將地圖平面對象映射到具有地圖構圖的組合對象

Question

當通過第二個映射函數將映射信息明確地映射到Mapstruct時，爲什麼不自動生成映射代碼？

Flat   Composed 
----   -------- 
- String a - String a 
       - Sub 
- String b  - String b 
- String c  - String c 

。

爲了清楚

@Mapper(uses = SubToFlatMapper.class) 
public interface ComposedToFlatMapper { 

    Flat map(Composed c); // Unmapped target properties: "b, c". 
          // Not OK because Sub mapping is defined in 
          // SubToFlatMapper and is used here :(
} 

@Mapper 
public interface SubToFlatMapper { 

    Flat map(Sub s); // Unmapped target properties: "a". 
        // (OK because 'a' is not in Sub) 
} 

用法：

Composed composed = new Composed(); 
Sub sub = new Sub(); 

composed.setA("A"); 
composed.setSub(sub); 

sub.setB("B"); 
sub.setC("C"); 

Flat flat = ComposedToFlatMapper.INSTANCE.map(composed); 

// flat.getA() is "A", OK! 
// flat.getB() is null, unexpected, should be "B" 
// flat.getC() also null, should be "C". 

"B"和"C"未映射到flat由於生成代碼不會創建，該方法。

我以爲我已經提供了所需的映射信息到生成器。當它看到Flat map(Sub s)方法時，它應該在ComposedToFlatMapperImpl內生成。

編輯：

之前我沒有提到這一點，但在原來的問題，我還可以使用update - 方法這是更靈活。

現在，假設有一個新的Sub2類型。 ：

@Mapper(uses = {SubToFlatMapper.class, Sub2ToFlatMapper.class}) 
public interface ComposedToFlatMapper { 

    Flat map(Composed c); 

    void update(Composed source, @MappingTarget Flat target); 
} 

@Mapper 
public interface SubToFlatMapper { 

    Flat map(Sub s); 

    void update(Sub source, @MappingTarget Flat target); 
} 

@Mapper 
public interface Sub2ToFlatMapper { 

    Flat map(Sub2 s); 

    void update(Sub2 source, @MappingTarget Flat target); 
} 

。

Flat   Composed 
----   -------- 
- String a - String a 
       - Sub 
- String b  - String b 
- String c  - String c 
       - Sub2 
       - String d 
       - String e 

由於Flat沒有Sub2引用只是忽略它。不需要生成映射代碼。

Answer 1

可以簡單地壓平嵌套源結構，如下所示：

@Mapper 
public interface ComposedToFlatMapper { 

    @Mapping(target="b", source="sub.b") 
    @Mapping(target="c", source="sub.c") 
    Flat map(Composed c); 
}