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
主要有两种选择……
Object mapping. When you deserialize JSON data to a number of instances of: 1.1. Some predefined classes, like Maps. In this case, you don't have to design your own POJO classes. Some libraries: org.json.simple https://www.mkyong.com/java/json-simple-example-read-and-write-json/ 1.2. Your own POJO classes. You have to design your own POJO classes to present JSON data, but this may be helpful if you are going to use them into your business logic as well. Some libraries: Gson, Jackson (see http://tutorials.jenkov.com/java-json/index.html)
映射的主要缺点是它会导致大量内存分配(以及GC压力)和CPU占用。
面向流的解析。例如，Gson和Jackson都支持这种轻量级解析。另外，您还可以查看一个自定义的、快速的、无gc的解析器示例https://github.com/anatolygudkov/green-jelly。在需要解析大量数据和对延迟敏感的应用程序中，更倾向于使用这种方式。
您可以使用JsonNode来表示JSON字符串的结构化树。它是无处不在的杰克逊图书馆的一部分。
ObjectMapper mapper = new ObjectMapper();
JsonNode yourObj = mapper.readTree("{\"k\":\"v\"}");
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}]}
阅读下面的博文，Java中的JSON。
这篇文章有点老了，但我仍然想回答你的问题。
步骤1:创建数据的POJO类。
步骤2:现在使用JSON创建一个对象。
Employee employee = null;
ObjectMapper mapper = new ObjectMapper();
try {
employee = mapper.readValue(newFile("/home/sumit/employee.json"), Employee.class);
}
catch(JsonGenerationException e) {
e.printStackTrace();
}
如需进一步参考，请参阅以下链接。
请像这样做:
JSONParser jsonParser = new JSONParser();
JSONObject obj = (JSONObject) jsonParser.parse(contentString);
String product = (String) jsonObject.get("productId");
