Is it possible to create a MAP<> within another MAP<>?

Question

It is possible to create a MAP<> within another MAP<>? like this code:

private Map<String, Map<String,Object>>  mapTESTE = new HashMap<String, Map<String,Object>>();

If yes:

How I enter values in the second MAP<>?

is a good practice?

I need something like this because both the key of the first MAP<> and the key of the second should not be repeated.

Other ways to implement something like this are welcome

_____**Edit**_____

I need to update data every five minutes

the structure is something like this: //

id1:

     nº do produto1 || dados do produto1

     nº do produto2 || dados do produto2
             .......
     nº do produtoX || dados do produtoX

id2:

     nº do produto1 || dados do produto1

     nº do produto2 || dados do produto2
             .......
     nº do produtoX || dados do produtoX

id3:

     nº do produto1 || dados do produto1

     nº do produto2 || dados do produto2
             .......
     nº do produtoX || dados do produtoX

id’s do not repeat, this is when an equal arrives I have to update the values inside the MAP<>

and the product numbers cannot be repeated either

Basically the values that suffer most updates is the product data.

Possibly this is not the best way to do this, but I confess that I especially like to use MAP because of automatically updated values by the latest and do not let repeat keys

Answer 1

It is possible yes, you entered normally like this:

Map<String, Map<String,Object>> mapTESTE = new HashMap<String, Map<String,Object>>();
Map<String, Object> segundoMap = new HashMap<String, Object>();
segundoMap.put("key1.1", "value1.1");
mapTESTE.put("key1", segundoMap);

Or so on a loop:

Map<String, Map<String,Object>> mapTESTE = new HashMap<String, Map<String,Object>>();
int size1 = 3;
int size2 = 5;
for (int i = 0; i < size1; i++) {
    Map<String, Object> segundoMap = new HashMap<String, Object>();
    for (int j = 0; j < size2; j++) {
        // key nesse formato: key-i-j, onde 'i' é o indice do primeiro map e 'j' é o indice do segundo map
        segundoMap.put("key-" + i + " - "+ j, "value-" + i + " - "+ j);
    }
    // key nesse formato: key-i, onde 'i' é o indice do primeiro map
    mapTESTE.put("key-" + i, segundoMap);
}

If it’s a good practice?

Yes is a good way to maintain collections with unique keys, it should only be analyzed if there is a need to maintain a collection in this structure or it is possible to maintain a simpler collection as ArrayList<T>. In your case you cited the need for the two keys to be single, so I believe this is the correct way.

As you quoted in your edit, to edit the content of the sub Map do something like this:

mapTESTE.get("key1").put("key1.1", "novo valor");

I don’t know if this is really what you want, but in a simple way this is how you do it.

Answer 2

Yes, it is possible. To access the second Map, simply get a reference for it (using the first key):

Map<String,Object> interno = new HashMap<String,Object>()
mapTESTE.put(primeiraChave, interno);
interno.put(segundaChave, object);

...

// Obtém o valor
Object object = mapTESTE.get(primeiraChave).get(segundaChave);

// Atualiza o valor
mapTESTE.get(primeiraChave).put(segundaChave, novoValor);

// Atualiza a segunda chave
Object valor = mapTESTE.get(primeiraChave).remove(segundaChave);
mapTESTE.get(primeiraChave).put(novaChave, valor);

// Atualiza a primeira chave (para todos os valores da segunda)
Map<String,Object> interno = mapTESTE.remove(primeiraChave);
mapTESTE.put(novaChave, interno);

Whether or not this is the best option depends on whether or not there is a hierarchy between the keys. As the example above showed, tampering with the first key affects all values independent of the second key. Sometimes this is exactly what you want, but in other situations it may not be.

If the keys are independent, then it is preferable to create an object composing the two keys:

class Chaves {
    String primeira;
    String segunda;
    // Implementar equals e hashCode (importante)
}

And use this object as your map key:

Map<Chaves,Object> mapTESTE

You can even create auxiliary maps to search for one key or another:

Map<String,Chaves> buscaPorPrimeiraChave;
Map<String,Chaves> buscaPorSegundaChave;

(Maybe there is some class in Java that serves for this, but if it exists I don’t remember; arrays are not an option, because their equality operation does not take into account the contents of this array - only whether or not it is the same object)

Answer 3

Yes, it is possible.

To put a value on the map inside the first map, do so:

Map<String, Map<String, String>> teste = new HashMap<String, Map<String, String>>();
teste.put("primeiro", new HashMap<String, String>());
teste.get("primeiro").put("primeiro chave do sub map", "primeiro valor do sub map");

Whether this is good practice or not is hard to say depends on what you are looking for may be a good solution. An alternative would be to create a class that contains two attributes to be used as the value of your first map.

class MeuMapa {
    private Set<String> primeiro;
    private String segundo;
    public MeuMapa(Set<String> primeiro, String segundo) {
        this.primeiro = primeiro;
        this.segundo = segundo;
    }
    public Set<String> getPrimeiro() { return primeiro; }
    public void setPrimeiro(Set<String> primeiro) { this.primeiro = primeiro; }
    public String getSegundo() { return segundo; }
    public void setSegundo(String segundo) { this.segundo = segundo; }
    //implemente o hashCode() e o equals() aqui
    //o próprio eclipse faz isso de forma automática
}

Note that the first attribute is a Set, which does not allow duplicates, so as you said that neither the first value of your map, nor the first value of your second map can be repeated I believe this meets.

To use it would look like this:

Map<String, MeuMapa> mapa = new HashMap<>();
MeuMapa meuMapa = new MeuMapa(new HashSet<String>(), "segundo");
mapa.put("primeiro", meuMapa);      

But it all depends on what your need is, you can go one way or the other, both can be considered "right".