Mixing
MapStruct: How to map all attributes to first element of a list?
I need a mapping to achieve this:
@Mapping(source = "a", target = "result.transaction[0].a"),
@Mapping(source = "b", target = "result.transaction[0].b"),
@Mapping(source = "c", target = "result.transaction[0].c"),
...
Response dataToResponse(DataModel model);
But this syntax does not work (btw: This works with Spring Bean wrapper).
A solution like this is just a half-cooked solution:
@AsList
public <T> List<T> asList( T in ) {
List<T> result = new ArrayList<T>();
if ( in!=null ) {
result.add(in);
}
return result;
}
This only works for exactly one attribute since it always creates a new list for each attribute. I don't need to map each attribute to the first element of a new list. The list must be reused but I don't know how this works. What is the proper way to achieve that?
I thought about something like this:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
...
Transaction dataToTransaction(DataModel model);
and then...
@Mapping([use Transaction from b4], target = "result");
But how can I pass the already mapped fields from above?
(I'm using the latest final release 1.1.0.Final)
mapstruct
asked Sep 27 '17 at 12:43
BevorBevor
4,081854104
add a comment |
I need a mapping to achieve this:
@Mapping(source = "a", target = "result.transaction[0].a"),
@Mapping(source = "b", target = "result.transaction[0].b"),
@Mapping(source = "c", target = "result.transaction[0].c"),
...
Response dataToResponse(DataModel model);
But this syntax does not work (btw: This works with Spring Bean wrapper).
A solution like this is just a half-cooked solution:
@AsList
public <T> List<T> asList( T in ) {
List<T> result = new ArrayList<T>();
if ( in!=null ) {
result.add(in);
}
return result;
}
This only works for exactly one attribute since it always creates a new list for each attribute. I don't need to map each attribute to the first element of a new list. The list must be reused but I don't know how this works. What is the proper way to achieve that?
I thought about something like this:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
...
Transaction dataToTransaction(DataModel model);
and then...
@Mapping([use Transaction from b4], target = "result");
But how can I pass the already mapped fields from above?
(I'm using the latest final release 1.1.0.Final)
mapstruct
asked Sep 27 '17 at 12:43
BevorBevor
4,081854104
add a comment |
I need a mapping to achieve this:
@Mapping(source = "a", target = "result.transaction[0].a"),
@Mapping(source = "b", target = "result.transaction[0].b"),
@Mapping(source = "c", target = "result.transaction[0].c"),
...
Response dataToResponse(DataModel model);
But this syntax does not work (btw: This works with Spring Bean wrapper).
A solution like this is just a half-cooked solution:
@AsList
public <T> List<T> asList( T in ) {
List<T> result = new ArrayList<T>();
if ( in!=null ) {
result.add(in);
}
return result;
}
This only works for exactly one attribute since it always creates a new list for each attribute. I don't need to map each attribute to the first element of a new list. The list must be reused but I don't know how this works. What is the proper way to achieve that?
I thought about something like this:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
...
Transaction dataToTransaction(DataModel model);
and then...
@Mapping([use Transaction from b4], target = "result");
But how can I pass the already mapped fields from above?
(I'm using the latest final release 1.1.0.Final)
mapstruct
asked Sep 27 '17 at 12:43
BevorBevor
4,081854104
I need a mapping to achieve this:
@Mapping(source = "a", target = "result.transaction[0].a"),
@Mapping(source = "b", target = "result.transaction[0].b"),
@Mapping(source = "c", target = "result.transaction[0].c"),
...
Response dataToResponse(DataModel model);
But this syntax does not work (btw: This works with Spring Bean wrapper).
A solution like this is just a half-cooked solution:
@AsList
public <T> List<T> asList( T in ) {
List<T> result = new ArrayList<T>();
if ( in!=null ) {
result.add(in);
}
return result;
}
This only works for exactly one attribute since it always creates a new list for each attribute. I don't need to map each attribute to the first element of a new list. The list must be reused but I don't know how this works. What is the proper way to achieve that?
I thought about something like this:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
...
Transaction dataToTransaction(DataModel model);
and then...
@Mapping([use Transaction from b4], target = "result");
But how can I pass the already mapped fields from above?
(I'm using the latest final release 1.1.0.Final)
mapstruct
mapstruct
asked Sep 27 '17 at 12:43
BevorBevor
4,081854104
asked Sep 27 '17 at 12:43
BevorBevor
4,081854104
edited Sep 28 '17 at 6:40
Bevor
asked Sep 27 '17 at 12:43
BevorBevor
4,081854104
asked Sep 27 '17 at 12:43
BevorBevor
4,081854104
asked Sep 27 '17 at 12:43
BevorBevor
4,081854104
4,081854104
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Obviously there is no clean solution for that. So I had to workaround it by exclude the following mapping into a separate mapper:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);
In the main mapper, I execute the separate mapper and convert it into a list by expression:
@Mapping(expression = "java(Arrays.asList(SubMapper.INSTANCE.dataToTransaction(model)))", target = "result.transactions")
answered Sep 27 '17 at 15:50
BevorBevor
4,081854104
add a comment |
Just came across this question. A bit more elegant solution would be something like this:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);
- this is the same as in the previous answer
What remains to do is to map single transaction to a one-element-list:
default List<Transaction> mapTransactionToList(Transaction source) {
return ImmutableList.of(source);
}
Now you can simply map model to List<Transaction> in your @Mapping definition and MapStruct should figure out what to do.
