真正动态的JPA CriteriaBuilder

Question

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 i have -> can not address them
query.where(predicates.get(0), predicates.get(1), ...);

// doesn't work, because it is a list of predicates
query.where(predicates);

// doesn't work, because the actual predicate overwrites the old predicate
for (Predicate pre : predicates) {
     query.where(pre)
}

我试图构建一个Predicate包含所有其他谓词的big ,并将其添加到query.where(),但谓词再次覆盖旧值.看起来没有可能添加一个Predicate而不是改变谓词:-(

真正的项目更复杂,因为有些对需要一个equal和其他一个like.这还不够:还有一个额外的声明or包括像type : 1;4;7.这里的值必须拆分并创建一个语句,如:

<rest of statement> AND (type = 1 OR type = 4 OR type = 7)

更新和解决方案 有两个列表,第一个AND列表运行良好.第二个列表包含exspected之类的OR语句:

final List<Predicate> andPredicates = new ArrayList<Predicate>();
final List<Predicate> orPredicates = new ArrayList<Predicate>();
for (final Entry<String, String> entry : statements.entrySet()) {
    final String colIndex = entry.getKey();
    final String colValue = entry.getValue();
    if (colIndex != null && colValue != null) {

        if (!colValue.contains(";")) {
            if (equals) {
                andPredicates.add(cb.equal(userRoot.get(colIndex), colValue));
            } else {
                andPredicates.add(cb.like(userRoot.<String> get(colIndex), "%" + colValue + "%"));
            }
        } else {
            String[] values = colValue.split(";");
            for (String value : values) {
                orPredicates.add(cb.or(cb.equal(userRoot.get(colIndex), value)));
            }
        }       
    }
}

// Here goes the magic to combine both lists
if (andPredicates.size() > 0 && orPredicates.size() == 0) {
    // no need to make new predicate, it is already a conjunction
    query.where(andPredicates.toArray(new Predicate[andPredicates.size()]));
} else if (andPredicates.size() == 0 && orPredicates.size() > 0) {
    // make a disjunction, this part is missing above
    Predicate p = cb.disjunction();
    p = cb.or(orPredicates.toArray(new Predicate[orPredicates.size()]));
    query.where(p);
} else {
    // both types of statements combined
    Predicate o = cb.and(andPredicates.toArray(new Predicate[andPredicates.size()]));
    Predicate p = cb.or(orPredicates.toArray(new Predicate[orPredicates.size()]));
    query.where(o, p);
}

query.where(predicates.toArray(new Predicate[predicates.size()]));
users = em.createQuery(query).getResultList();

Answer 1

您可以将一个谓词数组传递给CriteriaBuilder,决定equal或like当你去.为此,构建一个列表并将列表的内容打包到一个and语句中的数组中.像这样:

final List<Predicate> predicates = new ArrayList<Predicate>();

for (final Entry<String, String> e : myPredicateMap.entrySet()) {

    final String key = e.getKey();
    final String value = e.getValue();

    if ((key != null) && (value != null)) {

        if (value.contains("%")) {
            predicates.add(criteriaBuilder.like(root.<String> get(key), value));
        } else {
            predicates.add(criteriaBuilder.equal(root.get(key), value));
        }
    }
}

query.where(criteriaBuilder.and(predicates.toArray(new Predicate[predicates.size()])));
query.select(count);

如果您需要在and和之间进行distingiush or,请使用两个列表.

Answer 2

一种选择是使用具有可变数量的参数的方法可以采用数组的事实:

query.where(predicates.toArray(new Predicate[predicates.size()])); 

或者,您可以将它们组合成一个谓词(请注意,如果您不这样做,则不需要像示例中那样创建连接);:

Predicate where = cb.conjunction();
while (column.hasNext()) {
    ...
    where = cb.and(where, cb.equal(userRoot.get(colIndex), colValue));
}

query.where(where);

Answer 3

我一直认为创造像这样的解决方案就像重新发明自行车一样.这里https://github.com/sasa7812/jfbdemo是我的解决方案.它在EclipseLink和Hibernate上进行了测试(EclipseLink在生产中,我们在几个项目中用于简单案例).有时你只需要一个快速的解决方案来制作一个带有排序和过滤的dataTable,没什么好看的.它能够对连接的字段进行过滤和排序,甚至可以对集合进行过滤和排序.Project包含Primefaces上的演示,显示了FilterCriteriaBuilder的功能.在它的核心你只需要这个:

 public List<T> loadFilterBuilder(int first, int pageSize, Map<String, Boolean> sorts, List<FieldFilter> argumentFilters, Class<? extends AbstractEntity> entityClass) {
    FilterCriteriaBuilder<T> fcb = new FilterCriteriaBuilder<>(getEntityManager(), (Class<T>) entityClass);
    fcb.addFilters(argumentFilters).addOrders(sorts);
    Query q = getEntityManager().createQuery(fcb.getQuery());
    q.setFirstResult(first);
    q.setMaxResults(pageSize);
    return q.getResultList();
}

从数据库中获取结果.

我真的需要有人告诉我它很有用,并且用于某个地方继续我在这个库上的工作.