从字符串到 JSON 对象的转换

我正在开发一个 Android 应用程序。在我的应用程序中，我必须将一个字符串转换为 JSON 对象，然后解析这些值。我在 Stackoverflow 检查了一个解决方案，在这里发现了类似的问题

解决方案是这样的

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

我在代码中使用相同的方式

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


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

在替换之后，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);时

我得到了下面的 JSON 异常

异常: 字符1处的预期文本值

Please help me to solve my issue.

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最佳答案

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 + "\"");
}
小开

试试这个:

String json = "{'phonetype':'N95','cat':'WP'}";
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这是密码，你可以选择
(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();
}
}
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This method works

    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 + "\"");
}
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为了从 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;
}
}

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);
}

例如:

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;
}
}
小开

也许下面更好。

JSONObject jsonObject=null;
try {
jsonObject=new JSONObject();
jsonObject.put("phonetype","N95");
jsonObject.put("cat","wp");
String jsonStr=jsonObject.toString();
} catch (JSONException e) {
e.printStackTrace();
}
小开

您只需要下面的代码行:

   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());
}
小开

just try this , 最后，这个方法对我有效:

//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);
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利用科特林

    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) {
}

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