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

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!

其他回答

您可以使用Gson库来解析JSON字符串。

Gson gson = new Gson();
JsonObject jsonObject = gson.fromJson(jsonAsString, JsonObject.class);

String pageName = jsonObject.getAsJsonObject("pageInfo").get("pageName").getAsString();
String pagePic = jsonObject.getAsJsonObject("pageInfo").get("pagePic").getAsString();
String postId = jsonObject.getAsJsonArray("posts").get(0).getAsJsonObject().get("post_id").getAsString();

你也可以循环"posts"数组,如下所示:

JsonArray posts = jsonObject.getAsJsonArray("posts");
for (JsonElement post : posts) {
  String postId = post.getAsJsonObject().get("post_id").getAsString();
  //do something
}

Quick-json解析器非常简单,灵活,快速,可定制。试一试

特点:

Compliant with JSON specification (RFC4627) High-Performance JSON parser Supports Flexible/Configurable parsing approach Configurable validation of key/value pairs of any JSON Hierarchy Easy to use # Very small footprint Raises developer friendly and easy to trace exceptions Pluggable Custom Validation support - Keys/Values can be validated by configuring custom validators as and when encountered Validating and Non-Validating parser support Support for two types of configuration (JSON/XML) for using quick-JSON validating parser Requires JDK 1.5 No dependency on external libraries Support for JSON Generation through object serialisation Support for collection type selection during parsing process

它可以这样使用:

JsonParserFactory factory=JsonParserFactory.getInstance();
JSONParser parser=factory.newJsonParser();
Map jsonMap=parser.parseJson(jsonString);

请像这样做:

JSONParser jsonParser = new JSONParser();
JSONObject obj = (JSONObject) jsonParser.parse(contentString);
String product = (String) jsonObject.get("productId");

我相信最好的做法应该是通过仍在开发中的官方Java JSON API。

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。