我有以下JSON文本。我如何解析它以获得pageName, pagePic, post_id等的值?
{
"pageInfo": {
"pageName": "abc",
"pagePic": "http://example.com/content.jpg"
},
"posts": [
{
"post_id": "123456789012_123456789012",
"actor_id": "1234567890",
"picOfPersonWhoPosted": "http://example.com/photo.jpg",
"nameOfPersonWhoPosted": "Jane Doe",
"message": "Sounds cool. Can't wait to see it!",
"likesCount": "2",
"comments": [],
"timeOfPost": "1234567890"
}
]
}
当前回答
您需要使用JsonNode和来自jackson库的ObjectMapper类来获取Json树的节点。在pom.xml中添加以下依赖项以获得对Jackson类的访问权。
<!-- https://mvnrepository.com/artifact/com.fasterxml.jackson.core/jackson-databind -->
<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-databind</artifactId>
<version>2.9.5</version>
</dependency>
你应该尝试下面的代码，这将工作:
import com.fasterxml.jackson.core.JsonGenerationException;
import com.fasterxml.jackson.databind.JsonMappingException;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;
class JsonNodeExtractor{
public void convertToJson(){
String filepath = "c:\\data.json";
ObjectMapper mapper = new ObjectMapper();
JsonNode node = mapper.readTree(filepath);
// create a JsonNode for every root or subroot element in the Json String
JsonNode pageInfoRoot = node.path("pageInfo");
// Fetching elements under 'pageInfo'
String pageName = pageInfoRoot.path("pageName").asText();
String pagePic = pageInfoRoot.path("pagePic").asText();
// Now fetching elements under posts
JsonNode postsNode = node.path("posts");
String post_id = postsNode .path("post_id").asText();
String nameOfPersonWhoPosted = postsNode
.path("nameOfPersonWhoPosted").asText();
}
}
其他回答
如果你的数据很简单，你不想要外部依赖，可以使用以下几行代码:
/**
* A very simple JSON parser for one level, everything quoted.
* @param json the json content.
* @return a key => value map.
*/
public static Map<String, String> simpleParseJson(String json) {
Map<String, String> map = new TreeMap<>();
String qs[] = json.replace("\\\"", "\u0001").replace("\\\\", "\\").split("\"");
for (int i = 1; i + 3 < qs.length; i += 4) {
map.put(qs[i].replace('\u0001', '"'), qs[i + 2].replace('\u0001', '"'));
}
return map;
}
这些数据
{"name":"John", "age":"30", "car":"a \"quoted\" back\\slash car"}
生成一个包含
{age=30, car=a "quoted" back\slash car, name=John}
这也可以升级为使用未加引号的值…
/**
* A very simple JSON parser for one level, names are quoted.
* @param json the json content.
* @return a key => value map.
*/
public static Map<String, String> simpleParseJson(String json) {
Map<String, String> map = new TreeMap<>();
String qs[] = json.replace("\\\"", "\u0001").replace("\\\\", "\\").split("\"");
for (int i = 1; i + 1 < qs.length; i += 4) {
if (qs[i + 1].trim().length() > 1) {
String x = qs[i + 1].trim();
map.put(qs[i].replace('\u0001', '"'), x.substring(1, x.length() - 1).trim().replace('\u0001', '"'));
i -= 2;
} else {
map.put(qs[i].replace('\u0001', '"'), qs[i + 2].replace('\u0001', '"'));
}
}
return map;
}
为了解决复杂的结构，它变得很难看… ．.． 对不起! !．.． 但我忍不住把它编码了^^ 这将解析给定的JSON以及更多内容。它产生嵌套的映射和列表。
/**
* A very simple JSON parser, names are quoted.
*
* @param json the json content.
* @return a key => value map.
*/
public static Map<String, Object> simpleParseJson(String json) {
Map<String, Object> map = new TreeMap<>();
String qs[] = json.replace("\\\"", "\u0001").replace("\\\\", "\\").split("\"");
int index[] = { 1 };
recurse(index, map, qs);
return map;
}
/**
* Eierlegende Wollmilchsau.
*
* @param index index into array.
* @param map the current map to fill.
* @param qs the data.
