION")); // HERE IS ERROR
    } else {
        criteria.select(root);
    }
    return manager.createQuery(criteria).getResultList();
}

错误是: The method select(Selection<? extends T>) in the type CriteriaQuery<T> is not applicable for the arguments (Path<Object>).我该怎么改变？我想得到一个T只有ID和VERSION字段的类型对象,而其他所有的都是null.

类型Textends AbstractEntity包含这两个字段.

entityClazz是T.class.

您能帮我解决一下如何将以下代码转换为使用条件构建器的"in"运算符吗？我需要使用"in"使用列表/数组的用户名进行过滤.

我也尝试使用JPA CriteriaBuilder进行搜索 - "in"方法但是找不到好的结果.如果你能给我这个主题的参考网址,我也非常感谢.谢谢.

这是我的代码:

//usersList is a list of User that I need to put inside IN operator 

CriteriaBuilder builder = getJpaTemplate().getEntityManagerFactory().getCriteriaBuilder();
CriteriaQuery<ScheduleRequest> criteria = builder.createQuery(ScheduleRequest.class);

Root<ScheduleRequest> scheduleRequest = criteria.from(ScheduleRequest.class);
criteria = criteria.select(scheduleRequest);

List<Predicate> params = new ArrayList<Predicate>();

List<ParameterExpression<String>> usersIdsParamList = new ArrayList<ParameterExpression<String>>();

for (int i = 0; i < usersList.size(); i++) {
ParameterExpression<String> usersIdsParam = builder.parameter(String.class);
params.add(builder.equal(scheduleRequest.get("createdBy"), usersIdsParam) );
usersIdsParamList.add(usersIdsParam);
}

criteria = criteria.where(params.toArray(new Predicate[0]));

TypedQuery<ScheduleRequest> query = getJpaTemplate().getEntityManagerFactory().createEntityManager().createQuery(criteria);

for (int i = 0; i < usersList.size(); …

我正在尝试在我的新项目中使用Criteria API:

public List<Employee> findEmps(String name) {
    CriteriaBuilder cb = em.getCriteriaBuilder();
    CriteriaQuery<Employee> c = cb.createQuery(Employee.class);
    Root<Employee> emp = c.from(Employee.class);
    c.select(emp);
    c.distinct(emp);
    List<Predicate> criteria = new ArrayList<Predicate>();

    if (name != null) {
        ParameterExpression<String> p = cb.parameter(String.class, "name");
        criteria.add(cb.equal(emp.get("name"), p));
    }

    /* ... */

    if (criteria.size() == 0) {
        throw new RuntimeException("no criteria");
    } else if (criteria.size() == 1) {
        c.where(criteria.get(0));
    } else {
        c.where(cb.and(criteria.toArray(new Predicate[0])));
    }

    TypedQuery<Employee> q = em.createQuery(c);

    if (name != null) {
        q.setParameter("name", name);
    }

    /* ... */ …

我需要创建一个"真正的"动态JPA CriteriaBuilder.我得到Map<String, String>了陈述.看起来像:

name : John
surname : Smith
email : email@email.de

...more pairs possible

这是我实现的:

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<User> query = cb.createQuery(User.class);
Root<User> userRoot = query.from(User.class);
query.select(userRoot);

List<Predicate> predicates = new ArrayList<Predicate>();
Iterator<String> column = statements.keySet().iterator();
while (column.hasNext()) {

    // get the pairs
    String colIndex = column.next();
    String colValue = statements.get(colIndex);

    // create the statement
    Predicate pAnd = cb.conjunction();
    pAnd = cb.and(pAnd, cb.equal(userRoot.get(colIndex), colValue));
    predicates.add(pAnd);
}

// doesn't work, i don't know how many predicates …

将JPA 2与EclipseLink实现一起使用.

我正在尝试构建一个动态查询,它应该会在给定日期之后保留一些记录.

CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery<Event> criteria = builder.createQuery(Event.class);
Root<Event> root = criteria.from(Event.class);
criteria.select(root);
criteria.distinct(true);
List<Predicate> predicates = new ArrayList<Predicate>();
//...
if (dateLimit != null){
    ParameterExpression<Date> param = builder.parameter(Date.class, "dateLimit");
    predicates.add(builder.lessThanOrEqualTo(root.get("dateCreated"), param));
}

lessThanOrEqualTo()并且le()是仅有的两个方法,在API中看起来像可以帮助我在这种情况下.这个警告是由日食引发的:

Bound mismatch: The generic method lessThanOrEqualTo(Expression<? extends Y>, Expression<? extends Y>)
of type CriteriaBuilder is not applicable for the arguments (Path<Object>, ParameterExpression<Date>).
The inferred type Object is not a valid substitute for the bounded parameter
<Y extends Comparable<? super Y>>

我可以想象我没有采取正确的方法解决这个问题,但我无法找到任何可能解决方案的提示或指示.

我曾尝试多次使用子查询和IN表达式编写查询语句.但我从来没有成功过.

我总是得到异常,"关键字'IN'附近的语法错误",查询语句是这样构建的,

SELECT t0.ID, t0.NAME
FROM EMPLOYEE t0
WHERE IN (SELECT ?
          FROM PROJECT t2, EMPLOYEE t1
          WHERE ((t2.NAME = ?) AND (t1.ID = t2.project)))

我知道'IN'失败前的这个词.

你有没有写过这样的问题？有什么建议吗？

我遇到了一个简单的问题; 奋力如何调用order by上join编辑实体.基本上我正在努力实现以下目标JPA Criteria:

select distinct d from Department d 
left join fetch d.children c 
left join fetch c.appointments a
where d.parent is null 
order by d.name, c.name

我有以下内容:

CriteriaBuilder cb = getEntityManager().getCriteriaBuilder();
CriteriaQuery<Department> c = cb.createQuery(Department.class);
Root<Department> root = c.from(Department.class);
Fetch<Department, Department> childrenFetch = root.fetch(
    Department_.children, JoinType.LEFT);
childrenFetch.fetch(Department_.appointments, JoinType.LEFT);

c.orderBy(cb.asc(root.get(Department_.name)));
c.distinct(true);
c.select(root);
c.where(cb.isNull(root.get(Department_.parent)));

如何实现order by d.name, c.name用Criteria API？我尝试使用Expression,Path但是没有用.任何指针将不胜感激.

我正在尝试使用编写NOT IN约束JPA Criteria.我尝试过这样的事情:

builder.not(builder.in(root.get(property1)));

虽然我知道它不会起作用.在上面的语法中,如何添加property1要检查的集合/列表？