在Android中未正确解析和拆分JSON
我试图解析一个包含未知数据的在Android中未正确解析和拆分JSON,android,json,Android,Json,我试图解析一个包含未知数据的JSON数组。JSON的JSON就是这样通过的 { "info": [ { "1": "4", "Store #": " 0560", "How many microwave circuits did you run?": " 3", "How many new ovens did you deliver to the store?": " 1",
JSON数组。JSON的JSON
就是这样通过的
{
"info": [
{
"1": "4",
"Store #": " 0560",
"How many microwave circuits did you run?": " 3",
"How many new ovens did you deliver to the store?": " 1",
"How many new racks did you deliver to the store?": " 5",
"Voltage readings on Turbo Chef 1": " 64",
"Voltage readings on Turbo Chef 2": " 54",
...
}
],
"success": 1
}
try {
// Checking for SUCCESS TAG
int success = json.getInt(TAG_SUCCESS);
if (success == 1) {
info = json.getJSONArray(TAG_INFO);
for (int i = 0; i < info.length(); i++) {
if (info != null) {
clientList.add(info.get(i).toString());
}
}
for (String s : clientList) {
Log.v("CHECKING S", s);
String[] str_arr= s.split("[:,]");
Log.v("CHECKING VALUES 0", str_arr[0]);
mQuestions.add(str_arr[0]);
Log.v("CHECKING VALUES 1", str_arr[1]);
}
for (String content : mQuestions) {
Log.v("QUESTIONS", content);
}
}
现在，因为在收到数据之前我不知道数据是什么，所以我不能使用传统的方法，比如
for (int i = 0; i < info.length(); i++) {
JSONObject c = info.getJSONObject(i);
String store = c.getString("Store #");
}
那么我做错了什么？如果你还需要密码，我会把它贴出来。谢谢你的帮助
编辑
根据我收到的答案，我对代码进行了修改。现在看起来像这样
{
"info": [
{
"1": "4",
"Store #": " 0560",
"How many microwave circuits did you run?": " 3",
"How many new ovens did you deliver to the store?": " 1",
"How many new racks did you deliver to the store?": " 5",
"Voltage readings on Turbo Chef 1": " 64",
"Voltage readings on Turbo Chef 2": " 54",
...
}
],
"success": 1
}
try {
// Checking for SUCCESS TAG
int success = json.getInt(TAG_SUCCESS);
if (success == 1) {
info = json.getJSONArray(TAG_INFO);
for (int i = 0; i < info.length(); i++) {
if (info != null) {
clientList.add(info.get(i).toString());
}
}
for (String s : clientList) {
Log.v("CHECKING S", s);
String[] str_arr= s.split("[:,]");
Log.v("CHECKING VALUES 0", str_arr[0]);
mQuestions.add(str_arr[0]);
Log.v("CHECKING VALUES 1", str_arr[1]);
}
for (String content : mQuestions) {
Log.v("QUESTIONS", content);
}
}
您需要将s.split（“：”）
结果分配给字符串数组，如下所示：
for (String s : clientList) {
Log.v("CHECKING S", s);
String[] str_arr= s.split(":"); //<<<<<
Log.v("CHECKING S SPLIT", str_arr);
Log.v("CHECKING VALUES 0", str_arr[0]);
mQuestions.add(str_arr[0]);
Log.v("CHECKING VALUES 1", str_arr[1]);
mAnswers.add(str_arr[1]);
}
for（字符串s:clientList）{
Log.v（“检查S”，S）；
String[]str_arr=s.split（“：”）；//您需要将s.split（“：”）
结果分配给字符串数组，如下所示：
for (String s : clientList) {
Log.v("CHECKING S", s);
String[] str_arr= s.split(":"); //<<<<<
Log.v("CHECKING S SPLIT", str_arr);
Log.v("CHECKING VALUES 0", str_arr[0]);
mQuestions.add(str_arr[0]);
Log.v("CHECKING VALUES 1", str_arr[1]);
mAnswers.add(str_arr[1]);
}
for（字符串s:clientList）{
Log.v（“检查S”，S）；
String[]str_arr=s.split（“：”）；//您好，我已经为您完成了这项工作。我只是在粘贴下面的完整代码。尝试这种json，它肯定会对您有用
我将该方法称为：
String data = "{'info': [{'1': '4','Store #': ' 0560','How many microwave circuits did you run?': ' 3','How many new ovens did you deliver to the store?': ' 1','How many new racks did you deliver to the store?': ' 5','Voltage readings on Turbo Chef 1': ' 64','Voltage readings on Turbo Chef 2': ' 54'}],'success': 1}";
try {
JSONObject jsonData = new JSONObject(data);
populateFromResponse(jsonData);
} catch (JSONException e) {
System.out.println(" Exception occured" + e.getMessage());
e.printStackTrace();
}
}
现在，populateFromResponse（）方法的实现如下所示：
private static final String TAG_SUCCESS = "success"; private static final String TAG_INFO = "info"; private ArrayList<JSONObject> clientList; private ArrayList<String> mQuestions; private ArrayList<String> mAnswers;
