How to take url output(JSON) as Java Variable

Question

I am trying to use the value of rate in the below JSON object as my variable.

URL is : http://rate-exchange-1.appspot.com/currency?from=USD&to=INR

Output of above URL is like " {"to": "INR", "rate": 64.806700000000006, "from": "USD"} ". I am assuming it as JSON object. how to get the value of 64, shall i get it by parsing?

http://stackoverflow.com/questions/4308554/simplest-way-to-read-json-from-a-url-in-java — Glen Pierce, Apr 02 '17 at 23:43
Possible duplicate of [How to parse JSON in Java](http://stackoverflow.com/questions/2591098/how-to-parse-json-in-java) — bhspencer, Apr 02 '17 at 23:50

score 1 · Answer 1 · answered Apr 02 '17 at 23:46

1

Q: how to get the value of 64, shall i get it by parsing?

A: Yes.

SUGGESTION:

You can also deserialize it into a Java object.

There are many libraries that support this, including:

JSONOjbect (Android)
GSon
Jackson

answered Apr 02 '17 at 23:46

paulsm4

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easeccy · Answer 2 · 2017-04-02T23:49:50.593

1

You can use JSONObject to convert strings into objects.

String responseString = "{'to': 'INR', 'rate': 64.806700000000006, 'from': 'USD'}";
JSONObject responseObject = new JSONObject(responseString);
Log.d("Response",responseObject.getString("rate")); //64.806700000000006

edited Apr 02 '17 at 23:49

answered Apr 02 '17 at 23:46

easeccy

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You have the wrong language here. The OP is using Java not JavaScript – bhspencer Apr 02 '17 at 23:48
2

This requires that you download a JSON library (like Jackson, JSON, etc). In this case, the library is [JSON]: https://mvnrepository.com/artifact/org.json/json – paulsm4 Apr 02 '17 at 23:53
I'm using JSONObject without downloading any library in Android Studio 2.3 . – easeccy Apr 02 '17 at 23:59
1

The OP only says they are using Java not Android studio. JSONObject is not part of the standard Java libraries. – bhspencer Apr 03 '17 at 00:13

score 1 · Answer 3 · answered Apr 03 '17 at 00:17

There're many options to deserialize JSON as mentioned in other answers, most of the time it would be better to define a corresponding Java class and then do serialization/deserialization. One example implementation with Gson would be:

public class Main {
    public static void main(String[] args) {
        Gson gson = new Gson();
        String jsonString = "{\"to\": \"INR\", \"rate\": 64.806700000000006, \"from\": \"USD\"}";
        CurrencyRate currencyRate = gson.fromJson(jsonString, CurrencyRate.class);
    }

    class CurrencyRate {
        private String from;
        private String to;
        private BigDecimal rate;

        public String getFrom() {
            return from;
        }

        public void setFrom(String from) {
            this.from = from;
        }

        public String getTo() {
            return to;
        }

        public void setTo(String to) {
            this.to = to;
        }

        public BigDecimal getRate() {
            return rate;
        }

        public void setRate(BigDecimal rate) {
            this.rate = rate;
        }
    }
}

And Gson is Thread Safe, so it's OK to init only one Gson instance and share it among all threads.

How to take url output(JSON) as Java Variable

3 Answers3