android JSONException 索引 1 超出范围 [0..1](在 1 个循环内解析 2 json 数组)

android JSONException index 1 out of range [0..1] (Parse 2 json arrays inside 1 loop)

我有这样的代码,jArrAnswer的值是

[{"answer":"Yes"},{"answer":"No"},{"answer":"maybe"},{"answer":"yrg"}]

jArrAnswer.length() 的结果是 4

但为什么我得到错误

org.json.JSONException: Index 1 out of range [0..1).

 try {
        JSONArray jArrAnswerid = new JSONArray(answerid);
        JSONArray jArrAnswer = new JSONArray(answer);
        for (int i = 0; i < jArrAnswer.length(); i++) {
            JSONObject jObjAnswerid = jArrAnswerid.getJSONObject(i);
            JSONObject jObjAnswer = jArrAnswer.getJSONObject(i);
            String ansid = jObjAnswerid.getString("answerid");
            String ans= jObjAnswer.getString("answer");
            GroupModel item2 = new GroupModel(String.valueOf(i + 1), ans, ansid);
        }
} catch (Exception e) {
        Log.w("asdf", e.toString());
}

您在 jArrAnswer 上迭代 for loop,同时在 jArrAnswerid 上获取索引 i

检查并确保 jArrAnswerid.size() 等于 jArrAnswer.size()

打印 jArrAnswerid.size() 并检查。

试试这个

try {
        JSONArray jArrAnswer = new JSONArray(answer);
        for (int i = 0; i < jArrAnswer.length(); i++) {
            JSONObject jObjAnswer = jArrAnswer.getJSONObject(i);
            String ansid = jObjAnswer.getString("answerid");
            String ans= jObjAnswer.getString("answer");
        }
     } catch (Exception e) {
        Log.w("asdf", e.toString());
     }

前提是 "answer" 是您的 json 数组响应

尝试

String json = "[{\"answer\":\"Yes\",\"answerid\":\"1\"},{\"answer\":\"No\",\"answerid\":\"2\"},{\"answer\":\"maybe\",\"answerid\":\"3\"},{\"answer\":\"yrg\",\"answerid\":\"4\"}]";

try {
    JSONArray jsonArray = new JSONArray(json);
    if(jsonArray != null) {
        for (int i = 0; i < jsonArray.length(); i++) {
            JSONObject jsonObject = jsonArray.getJSONObject(i);
            String answerId = jsonObject.getString("answerid");
            String answer = jsonObject.getString("answer");
            //Use answerId and answer
        }
    }
} catch(JSONException e) {
    e.printStackTrace();
}