从字符串转换为json对象android

时间:2013-08-12 17:17:46

标签: android json string-conversion jsonexception

我正在开发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中检查了一个解决方案,并在link

中找到了类似的问题

解决方案就像这样

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

我在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("\"", "\\\"");

在替换之后我得到了结果

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

执行JSONObject jsonObj = new JSONObject(mybizData);

我收到了以下json异常

org.json.JSONException: Expected literal value at character 1 of

请帮我解决我的问题。

9 个答案:

答案 0 :(得分:186)

删除斜杠:

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 + "\"");
}

答案 1 :(得分:18)

它的工作

    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 + "\"");
    }

答案 2 :(得分:7)

试试这个:

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

答案 3 :(得分:4)

从String中获取JSONObject或JSONArray我创建了这个类:

public static class JSON {

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

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

这里得到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);
}

示例:

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;
    }
}

答案 4 :(得分:3)

您只需要如下几行代码:

<text-prop>

答案 5 :(得分: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) {
    }

答案 6 :(得分:0)

这是代码,你可以决定哪个 (已同步)StringBuffer或 使用更快的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();
            }
        }

答案 7 :(得分:0)

可能会低于更好。

npm install

答案 8 :(得分:0)

只需尝试一下, 终于对我有用了:

//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);