使用gson反序列化对象的特定JSON字段 [英] Using gson to deserialize specific JSON field of an object
问题描述
我有以下JSON字符串:
{
ms:images,5160.1,
turl:http://ts1.mm.bing.net/th?id=I4693880201938488&pid=1.1,
height:178,
width :300,
imgurl:http://www.attackingsoccer.com/wp-content/uploads/2011/07/World-Cup-2012-Draw.jpg,
抵消:0,
t:2014年世界杯资格赛 - 欧洲平局2012世界杯平局......,
w:719,
h :427,
ff:jpeg,
fs:52,
durl:www.attackingsoccer.com/2011/07/world -cup-2012-qualification-europe ...,
surl:http://www.attackingsoccer.com/2011/07/world-cup-2012-qualification-europe-draw/world- cup-2012-draw /,
mid:D9E91A0BA6F9E4C65C82452E2A5604BAC8744F1B,
k:6,
ns:API.images
}
我需要存储 imgurl
在一个单独的字符串中。
这就是我所持续的l现在,但这只是给了我整个JSON字符串,而不是特定的imgurl字段。
pre $ g $ gson = new Gson() ;
Data data = new Data();
data = gson.fromJson(toExtract,Data.class);
System.out.println(data);
toExtract
是JSON字符串。
这里是我的数据类:
public class Data
{
public List< urlString> ; myurls;
}
class urlString
{
String imgurl;
}
解析这样一个简单的结构,不需要专门的课程。
解决方案1:
你可以这样做:
JsonParser parser = new JsonParser();
JsonObject obj = parser.parse(toExtract).getAsJsonObject();
String imgurl = obj.get(imgurl)。getAsString();
这使用原始解析到 JsonObject 。
解决方案2 : 或者,您可以使用
属性data = gson.fromJson(toExtract,Properties.class);
并读取您的网址
String imgurl = data.getProperty(imgurl);
I have the following JSON string:
{
"ms": "images,5160.1",
"turl": "http://ts1.mm.bing.net/th?id=I4693880201938488&pid=1.1",
"height": "178",
"width": "300",
"imgurl": "http://www.attackingsoccer.com/wp-content/uploads/2011/07/World-Cup-2012-Draw.jpg",
"offset": "0",
"t": "World Cup 2014 Qualification – Europe Draw World Cup 2012 Draw ...",
"w": "719",
"h": "427",
"ff": "jpeg",
"fs": "52",
"durl": "www.attackingsoccer.com/2011/07/world-cup-2012-qualification-europe...",
"surl": "http://www.attackingsoccer.com/2011/07/world-cup-2012-qualification-europe-draw/world-cup-2012-draw/",
"mid": "D9E91A0BA6F9E4C65C82452E2A5604BAC8744F1B",
"k": "6",
"ns": "API.images"
}
I need to store the value of imgurl
in a separate string.
This is what I have till now, but this just gives me the whole JSON string instead of the specific imgurl field.
Gson gson = new Gson();
Data data = new Data();
data = gson.fromJson(toExtract, Data.class);
System.out.println(data);
toExtract
is the JSON string.
Here is my data class:
public class Data
{
public List<urlString> myurls;
}
class urlString
{
String imgurl;
}
When parsing such a simple structure, no need to have dedicated classes.
Solution 1 :
To get the imgurURL from your String with gson, you can do this :
JsonParser parser = new JsonParser();
JsonObject obj = parser.parse(toExtract).getAsJsonObject();
String imgurl = obj.get("imgurl").getAsString();
This uses a raw parsing into a JsonObject.
Solution 2 :
Alternatively, you could extract your whole data in a Properties
instance using
Properties data = gson.fromJson(toExtract, Properties.class);
and read your URL with
String imgurl = data.getProperty("imgurl");
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