我正在尝试为我的crud服务添加指定我需要的嵌套关系的可能性,因此我不必从数据库中读取所有内容。
以我为例,我有这些实体
Company.java
private List<Department> departments;
private SalaryCode salaryCode;
Department.java
private List<Employee> employees;
private Company company;
private SalaryCode salaryCode;
Employee.java
private Department department;
private SalaryCode salaryCode
现在我的Criteria查询是这样的:
Session session = sessionFactory.openSession();
CriteriaBuilder builder = session.getCriteriaBuilder();
CriteriaQuery<T> criteriaQuery = builder.createQuery(clazz);
Root<T> root = criteriaQuery.from(clazz);
//nestedRelationships is a varargs passed as parameters
for(String nestedRelationship : nestedRelationships) {
root.fetch(nestedRelationship, JoinType.LEFT);
}
List<T> result = session.createQuery(criteriaQuery.select(root)).list();
问题是如果我将“department”指定为nestedRelationship并查询Employee实体它运行良好但是当我尝试指定“department.salaryCode”时它不起作用“无法找到具有给定名称的属性” 。 当然我先取“部门”,然后取“department.salaryCode”。
是否支持?如果是,它是如何工作的,如果不支持我该怎么办?
答案 0 :(得分:0)
是的,它受到支持。你需要使用联接。
Root<Company> root = criteriaQuery.from(Company.class);
Join<Company,Department> joinDepartment = root.join( Company_.departments );
Join<Department,SalaryCode> joinSalaryCode = joinDepartment.join( Department_.salaryCode );
要生成元模型类(例如 Department _ ),请查看here。
答案 1 :(得分:0)
我通过使用Root元素
制作算法找到了解决方案protected void fetch(Root<T> root, String... joins) {
//Sort the joins so they are like this :
//A
//A.F
//B.E
//B.E.D
//B.G
Arrays.sort(joins);
Map<String, Fetch> flattenFetches = new HashMap<>();
for (String join : joins) {
try {
if (join.contains(".")) {
String[] subrelations = join.split("\\.");
Fetch lastRelation = null;
int i;
for (i = subrelations.length - 1; i >= 0; i--) {
String subJoin = String.join(".", Arrays.copyOf(subrelations, i));
if (flattenFetches.containsKey(subJoin)) {
lastRelation = flattenFetches.get(subJoin);
break;
}
}
if (lastRelation == null) {
lastRelation = root.fetch(subrelations[0], JoinType.LEFT);
flattenFetches.put(subrelations[0], lastRelation);
i = 1;
}
for (; i < subrelations.length; i++) {
String relation = subrelations[i];
String path = String.join(".", Arrays.copyOf(subrelations, i + 1));
if (i == subrelations.length - 1) {
Fetch fetch = lastRelation.fetch(relation, JoinType.LEFT);
flattenFetches.put(path, fetch);
} else {
lastRelation = lastRelation.fetch(relation, JoinType.LEFT);
flattenFetches.put(path, lastRelation);
}
}
} else {
Fetch fetch = root.fetch(join, JoinType.LEFT);
flattenFetches.put(join, fetch);
}
} catch (Exception e) {
e.printStackTrace();
}
}
}
并使用它我只需要做例如:
employeeController.getAll("punches", "currentSchedule.shifts", "defaultDepartment.currentSchedule.shifts",
"defaultDepartment.company.currentSchedule.shifts", "bankExtras")
我想评论算法,但我没有时间,而且很容易理解