Java 8 Stream with 2 Lists - list

Question

I want to build a JSON from two lists. I need to use the corresponding elements from both lists to create a single JSON object.
My problem could be solved with ordinary loop like this:
List<Class1> items = baseManager.findObjectsByNamedQuery(Class1.class, "Class1.findAll", new Object[]{});
for(int i=0 ; i<items.size();i++){
List<Class2> items2 = baseManager.findObjectsByNamedQuery(Class2.class, "Class2.findByCreatedBy" ,new Object[] {items.get(i).getCreatedBy()});
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");
JsonObjectBuilder jpb = Json.createObjectBuilder()
.add("createdBy",items.get(i).getCreatedBy())
.add("phone",items2.get(0).getPhone())
groupsBuilder.add(jpb);
}
Is it possible to solve it using Java 8 Stream API?

Answer 1

There are still some things unclear. Like why you are insisting on creating that SimpleDateFormat instance that you are not using anywhere. Or whether there is a significance in calling getCreatedBy() multiple times. Assuming that it is not necessary, the following code is equivalent
baseManager.findObjectsByNamedQuery(Class1.class, "Class1.findAll", new Object[]{})
.stream()
.map(item -> item.getCreatedBy())
.map(createdBy -> Json.createObjectBuilder()
.add("createdBy", createdBy)
.add("phone", baseManager.findObjectsByNamedQuery(
Class2.class, "Class2.findByCreatedBy", new Object[] {createdBy})
.get(0).getPhone())
)
.forEach(jpb -> groupsBuilder.add(jpb));
It’s still unclear to me whether (or why) findObjectsByNamedQuery is not a varargs method. It would be quite natural to be a varargs method, not requiring these explicit new Object[] { … } allocations.

Answer 2

With pure Java8 Stream API:
public void convertItemsToJSon(List<Item> items) {
...
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");
Map<Item, List<Class2>> sqlItems = items
.stream()
.collect(Collectors.toMap(Function.identity(), (item) -> baseManager.findObjectsByNamedQuery(Class2.class, "Class2.findByCratedBy", new Object[]{item.getCreatedBy()})));
sqlItems.entrySet()
.stream()
.map(sqlItem -> buildJson(sqlItem.getKey(), sqlItem.getValue()))
.forEach(groupsBuilder::add);
...
}
private JsonObjectBuilder buildJson(Item item, List<Class2> class2Items) {
return Json.createObjectBuilder().add("createdBy", item.getCreatedBy());
}
With StreamEx library
public void convertItemsToJSonStreamEx(List<Item> items) {
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");
...
StreamEx.of(items)
.cross(item -> baseManager.findObjectsByNamedQuery(Class2.class, "Class2.findByCratedBy", new Object[]{item.getCreatedBy()}).stream())
.mapKeys(item -> Json.createObjectBuilder().add("createdBy", item.getCreatedBy()))
.mapKeyValue(this::addField)
.forEach(groupsBuilder::add);
...
}
private JsonObjectBuilder addField(JsonObjectBuilder json, Class2 class2) {
// You logic how to convert class2 to field in JSON
return json;
}

Answer 3

Thanks for your help and solution. Most helpful was the first response from Vlad Bochenin The code is here:
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");
items.stream().map(item -> {
List<Class2> items2 = baseManager.findObjectsByNamedQuery(Class2.class, "Class2.findByCreatedBy", new Object[]{item.getCreatedBy()});
JsonObjectBuilder jpb = Json.createObjectBuilder()
.add("createdBy", item.getCreatedBy())
.add("phone", items2.get(0).getPhone())
return jpb;
}).forEach(groupsBuilder::add);