I think this looks better and is less error-prone than "expression" based solution.
Please note that you can include the code in your Mapper interface.
answered Nov 21 '18 at 9:42
Antonín KarásekAntonín Karásek
115
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Obviously there is no clean solution for that. So I had to workaround it by exclude the following mapping into a separate mapper:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);
In the main mapper, I execute the separate mapper and convert it into a list by expression:
@Mapping(expression = "java(Arrays.asList(SubMapper.INSTANCE.dataToTransaction(model)))", target = "result.transactions")
answered Sep 27 '17 at 15:50
BevorBevor
4,081854104
add a comment |
Obviously there is no clean solution for that. So I had to workaround it by exclude the following mapping into a separate mapper:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);
In the main mapper, I execute the separate mapper and convert it into a list by expression:
@Mapping(expression = "java(Arrays.asList(SubMapper.INSTANCE.dataToTransaction(model)))", target = "result.transactions")
answered Sep 27 '17 at 15:50
BevorBevor
4,081854104
add a comment |
Obviously there is no clean solution for that. So I had to workaround it by exclude the following mapping into a separate mapper:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);
In the main mapper, I execute the separate mapper and convert it into a list by expression:
@Mapping(expression = "java(Arrays.asList(SubMapper.INSTANCE.dataToTransaction(model)))", target = "result.transactions")
answered Sep 27 '17 at 15:50
BevorBevor
4,081854104
Obviously there is no clean solution for that. So I had to workaround it by exclude the following mapping into a separate mapper:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);
In the main mapper, I execute the separate mapper and convert it into a list by expression:
@Mapping(expression = "java(Arrays.asList(SubMapper.INSTANCE.dataToTransaction(model)))", target = "result.transactions")
answered Sep 27 '17 at 15:50
BevorBevor
4,081854104
answered Sep 27 '17 at 15:50
BevorBevor
4,081854104
answered Sep 27 '17 at 15:50
BevorBevor
4,081854104
answered Sep 27 '17 at 15:50
BevorBevor
4,081854104
4,081854104
add a comment |
add a comment |
Just came across this question. A bit more elegant solution would be something like this:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);
- this is the same as in the previous answer
What remains to do is to map single transaction to a one-element-list:
default List<Transaction> mapTransactionToList(Transaction source) {
return ImmutableList.of(source);
}
Now you can simply map model to List<Transaction> in your @Mapping definition and MapStruct should figure out what to do.
I think this looks better and is less error-prone than "expression" based solution.
Please note that you can include the code in your Mapper interface.
answered Nov 21 '18 at 9:42
Antonín KarásekAntonín Karásek
115
add a comment |
Just came across this question. A bit more elegant solution would be something like this:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);
- this is the same as in the previous answer
What remains to do is to map single transaction to a one-element-list:
default List<Transaction> mapTransactionToList(Transaction source) {
return ImmutableList.of(source);
}
Now you can simply map model to List<Transaction> in your @Mapping definition and MapStruct should figure out what to do.
I think this looks better and is less error-prone than "expression" based solution.
Please note that you can include the code in your Mapper interface.
answered Nov 21 '18 at 9:42
Antonín KarásekAntonín Karásek
115
add a comment |
Just came across this question. A bit more elegant solution would be something like this:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);
- this is the same as in the previous answer
What remains to do is to map single transaction to a one-element-list:
default List<Transaction> mapTransactionToList(Transaction source) {
return ImmutableList.of(source);
}
Now you can simply map model to List<Transaction> in your @Mapping definition and MapStruct should figure out what to do.
I think this looks better and is less error-prone than "expression" based solution.
Please note that you can include the code in your Mapper interface.
answered Nov 21 '18 at 9:42
Antonín KarásekAntonín Karásek
115
Just came across this question. A bit more elegant solution would be something like this:
@Mapping(source = "a", target = "transaction.a"),
@Mapping(source = "b", target = "transaction.b"),
@Mapping(source = "c", target = "transaction.c"),
Transaction dataToTransaction(DataModel model);
- this is the same as in the previous answer
What remains to do is to map single transaction to a one-element-list:
default List<Transaction> mapTransactionToList(Transaction source) {
return ImmutableList.of(source);
}
Now you can simply map model to List<Transaction> in your @Mapping definition and MapStruct should figure out what to do.
I think this looks better and is less error-prone than "expression" based solution.
Please note that you can include the code in your Mapper interface.
answered Nov 21 '18 at 9:42
Antonín KarásekAntonín Karásek
115
answered Nov 21 '18 at 9:42
Antonín KarásekAntonín Karásek
115
answered Nov 21 '18 at 9:42
Antonín KarásekAntonín Karásek
115
answered Nov 21 '18 at 9:42
Antonín KarásekAntonín Karásek
115
115
add a comment |
add a comment |
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