Question

错误：

org.json.JSONException：java.lang.String类型的值[{“username”：“ghs”}]无法转换为JSONObject W / System.err：at org.json.JSON.typeMismatch（JSON.java:111）

Android端代码：

$ g++ --version
g++ (GCC) 6.2.1 20160830
Copyright (C) 2016 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

$ clang++ --version
clang version 3.9.0 (tags/RELEASE_390/final)
Target: x86_64-unknown-linux-gnu
Thread model: posix
InstalledDir: /usr/bin

$ g++ -std=c++11 order_of_ops.cpp -o a.out && ./a.out; echo $?
27

$ clang++ -std=c++11 order_of_ops.cpp -o a.out && ./a.out; echo $?
81

php结束：

public void listdrawer(){
            onlineList_uname = new ArrayList<>();
        try {
            JSONObject jsonResponse = new JSONObject(jsonResult);
            JSONArray jsonMainNode = jsonResponse.getJSONArray("username");
            for (int i = 0; i < jsonMainNode.length(); i++) {
                JSONObject jsonChildNode = jsonMainNode.getJSONObject(i);
                String name = jsonChildNode.optString("username");
                onlineList_uname.add(i,name);
            }
        } catch (JSONException e) {
            e.printStackTrace();
        }
    }

Answer 1

如果您的jsonResult = [{"username":"g4"},{"username":"ghs"},{"username":"g"},{"username":"e"},{"username":"a"}]

<强>代码：

ArrayList<String> onlineList_uname = new ArrayList<>();
String jsonResult = "[{\"username\":\"g4\"},{\"username\":\"ghs\"},{\"username\":\"g\"},{\"username\":\"e\"},{\"username\":\"a\"}]"; // = [{"username":"g4"},{"username":"ghs"},{"username":"g"},{"username":"e"},{"username":"a"}]

try {
  JSONArray jsonResponse = new JSONArray(jsonResult);

  for (int i = 0; i < jsonResponse.length(); i++) {
    JSONObject jObj = jsonResponse.getJSONObject(i);
    String name = jObj.getString("username");
    onlineList_uname.add(i, name);
  }
  for (int j = 0; j < onlineList_uname.size(); j++) {
    Log.i("LOG", "username#" + j + " " + onlineList_uname.get(j));
  }
} catch (JSONException e) {
  e.printStackTrace();
}

结果：

用户名＃0 g4

用户名＃1 ghs

用户名＃2 g

用户名＃3 e

用户名＃4 a

Answer 2

尝试：

jsonChildNode.optString("username").toString();

我认为之前我遇到过同样的问题，并且toString方法解决了它。

Android-Json到arraylist，org.json.JSONException org.json.JSON.typeMismatch（JSON.java:111）

2 个答案: