问题
I have a JSON string of the following format:
{
"foo": "small_vale"
"baz": "large_value"
"bar": "another_large_value"
}
How can I efficiently extract foo
while ignoring the rest of the fields?
Basically, I'm using Gson and I defined a "lean class" like that:
MyClass {
private String foo;
}
If I'm ensuring that foo
appears first in the JSON string, would Gson still scan the whole string, or is it smart enough to stop?
Should I use JsonPath instead?
回答1:
To answer on this question we need to see how do you parse JSON
. I assume that you are using simplest:
Test test = gson.fromJson(new FileReader(jsonFile), Test.class);
If this is your case, answer for your question is Gson
is not smart enough to do that. If you check implementation of this method, you will find out:
public <T> T fromJson(Reader json, Class<T> classOfT) throws JsonSyntaxException, JsonIOException {
JsonReader jsonReader = newJsonReader(json);
Object object = fromJson(jsonReader, classOfT);
assertFullConsumption(object, jsonReader);
return Primitives.wrap(classOfT).cast(object);
}
Before method returns value, it checks whether whole JSON
was consumed and in case not, JsonIOException
is thrown. Gson
internally uses TypeAdapter
implementation for given type. For your custom MyClass
it will use ReflectiveTypeAdapterFactory.Adapter
class which will consume the whole JSON
payload. To avoid this situation you can write your own TypeAdapter
:
class TestTypeAdapter extends TypeAdapter<Test> {
@Override
public void write(JsonWriter out, Test value) throws IOException {
throw new IllegalStateException("Implement me!");
}
@Override
public Test read(JsonReader in) throws IOException {
if (in.peek() == JsonToken.NULL) {
in.nextNull();
return null;
}
Test test = new Test();
try {
in.beginObject();
while (in.hasNext()) {
String name = in.nextName();
if (name.equals("foo")) {
test.setFoo(in.nextString());
break;
}
}
} catch (IllegalStateException e) {
throw new JsonSyntaxException(e);
}
return test;
}
}
Simple usage:
import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
import com.google.gson.JsonSyntaxException;
import com.google.gson.TypeAdapter;
import com.google.gson.stream.JsonReader;
import com.google.gson.stream.JsonToken;
import com.google.gson.stream.JsonWriter;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
public class GsonApp {
public static void main(String[] args) throws Exception {
File jsonFile = new File("./resource/test.json").getAbsoluteFile();
Gson gson = new GsonBuilder().create();
Test test = gson.fromJson(new FileReader(jsonFile), Test.class);
Test test1 = new TestTypeAdapter().fromJson(new FileReader(jsonFile));
System.out.println(test);
System.out.println(test1);
}
}
class Test {
private String foo;
public String getFoo() {
return foo;
}
public void setFoo(String foo) {
this.foo = foo;
}
@Override
public String toString() {
return "Test{" +
"foo='" + foo + '\'' +
'}';
}
}
Above code prints:
Test{foo='small_value'}
Test{foo='small_value'}
As, you can see, in both cases we parsed our small value
. You can test this code and calculate how faster custom TypeAdapter
is for your JSON
payload.
But in case, you have much complex situation and you need to parse much more JSON
to find your value, try to use JSONPath
solution. You can start from this question: how to parse a huge JSON file without loading it in memory.
回答2:
Your json file is not valid. Commas are missing. It should look like this:
{
"foo":"small_value",
"baz":"large_value",
"bar":"another_large_value"
}
This blog post says that Jackson or simple JSON are the fastest ways to parse big json data. Se the chapter "Big File Results" for reference.
Example code for Jackson: Jackson JSON – Read Specific JSON Key
It shows how to parse a json file ang get a value of a specific element.
//read json file data to String
byte[] jsonData = Files.readAllBytes(Paths.get("data.json"));
//create ObjectMapper instance
ObjectMapper objectMapper = new ObjectMapper();
//read JSON like DOM Parser
JsonNode rootNode = objectMapper.readTree(jsonData);
JsonNode fooNode = rootNode.path("foo");
System.out.println("foo value = "+fooNode.asText());
来源:https://stackoverflow.com/questions/54852415/parse-only-one-field-in-a-large-json-string