convert array to map using spring mvc with jpa

Question

From my service I am generating the JSON based on the language.I need to covert the array to the map which contains one value as key and another as value (lot of value in the arrays based on the database).

I have written the code I am getting the array and only "I need to convert into the list of key value pair like the given sample and write in the text file.

JSON from my service

[
[
    "task",
    "Comments"
],
[
    "CUSTOM_43_01",
    "Email"
],
[
    "CUSTOM_44_02",
    "Mobile"
],..........
]

I think it would happen if you returned a Map instead of a LIst. — duffymo
– duffymo, Commented Jan 7, 2019 at 17:45
As per the previous comments, A map would get you closer to your result. Also a map literal would separate key/value with a colon instead of an equal. See stackoverflow.com/q/443499/744133. — YoYo
– YoYo, Commented Jan 8, 2019 at 3:32

Hadi · Accepted Answer · 2019-01-07 17:58:36Z

1

Try to convert your result from repository to expected result in the service like bellow:

public Map<String,String> getAllLangaugeDataForEn(Integer appId, String language) {
  Map<String,String> result = new HashMap();
  if (language.equalsIgnoreCase("EN")) {
   result  = languageRepository.getEnLanguageList(appId)
            .stream()
            .map(language-> new AbstractMap.SimpleEntry<>(language.getX(),language.getY()))
            .collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue));
   }
     return result;
}

language.getX() and language.getY() are LanguageDTO getters methods(LanguageDTO properties)!

answered Jan 7, 2019 at 17:58

Hadi

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5 Comments

Shashank Ranjan Over a year ago

I am getting The method stream() is undefined for the type Map<String,String> error

Shashank Ranjan Over a year ago

I have added my DTO

Hadi Over a year ago

well, you tagged java-8 due to this reason I put this answer. what is your java version?

Shashank Ranjan Over a year ago

java version is java -8

Hadi Over a year ago

thelanguageRepository.getEnLanguageList(appId) result is list so stream() is defined for it. can you display or say in detail your error?

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