JSON解析报错 怎么解决？贴出错误和代码

{"res":"01","info":{"SongName":"L\u1ee1 y\u00eaubbbbbbbb","WordNum":"2","PyCode":"LY","Stroke":"0","SingerName1":"Hu\u1ef3nh Nh\u1eadt Huy","SingerName2":"","FileName":"2","Lang":"1","SongTypeID":"3","SingerID1":"123","SingerID2":"456","MType":"1","yTrack":"0","bTrack":"1","yVolume":"92","bVolume":"92","NewSong":"1","style":"9","SongNameSpell":"aaa","SingerNameSpell1":null,"SingerNameSpell2":null}}

这是我的JSON串
然后我的解析是这样做的

JSONObject obj = new JSONObject(response.toString());
JSONArray arr = obj.getJSONArray("info");
for (int i = 0; i < arr.length(); i++) {
JSONObject subObj = arr.getJSONObject(i);
String sn = subObj.getString("SongName");
String w = subObj.getString("WordNum");
String p = subObj.getString("PyCode");
String st = subObj.getString("Stroke");
String singer1 = subObj.getString("SingerName1");
。。。。。。。
然后打印出解析的JSON串
}

报错是
Value {"Lang":"1","yVolume":"92","MType":"1","SongNameSpell":"aaa","Stroke":"0","SongName":"Lỡ yêubbbbbbbb","SingerNameSpell1":null,"SingerNameSpell2":null,"yTrack":"0","SongTypeID":"3","PyCode":"LY","style":"9","bVolume":"92","FileName":"2","SingerName1":"Huỳnh Nhật Huy","WordNum":"2","SingerName2":"","SingerID2":"456","NewSong":"1","SingerID1":"123","bTrack":"1"}
at info of type org.json.JSONObject cannot be converted to JSONArray

请问这是什么情况 求解答啊。。

json字符串不是JSONArray的类型啊 当然转换不了成JSONArray

JSONObject 不能被转化为JSONArray ,

换个json工具 fastJson 解析就一行代码 JSONObject.parseObject("json字符串")

你这个里面都没有数组，所以不需要用JSONArray，直接像第一个解析那样用JSONObject就行。然后不需要for循环，直接就拿你需要数据的字段就可以了

{“res”:01, "info": {???????} } 。 都是对象 JSONObject arr = obj.getJSONObject("info"); ‘

{“res”:01, "info": [???????] } 。 如果是这样的话 你就没错。注意，区别不大，括号不同。这是数组
但是如果你要拿到全部数据会很麻烦。 可以用第三方的解析工具。 Gson、fastJson什么的会简单很多。

    db.execSQL("CREATE TABLE minfo (id INTEGER PRIMARY KEY AUTOINCREMENT, name VARCHAR, age SMALLINT, SingerName VARCHAR ,WordNum SMALLINT,ver SMALLINT,PyCode SMALLINT,Stroke SMALLINT,SingerName1 VARCHAR,SingerName2 VARCHAR,FileName SMALLINT,Lang VARCHAR,SongTypeID SMALLINT,SingerID1 SMALLINT,SingerID2 SMALLINT,MType SMALLINT,yTrack SMALLINT,bTrack SMALLINT,yVolume SMALLINT,bVolume SMALLINT,NewSong SMALLINT,style SMALLINT,SongNameSpell VARCHAR,SongNameSpell1 VARCHAR,SongNameSpell2 VARCHAR)");

cv1.put("name", "aaaa");
cv1.put("WordNum", "6");
cv1.put("ver", 6);
cv1.put("age", "24");
cv1.put("PyCode", "KD");
cv1.put("Stroke", 4);
cv1.put("SingerName1", "lb");
cv1.put("SingerName2", "ja");
cv1.put("Lang", "0");
cv1.put("FileName", "3");
cv1.put("SongTypeID", 1);
‘’‘’
。。。
这些我都写了 可是解析的时候
JSONObject obj = new JSONObject(response.toString());
JSONArray arr = obj.getJSONArray("info");
String l = obj.getString("Lang");
String w = obj.getString("SingerName1");
Log.i("解析后的数据 ","SingerName1"+w);
为什么 比如说我要打印SingerName1 或 Lang等等这些值 后台会提示System.err: org.json.JSONException: No value for SingerName1 呢？？

很明显：info这个字段对应的是JsonObject 而不是JsonArray,所以你转换就会抛异常啊

JSONObject obj = new JSONObject(response.toString());
JSONObject object = obj.getJSONObject("info");
String l = object.getString("Lang");
String w = object.getString("SingerName1");
Log.i("解析后的数据 ","SingerName1"+w);
这样的话就没问题了，要区分[]和{}的区别哟~