117

我正在开发一个 Android 应用程序。在我的应用程序中，我必须将字符串转换为 JSON 对象，然后解析值。我在 Stackoverflow 中检查了解决方案，并在此处找到了类似的问题链接

解决方案是这样的

       `{"phonetype":"N95","cat":"WP"}`
        JSONObject jsonObj = new JSONObject("{\"phonetype\":\"N95\",\"cat\":\"WP\"}");

我在我的代码中使用相同的方式。我的字符串是

{"ApiInfo":{"description":"userDetails","status":"success"},"userDetails":{"Name":"somename","userName":"value"},"pendingPushDetails":[]}

string mystring= mystring.replace("\"", "\\\"");

替换后我得到了这样的结果

{\"ApiInfo\":{\"description\":\"userDetails\",\"status\":\"success\"},\"userDetails\":{\"Name\":\"Sarath Babu\",\"userName\":\"sarath.babu.sarath babu\",\"Token\":\"ZIhvXsZlKCNL6Xj9OPIOOz3FlGta9g\",\"userId\":\"118\"},\"pendingPushDetails\":[]}

当我执行JSONObject jsonObj = new JSONObject(mybizData);

我收到以下 JSON 异常

org.json.JSONException：字符 1 处的预期文字值

请帮我解决我的问题。

4

9 回答 9

254

删除斜线：

String json = {"phonetype":"N95","cat":"WP"};

try {

    JSONObject obj = new JSONObject(json);

    Log.d("My App", obj.toString());

} catch (Throwable t) {
    Log.e("My App", "Could not parse malformed JSON: \"" + json + "\"");
}
于 2013-08-12T17:24:24.953 回答
36

这种方法有效

    String json = "{\"phonetype\":\"N95\",\"cat\":\"WP\"}";

    try {

        JSONObject obj = new JSONObject(json);

        Log.d("My App", obj.toString());
        Log.d("phonetype value ", obj.getString("phonetype"));

    } catch (Throwable tx) {
        Log.e("My App", "Could not parse malformed JSON: \"" + json + "\"");
    }
于 2016-05-14T20:39:32.113 回答
8

试试这个：

String json = "{'phonetype':'N95','cat':'WP'}";
于 2014-03-13T10:22:29.067 回答
8

您只需要如下代码行：

   try {
        String myjsonString = "{\"phonetype\":\"N95\",\"cat\":\"WP\"}";
        JSONObject jsonObject = new JSONObject(myjsonString );
        //displaying the JSONObject as a String
        Log.d("JSONObject = ", jsonObject.toString());
        //getting specific key values
        Log.d("phonetype = ", jsonObject.getString("phonetype"));
        Log.d("cat = ", jsonObject.getString("cat");
    }catch (Exception ex) {
         StringWriter stringWriter = new StringWriter();
         ex.printStackTrace(new PrintWriter(stringWriter));
         Log.e("exception ::: ", stringwriter.toString());
    }
于 2018-07-06T11:01:03.073 回答
5

试试这个，最后这对我有用：

//delete backslashes ( \ ) :
            data = data.replaceAll("[\\\\]{1}[\"]{1}","\"");
//delete first and last double quotation ( " ) :
            data = data.substring(data.indexOf("{"),data.lastIndexOf("}")+1);
            JSONObject json = new JSONObject(data);
于 2019-01-21T14:08:48.223 回答
4

要从字符串中获取 JSONObject 或 JSONArray，我创建了此类： 

public static class JSON {

     public Object obj = null;
     public boolean isJsonArray = false;

     JSON(Object obj, boolean isJsonArray){
         this.obj = obj;
         this.isJsonArray = isJsonArray;
     }
}

在这里获取 JSON：

public static JSON fromStringToJSON(String jsonString){

    boolean isJsonArray = false;
    Object obj = null;

    try {
        JSONArray jsonArray = new JSONArray(jsonString);
        Log.d("JSON", jsonArray.toString());
        obj = jsonArray;
        isJsonArray = true;
    }
    catch (Throwable t) {
        Log.e("JSON", "Malformed JSON: \"" + jsonString + "\"");
    }

    if (object == null) {
        try {
            JSONObject jsonObject = new JSONObject(jsonString);
            Log.d("JSON", jsonObject.toString());
            obj = jsonObject;
            isJsonArray = false;
        } catch (Throwable t) {
            Log.e("JSON", "Malformed JSON: \"" + jsonString + "\"");
        }
    }

    return new JSON(obj, isJsonArray);
}

例子：

JSON json = fromStringToJSON("{\"message\":\"ciao\"}");
if (json.obj != null) {

    // If the String is a JSON array
    if (json.isJsonArray) {
        JSONArray jsonArray = (JSONArray) json.obj;
    }
    // If it's a JSON object
    else {
        JSONObject jsonObject = (JSONObject) json.obj;
    }
}
于 2016-09-28T08:10:25.813 回答
1

使用 Kotlin

    val data = "{\"ApiInfo\":{\"description\":\"userDetails\",\"status\":\"success\"},\"userDetails\":{\"Name\":\"somename\",\"userName\":\"value\"},\"pendingPushDetails\":[]}\n"
    
try {
      val jsonObject = JSONObject(data)
      val infoObj = jsonObject.getJSONObject("ApiInfo")
    } catch (e: Exception) {
    }
于 2021-02-20T00:14:10.757 回答
0

这是代码，您可以决定使用哪个
（同步）StringBuffer 或更快的 StringBuilder。

基准测试显示 StringBuilder 更快。

public class Main {
            int times = 777;
            long t;

            {
                StringBuffer sb = new StringBuffer();
                t = System.currentTimeMillis();
                for (int i = times; i --> 0 ;) {
                    sb.append("");
                    getJSONFromStringBuffer(String stringJSON);
                }
                System.out.println(System.currentTimeMillis() - t);
            }

            {
                StringBuilder sb = new StringBuilder();
                t = System.currentTimeMillis();
                for (int i = times; i --> 0 ;) {
                     getJSONFromStringBUilder(String stringJSON);
                    sb.append("");
                }
                System.out.println(System.currentTimeMillis() - t);
            }
            private String getJSONFromStringBUilder(String stringJSONArray) throws JSONException {
                return new StringBuffer(
                       new JSONArray(stringJSONArray).getJSONObject(0).getString("phonetype"))
                           .append(" ")
                           .append(
                       new JSONArray(employeeID).getJSONObject(0).getString("cat"))
                      .toString();
            }
            private String getJSONFromStringBuffer(String stringJSONArray) throws JSONException {
                return new StringBuffer(
                       new JSONArray(stringJSONArray).getJSONObject(0).getString("phonetype"))
                           .append(" ")
                           .append(
                       new JSONArray(employeeID).getJSONObject(0).getString("cat"))
                      .toString();
            }
        }
于 2015-02-09T11:28:55.403 回答
0

可能下面会更好。

JSONObject jsonObject=null;
    try {
        jsonObject=new JSONObject();
        jsonObject.put("phonetype","N95");
        jsonObject.put("cat","wp");
        String jsonStr=jsonObject.toString();
    } catch (JSONException e) {
        e.printStackTrace();
    }
于 2017-12-14T06:37:04.567 回答