将 Json 数组转换为普通的 Java 列表
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原文地址: http://stackoverflow.com/questions/3395729/
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Convert Json Array to normal Java list
提问by Houston we have a problem
Is there a way to convert JSON Array to normal Java Array for android ListView data binding?
有没有办法将 JSON 数组转换为普通的 Java 数组，用于 android ListView 数据绑定？
采纳答案by Pentium10
ArrayList<String> list = new ArrayList<String>();
JSONArray jsonArray = (JSONArray)jsonObject;
if (jsonArray != null) {
int len = jsonArray.length();
for (int i=0;i<len;i++){
list.add(jsonArray.get(i).toString());
}
}
回答by Nick
If you don't already have a JSONArray object, call
如果您还没有 JSONArray 对象，请调用
JSONArray jsonArray = new JSONArray(jsonArrayString);
Then simply loop through that, building your own array. This code assumes it's an array of strings, it shouldn't be hard to modify to suit your particular array structure.
然后简单地循环遍历，构建自己的数组。此代码假定它是一个字符串数组，修改以适合您的特定数组结构应该不难。
List<String> list = new ArrayList<String>();
for (int i=0; i<jsonArray.length(); i++) {
list.add( jsonArray.getString(i) );
}
回答by Michal Tsadok
How about using java.util.Arrays?
使用 java.util.Arrays 怎么样？
List<String> list = Arrays.asList((String[])jsonArray.toArray())
回答by Guido Sabatini
Maybe it's only a workaround (not very efficient) but you could do something like this:
也许这只是一种解决方法（效率不高），但您可以执行以下操作：
String[] resultingArray = yourJSONarray.join(",").split(",");
String[] resultingArray = yourJSONarray.join(",").split(",");
Obviously you can change the ',' separator with anything you like (I had a JSONArrayof email addresses)
显然你可以,用你喜欢的任何东西来改变 ' ' 分隔符（我有一个JSONArray电子邮件地址）
回答by pchot
I know that question is about JSONArray but here's example I've found useful where you don't need to use JSONArray to extract objects from JSONObject.
我知道这个问题是关于 JSONArray 的，但我发现这里的示例很有用，您不需要使用 JSONArray 从 JSONObject 中提取对象。
import org.json.simple.JSONObject;
import org.json.simple.JSONValue;
String jsonStr = "{\"types\":[1, 2]}";
JSONObject json = (JSONObject) JSONValue.parse(jsonStr);
List<Long> list = (List<Long>) json.get("types");
if (list != null) {
for (Long s : list) {
System.out.println(s);
}
}
Works also with array of strings
也适用于字符串数组
回答by Joolah
Here is a better way of doing it: if you are getting the data from API. Then PARSE the JSON and loading it onto your listview:
这是一个更好的方法：如果您从 API 获取数据。然后解析 JSON 并将其加载到您的列表视图中：
protected void onPostExecute(String result) {
Log.v(TAG + " result);
if (!result.equals("")) {
// Set up variables for API Call
ArrayList<String> list = new ArrayList<String>();
try {
JSONArray jsonArray = new JSONArray(result);
for (int i = 0; i < jsonArray.length(); i++) {
list.add(jsonArray.get(i).toString());
}//end for
} catch (JSONException e) {
Log.e(TAG, "onPostExecute > Try > JSONException => " + e);
e.printStackTrace();
}
adapter = new ArrayAdapter<String>(ListViewData.this, android.R.layout.simple_list_item_1, android.R.id.text1, list);
listView.setAdapter(adapter);
listView.setOnItemClickListener(new OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
// ListView Clicked item index
int itemPosition = position;
// ListView Clicked item value
String itemValue = (String) listView.getItemAtPosition(position);
// Show Alert
Toast.makeText( ListViewData.this, "Position :" + itemPosition + " ListItem : " + itemValue, Toast.LENGTH_LONG).show();
}
});
adapter.notifyDataSetChanged();
adapter.notifyDataSetChanged();
...
回答by StaxMan
Instead of using bundled-in org.jsonlibrary, try using Hymanson or GSON, where this is a one-liner. With Hymanson, f.ex:
不要使用捆绑org.json库，而是尝试使用 Hymanson 或 GSON，这是单行的。与Hyman逊，f.ex：
List<String> list = new ObjectMapper().readValue(json, List.class);
// Or for array:
String[] array = mapper.readValue(json, String[].class);
回答by detman
Use can use a String[]instead of an ArrayList<String>:
使用可以使用 aString[]代替 an ArrayList<String>：
It will reduce the memory overhead that an ArrayList has
它将减少 ArrayList 的内存开销
Hope it helps!
希望能帮助到你！
String[] stringsArray = new String[jsonArray.length()];
for (int i = 0; i < jsonArray.length; i++) {
parametersArray[i] = parametersJSONArray.getString(i);
}
回答by ceph3us
we starting from conversion [ JSONArray -> List < JSONObject > ]
我们从转换开始 [ JSONArray -> List < JSONObject > ]
public static List<JSONObject> getJSONObjectListFromJSONArray(JSONArray array)
throws JSONException {
ArrayList<JSONObject> jsonObjects = new ArrayList<>();
for (int i = 0;
i < (array != null ? array.length() : 0);
jsonObjects.add(array.getJSONObject(i++))
);
return jsonObjects;
}
next create generic version replacing array.getJSONObject(i++) with POJO
接下来创建通用版本用 POJO 替换 array.getJSONObject(i++)
example :
例子 ：
public <T> static List<T> getJSONObjectListFromJSONArray(Class<T> forClass, JSONArray array)
throws JSONException {
ArrayList<Tt> tObjects = new ArrayList<>();
for (int i = 0;
i < (array != null ? array.length() : 0);
tObjects.add( (T) createT(forClass, array.getJSONObject(i++)))
);
return tObjects;
}
private static T createT(Class<T> forCLass, JSONObject jObject) {
// instantiate via reflection / use constructor or whatsoever
T tObject = forClass.newInstance();
// if not using constuctor args fill up
//
// return new pojo filled object
return tObject;
}
回答by Iman Marashi
You can use a String[]instead of an ArrayList<String>:
您可以使用 aString[]代替 an ArrayList<String>：
Hope it helps!
希望能帮助到你！
private String[] getStringArray(JSONArray jsonArray) throws JSONException {
if (jsonArray != null) {
String[] stringsArray = new String[jsonArray.length()];
for (int i = 0; i < jsonArray.length(); i++) {
stringsArray[i] = jsonArray.getString(i);
}
return stringsArray;
} else
return null;
}
