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"
}
]
}
JSON是这样的:
{
"pageInfo": {
"pageName": "abc",
"pagePic": "http://example.com/content.jpg"
}
}
Java类
class PageInfo {
private String pageName;
private String pagePic;
// Getters and setters
}
将此JSON转换为Java类的代码。
PageInfo pageInfo = JsonPath.parse(jsonString).read("$.pageInfo", PageInfo.class);
Maven
<dependency>
<groupId>com.jayway.jsonpath</groupId>
<artifactId>json-path</artifactId>
<version>2.2.0</version>
</dependency>
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;
}
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!