如何解析JSON对象的Andr​​oid工作室 [英] How to parse JSON Object Android Studio

问题描述

JSON对象

  {标题：九二，额定：PG-13，释放：年，蝙蝠侠归来九二6月19日 运行时：126分，类型：行动，导演：蒂姆·波顿，作家：鲍勃·凯恩（蝙蝠侠字符），丹尼尔·沃特斯（故事），萨姆·哈姆（故事），丹尼尔·沃特斯（剧本），演员：迈克尔·基顿，丹尼·德维托，米歇尔·菲佛，克里斯托弗·沃肯，语言：英语，国家：美国，英国，大奖：提名2奥斯卡另2胜＆安培; 15 nominations.\",\"Poster\":\"http://ia.media-imdb.com/images/M/MV5BODM2OTc0Njg2OF5BMl5BanBnXkFtZTgwMDA4NjQxMTE@._V1_SX300.jpg\",\"Metascore\":\"N/A\",\"imdbRating\":\"7.0\",\"imdbVotes\":\"199,878\",\"imdbID\":\"tt0103776\",\"Type\":\"movie\",\"Response\":\"True\"}
 

我试图解析在Android Studio中这个对象然而即时得到一个错误：

 类型org.json.JSONObject不能转换成JSONArray
 

这是code，我使用

  JSONArray mJsonArray =新JSONArray（jsonResult）;
的JSONObject movieObject = mJsonArray.getJSONObject（0）;字符串title = movieObject.getString（标题）;
 

解决方案

您JSON包含一个对象，而不是一个数组。替换

  JSONArray mJsonArray =新JSONArray（jsonResult）;
 

按

 的JSONObject movieObject =新的JSONObject（jsonResult）;
字符串title = movieObject.getString（标题）;
 

The JSON Object

{"Title":"Batman Returns","Year":"1992","Rated":"PG-13","Released":"19 Jun 1992","Runtime":"126 min","Genre":"Action","Director":"Tim Burton","Writer":"Bob Kane (Batman characters), Daniel Waters (story), Sam Hamm (story), Daniel Waters (screenplay)","Actors":"Michael Keaton, Danny DeVito, Michelle Pfeiffer, Christopher Walken","Language":"English","Country":"USA, UK","Awards":"Nominated for 2 Oscars. Another 2 wins & 15 nominations.","Poster":"http://ia.media-imdb.com/images/M/MV5BODM2OTc0Njg2OF5BMl5BanBnXkFtZTgwMDA4NjQxMTE@._V1_SX300.jpg","Metascore":"N/A","imdbRating":"7.0","imdbVotes":"199,878","imdbID":"tt0103776","Type":"movie","Response":"True"}

I'm trying to parse this object in android studio however im getting an error:

of type org.json.JSONObject cannot be converted to JSONArray

This is the code that I'm using

JSONArray mJsonArray = new JSONArray(jsonResult);
JSONObject movieObject = mJsonArray.getJSONObject(0);

String title = movieObject.getString("Title");

解决方案

Your json contains an object, not an array. Replace

JSONArray mJsonArray = new JSONArray(jsonResult);

by

JSONObject movieObject = new JSONObject(jsonResult);    
String title = movieObject.getString("Title");

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