问题
I have a fairly large JSON response in which I'm interested in single field - status
:
{
"title": "Some title",
"status": "pending",
"data": {
...
},
"meta": {
...
}
}
All I need to do is read the status
value of the JSON response as string. I would prefer to not have to build a POJO to model it, because in my application I just need to store the JSON in a database on a particular status or discard it.
The application already uses Jackson for other more complicated cases so I'd prefer to stick with that library. So far all the examples I've found try to map the JSON to an object.
回答1:
If the field required is a non-null text field, at the "first level" of the hierarchy (i.e., not any nested object) of a JSON
object small enough to fit in main memory, a simple way of retrieving its value is using a method like
public static String readField(String json, String name) throws IOException {
if (field != null) {
ObjectNode object = new ObjectMapper().readValue(json, ObjectNode.class);
JsonNode node = object.get(name);
return (node == null ? null : node.textValue());
}
return null;
}
ObjectNode
is a generic Jackson class, not a POJO. If multiple values are to be used, the ObjectMapper
should be cached (it is even thread-safe).
Running
System.out.println(readField(response, "status"));
using the JSON response string above, returns
pending
as expected. A similar solution can be found elsewhere in StackOverflow
.
For very large JSON objects (e.g., stored in files), the streaming approach of Jackson
should be used, as suggested in other answers.
回答2:
If the json response is not very large, you can use ObjectMapper.readTree
to deserialize the full json into JsonNode
and get a particular property in it.
For example:
ObjectMapper mapper = new ObjectMapper();
JsonNode jsonNode = mapper.readTree(json);
JsonNode statusNode = jsonNode.get("status");
String statusValue = statusNode.textValue();
You can also deserialize the full json into ObjectNode
by ObjectMapper.readValue
suggested in other answer.
If the json response is huge and deserialize the full json is a concern, you may consider to use the Jackson Streaming API, JsonParser
, to parse the json.
Following is an example:
String statusValue = null;
JsonFactory jsonFactory = new JsonFactory();
try (JsonParser parser = jsonFactory.createParser(json)) {
JsonToken startToken = parser.nextToken();
while (parser.nextToken() != JsonToken.END_OBJECT) {
String fieldName = parser.getCurrentName();
if ("status".equals(fieldName)) {
if (parser.nextToken() == JsonToken.VALUE_STRING) {
statusValue = parser.getValueAsString();
break;
}
} else {
parser.skipChildren();
}
}
}
JsonParser
break down the json into a sequence of tokens which you can iterate one by one. In the example, JsonParser.skipChildren
is used to skip over a complete object or an array such that only the first level of json hierarchy is checked.
来源:https://stackoverflow.com/questions/38732849/how-to-read-single-json-field-with-jackson