*/
private static void recurse(int[] index, Map<String, Object> map, String[] qs) {
int i = index[0];
for (;; i += 4) {
String end = qs[i - 1].trim(); // check for termination of an object
if (end.startsWith("}")) {
qs[i - 1] = end.substring(1).trim();
i -= 4;
break;
}
String key = qs[i].replace('\u0001', '"');
String x = qs[i + 1].trim();
if (x.endsWith("{")) {
x = x.substring(0, x.length() - 1).trim();
if (x.endsWith("[")) {
List<Object> list = new ArrayList<>();
index[0] = i + 2;
for (;;) {
Map<String, Object> inner = new TreeMap<>();
list.add(inner);
recurse(index, inner, qs);
map.put(key, list);
i = index[0];
String y = qs[i + 3]; // check for termination of array
if (y.startsWith("]")) {
qs[i + 3] = y.substring(1).trim();
break;
}
}
continue;
}
Map<String, Object> inner = new TreeMap<>();
index[0] = i + 2;
recurse(index, inner, qs);
map.put(key, inner);
i = index[0];
continue;
}
if (x.length() > 1) { // unquoted
String value = x.substring(1, x.length() - 1).trim().replace('\u0001', '"');
if ("[]".equals(value)) // handle empty array
map.put(key, new ArrayList<>());
else
map.put(key, value);
i -= 2;
} else {
map.put(key, qs[i + 2].replace('\u0001', '"'));
}
}
index[0] = i;
}
yield -如果你打印地图:
{pageInfo={pageName=abc, pagePic=http://example.com/content.jpg}, posts=[{actor_id=1234567890, comments=[], likesCount=2, message=Sounds cool. Can't wait to see it!, nameOfPersonWhoPosted=Jane Doe, picOfPersonWhoPosted=http://example.com/photo.jpg, post_id=123456789012_123456789012, timeOfPost=1234567890}]}
Gson很容易学习和实现，我们需要知道的是以下两种方法
toJson() -将Java对象转换为JSON格式 fromJson() -将JSON转换为Java对象
`
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import com.google.gson.Gson;
public class GsonExample {
public static void main(String[] args) {
Gson gson = new Gson();
try {
BufferedReader br = new BufferedReader(
new FileReader("c:\\file.json"));
//convert the json string back to object
DataObject obj = gson.fromJson(br, DataObject.class);
System.out.println(obj);
} catch (IOException e) {
e.printStackTrace();
}
}
}
`
除了其他答案，我推荐这个在线开源服务jsonschema2pojo.org，它可以从json或json模式快速生成Java类，用于GSON, Jackson 1。或者Jackson 2.x。例如，如果你有:
{
"pageInfo": {
"pageName": "abc",
"pagePic": "http://example.com/content.jpg"
}
"posts": [
{
"post_id": "123456789012_123456789012",
"actor_id": 1234567890,
"picOfPersonWhoPosted": "http://example.com/photo.jpg",
"nameOfPersonWhoPosted": "Jane Doe",
"message": "Sounds cool. Can't wait to see it!",
"likesCount": 2,
"comments": [],
"timeOfPost": 1234567890
}
]
}
GSON的jsonschema2pojo.org生成:
@Generated("org.jsonschema2pojo")
public class Container {
@SerializedName("pageInfo")
@Expose
public PageInfo pageInfo;
@SerializedName("posts")
@Expose
public List<Post> posts = new ArrayList<Post>();
}
@Generated("org.jsonschema2pojo")
public class PageInfo {
@SerializedName("pageName")
@Expose
public String pageName;
@SerializedName("pagePic")
@Expose
public String pagePic;
}
@Generated("org.jsonschema2pojo")
public class Post {
@SerializedName("post_id")
@Expose
public String postId;
@SerializedName("actor_id")
@Expose
public long actorId;
@SerializedName("picOfPersonWhoPosted")
@Expose
public String picOfPersonWhoPosted;
@SerializedName("nameOfPersonWhoPosted")
@Expose
public String nameOfPersonWhoPosted;
@SerializedName("message")
@Expose
public String message;
@SerializedName("likesCount")
@Expose
public long likesCount;
@SerializedName("comments")
@Expose
public List<Object> comments = new ArrayList<Object>();
@SerializedName("timeOfPost")
@Expose
public long timeOfPost;
}
我相信最好的做法应该是通过仍在开发中的官方Java JSON API。
由于还没有人提到它，这里是一个使用Nashorn (Java 8的JavaScript运行时部分，但在Java 11中已弃用)的解决方案的开始。
解决方案
private static final String EXTRACTOR_SCRIPT =
"var fun = function(raw) { " +
"var json = JSON.parse(raw); " +
"return [json.pageInfo.pageName, json.pageInfo.pagePic, json.posts[0].post_id];};";
public void run() throws ScriptException, NoSuchMethodException {
ScriptEngine engine = new ScriptEngineManager().getEngineByName("nashorn");
engine.eval(EXTRACTOR_SCRIPT);
Invocable invocable = (Invocable) engine;
JSObject result = (JSObject) invocable.invokeFunction("fun", JSON);
result.values().forEach(e -> System.out.println(e));
}
性能比较
我编写的JSON内容包含三个数组，分别为20、20和100个元素。我只想从第三个数组中获取100个元素。我使用下面的JavaScript函数来解析和获取我的条目。
var fun = function(raw) {JSON.parse(raw).entries};
使用Nashorn运行一百万次调用需要7.5~7.8秒
(JSObject) invocable.invokeFunction("fun", json);
org。Json需要20~21秒
new JSONObject(JSON).getJSONArray("entries");
杰克逊用时6.5~7秒
mapper.readValue(JSON, Entries.class).getEntries();
在这种情况下，Jackson的性能比Nashorn好，后者的性能比org.json好得多。 Nashorn API比org更难使用。json或Jackson的。根据您的需求，Jackson和Nashorn都是可行的解决方案。
