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這是我的java代碼來從服務器獲取響應。如何將json對象轉換爲java
try
{
HttpClient httpclient = new DefaultHttpClient();
HttpResponse response;
HttpPost httppost = new HttpPost("http://www.def.net/my_script.php");
response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
String s = EntityUtils.toString(entity);
JSONObject json = (JSONObject)new JSONParser().parse(s);
System.out.println("News title= " + json.get("news_title"));
System.out.println("Content= " + json.get("news_content"));
}
catch (Exception e)
{
e.printStackTrace();
}
這是什麼PHP腳本返回。
{ "desktop_app_newsfeed":
[
{ "news_id": "132",
"news_title": "test1",
"news_content": "Lorem ipsum dolor sit amet Donec...",
"news_date": "2013-07-18 10:38:20" },
{ "news_id": "1",
"news_title": "Hello world!",
"news_content":"Lorem ipsum dolor sit amet Donec...",
"news_date": "2013-04-22 17:54:05"
}
]
}
迭代得到如何分配給java中的變量。
您正在使用哪個JSON實現庫? – zsawyer
簡單的json對象庫。 – belekka
如果您的意思是JSON-simple - **您不清楚** - 解碼示例位於http://code.google.com/p/json-simple/wiki/DecodingExamples。你的代碼提到JSONObject,我認爲它可能在不同的庫中。您的Java代碼是否高於實際代碼,是否會進行編譯?如果是,它的功能是什麼? –