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
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}]}
其他回答
这让我惊讶于它是多么简单。你可以在默认的组织中传递一个包含JSON的String给JSONObject的构造函数。json包。
JSONArray rootOfPage = new JSONArray(JSONString);
完成了。滴麦克风。 这也适用于JSONObjects。在此之后，您可以使用对象上的get()方法查看对象的层次结构。
首先，您需要选择一个实现库来执行此操作。
用于JSON处理的Java API (JSR 353)提供了使用对象模型和流API来解析、生成、转换和查询JSON的可移植API。
参考实现在这里:https://jsonp.java.net/
下面是JSR 353的实现列表:
哪些API实现了JSR-353 (JSON)
为了帮助你决定…我也找到了这篇文章:
http://blog.takipi.com/the-ultimate-json-library-json-simple-vs-gson-vs-jackson-vs-json/
如果您选择Jackson，这里有一篇关于使用Jackson在JSON和Java之间转换的好文章:https://www.mkyong.com/java/how-to-convert-java-object-to-from-json-jackson/
希望能有所帮助!
为了便于示例，让我们假设您有一个只有名称的Person类。
private class Person {
public String name;
public Person(String name) {
this.name = name;
}
}
谷歌GSON (Maven)
我个人最喜欢的JSON对象序列化/反序列化。
Gson g = new Gson();
Person person = g.fromJson("{\"name\": \"John\"}", Person.class);
System.out.println(person.name); //John
System.out.println(g.toJson(person)); // {"name":"John"}
更新
如果你想获取单个属性，你可以很容易地使用谷歌库:
JsonObject jsonObject = new JsonParser().parse("{\"name\": \"John\"}").getAsJsonObject();
System.out.println(jsonObject.get("name").getAsString()); //John
Org。JSON (Maven)
如果您不需要对象反序列化，而只是获得一个属性，您可以尝试org。json(或查看上面的GSON示例!)
JSONObject obj = new JSONObject("{\"name\": \"John\"}");
System.out.println(obj.getString("name")); //John
杰克逊(Maven)
ObjectMapper mapper = new ObjectMapper();
Person user = mapper.readValue("{\"name\": \"John\"}", Person.class);
System.out.println(user.name); //John
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}]}
我们可以使用JSONObject类将JSON字符串转换为JSON对象， 和遍历JSON对象。使用下面的代码。
JSONObject jObj = new JSONObject(contents.trim());
Iterator<?> keys = jObj.keys();
while( keys.hasNext() ) {
String key = (String)keys.next();
if ( jObj.get(key) instanceof JSONObject ) {
System.out.println(jObj.getString(String key));
}
}
