3
我是JPA的新手,我遇到了問題。我怎樣才能使用JPA標準查詢API加入列?
假設我們有兩個,它們通過多對一關聯的,這意味着
表A中存儲在其內表B的主鍵相關
表。
當這兩個表映射到JPA實體
我對這種現狀的搜索問題。
我已經使用現有的代碼從RichFaces的演示,通過使用
JPA處理過濾和排序。此代碼使用輸入參數來創建條件查詢。
這是代碼:
private CriteriaQuery<T> createSelectCriteriaQuery() {
CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<T> criteriaQuery = criteriaBuilder.createQuery(entityClass);
Root<T> root = criteriaQuery.from(entityClass);
if (arrangeableState != null) {
List<Order> orders = createOrders(criteriaBuilder, root);
if (!orders.isEmpty()) {
criteriaQuery.orderBy(orders);
}
Expression<Boolean> filterCriteria = createFilterCriteria(criteriaBuilder, root);
if (filterCriteria != null) {
criteriaQuery.where(filterCriteria);
}
}
return criteriaQuery;
}
protected Expression<Boolean> createFilterCriteriaForField(String propertyName, Object filterValue, Root<T> root, CriteriaBuilder criteriaBuilder) {
String stringFilterValue = (String) filterValue;
if (Strings.isNullOrEmpty(stringFilterValue)) {
return null;
}
stringFilterValue = stringFilterValue.toLowerCase(arrangeableState.getLocale());
Path<String> expression = root.get(propertyName);
Expression<Integer> locator = criteriaBuilder.locate(criteriaBuilder.lower(expression), stringFilterValue, 1);
return criteriaBuilder.gt(locator, 0);
}
private Expression<Boolean> createFilterCriteria(CriteriaBuilder criteriaBuilder, Root<T> root) {
Expression<Boolean> filterCriteria = null;
List<FilterField> filterFields = arrangeableState.getFilterFields();
if (filterFields != null && !filterFields.isEmpty()) {
FacesContext facesContext = FacesContext.getCurrentInstance();
for (FilterField filterField : filterFields) {
String propertyName = (String) filterField.getFilterExpression().getValue(facesContext.getELContext());
Object filterValue = filterField.getFilterValue();
Expression<Boolean> predicate = createFilterCriteriaForField(propertyName, filterValue, root, criteriaBuilder);
if (predicate == null) {
continue;
}
if (filterCriteria == null) {
filterCriteria = predicate.as(Boolean.class);
} else {
filterCriteria = criteriaBuilder.and(filterCriteria, predicate.as(Boolean.class));
}
}
}
return filterCriteria;
}
代碼是好的,當我嘗試過濾柱(沒有加入列),但是當我嘗試
查詢上聯接的列,生成的查詢是不正確的,它會拋出異常。
所以我的問題是,我如何使用JPA的條件查詢API,雙方
聯接的列和非固定coulmns過濾行。
謝謝