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。
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);