面向 Java 開發人員的 db4o 指南: 數組和集合

注意，存儲整個家庭所做的工作仍然不比存儲單個 Person 對象所做的工作多。您可能還記得，在上一篇文章中，由于存儲的對象具有遞歸的性質，當把 bruce 引用傳遞給 db.set() 調用時，從 bruce 可達的所有對象都被存儲。不過眼見為實，讓我們看看當運行我那個簡單的探察測試時，實際上會出現什么情況。首先，我將測試當調用隨 Person 存儲的各種 Address 時，是否可以找到它們。然后，我將測試是否孩子們也被存儲。

清單 5. 搜索住房和家庭
                   
@Test public void testTheStorageOfAddresses()
    {
        List<Person> maleTates =
            db.query(new Predicate<Person>() {
                public boolean match(Person candidate) {
                    return candidate.getLastName().equals("Tate") &&
                            candidate.getGender().equals(Gender.MALE);
                }
            });
        Person bruce = maleTates.get(0);

        Address homeAndWork =
            new Address("5 Maple Drive", "Austin",
                "TX", "12345");
        Address vacation =
            new Address("10 Wanahokalugi Way", "Oahu",
                "HA", "11223");
               
        assertTrue(bruce.getHomeAddress().equals(homeAndWork));
        assertTrue(bruce.getWorkAddress().equals(homeAndWork));
        assertTrue(bruce.getVacationAddress().equals(vacation));
    }

    @Test public void testTheStorageOfChildren()
    {
        List<Person> maleTates =
            db.query(new Predicate<Person>() {
                public boolean match(Person candidate) {
                    return candidate.getLastName().equals("Tate") &&
                            candidate.getGender().equals(Gender.MALE);
                }
            });
        Person bruce = maleTates.get(0);
       
        int n = 0;
        for (Iterator<Person> children = bruce.getChildren();
             children.hasNext();
            )
        {
            Person child = children.next();
           
            System.out.println(child);
           
            if (n==0) assertTrue(child.getFirstName().equals("Kayla"));
            if (n==1) assertTrue(child.getFirstName().equals("Julia"));
           
            n++;
        }
    }
 

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