public void populateFromResponse(JSONObject json) throws JSONException {
// parse the response if (json != null) {
// Checking for SUCCESS TAG int success = json.getInt(TAG_SUCCESS); // checking for null if (success == 1) { JSONArray info = json.getJSONArray(TAG_INFO); if (info != null) {
for (int i = 0; i < info.length(); i++) {
// checking for null
if (info.get(i) != null
&& info.get(i) instanceof JSONObject)
if (clientList == null) {
clientList = new ArrayList<JSONObject>();
}
clientList.add(info.getJSONObject(i));
} }
for (JSONObject s : clientList) {
Log.v("CHECKING S", s.toString());
// Iterator containing all the keys
Iterator<String> iterator = s.keys();
while (iterator.hasNext()) {
String key = (String) iterator.next();
String value = s.getString(key);
print("CHECKING VALUES 0: " + key);
// checking for null
if (mQuestions == null) {
mQuestions = new ArrayList<String>();
}
mQuestions.add(key);
print("CHECKING VALUES 1: " + value);
// checking for null
if (mAnswers == null) {
mAnswers = new ArrayList<String>();
}
mAnswers.add(value);
}
print("CHECKING S SPLIT: " + s);
} } } else { print("response is null"); }
}
private void print(String string) { System.out.println("######-- " + string + " -----#########"); }
现在享受代码并快乐起来。
再见。您好，我已经为您完成了这项工作。我只是在粘贴下面的完整代码。尝试一下这种json，它肯定会对您有用
我将该方法称为：
String data = "{'info': [{'1': '4','Store #': ' 0560','How many microwave circuits did you run?': ' 3','How many new ovens did you deliver to the store?': ' 1','How many new racks did you deliver to the store?': ' 5','Voltage readings on Turbo Chef 1': ' 64','Voltage readings on Turbo Chef 2': ' 54'}],'success': 1}";
try {
JSONObject jsonData = new JSONObject(data);
populateFromResponse(jsonData);
} catch (JSONException e) {
System.out.println(" Exception occured" + e.getMessage());
e.printStackTrace();
}
}
现在，populateFromResponse（）方法的实现如下所示：
private static final String TAG_SUCCESS = "success"; private static final String TAG_INFO = "info"; private ArrayList<JSONObject> clientList; private ArrayList<String> mQuestions; private ArrayList<String> mAnswers;
public void populateFromResponse(JSONObject json) throws JSONException {
// parse the response if (json != null) {
// Checking for SUCCESS TAG int success = json.getInt(TAG_SUCCESS); // checking for null if (success == 1) { JSONArray info = json.getJSONArray(TAG_INFO); if (info != null) {
for (int i = 0; i < info.length(); i++) {
// checking for null
if (info.get(i) != null
&& info.get(i) instanceof JSONObject)
if (clientList == null) {
clientList = new ArrayList<JSONObject>();
}
clientList.add(info.getJSONObject(i));
} }
for (JSONObject s : clientList) {
Log.v("CHECKING S", s.toString());
// Iterator containing all the keys
Iterator<String> iterator = s.keys();
while (iterator.hasNext()) {
String key = (String) iterator.next();
String value = s.getString(key);
print("CHECKING VALUES 0: " + key);
// checking for null
if (mQuestions == null) {
mQuestions = new ArrayList<String>();
}
mQuestions.add(key);
print("CHECKING VALUES 1: " + value);
// checking for null
if (mAnswers == null) {
mAnswers = new ArrayList<String>();
}
mAnswers.add(value);
}
print("CHECKING S SPLIT: " + s);
} } } else { print("response is null"); }
}
private void print(String string) { System.out.println("######-- " + string + " -----#########"); }
现在享受代码并快乐起来。
再见。你在哪里向值添加项目？未知数据，你的意思是未知值吗？@SmartLemon是的，我永远不知道JSON
包含什么，或者它将包含多少值。你不知道在什么级别？你知道它是否会有一个信息数组吗？@SmartLemon它将始终具有信息数组，我创建了它，但是res它的t是未知的。此外，信息数组将始终有一个“success”标记它将是1或0，您在哪里向值添加项目？未知数据，您是指未知值吗？@SmartLemon是的，我永远不知道JSON
包含什么，或者它将包含多少值。您不知道在什么级别？您知道它是否将具有信息数组吗？@SmartLemon它将始终具有信息数组，我知道吃掉它，但它的其余部分是未知的。此外，信息数组将始终有一个“成功”的标记谢谢你的回答！你是最好的。而且，现在logcat显示我也需要在，
上拆分，因为每行JSON
的结尾都是，
谢谢你的回答！你是最好的。另外，现在logcat显示我也需要在，
上拆分，因为eac的结尾hJSON
行以，