更新时间:2022-01-02 05:06:24
阅读JsonArray和JsonObject文档后,我了解了如何解决此问题.
After reading JsonArray and JsonObject doc i understand how to sort out this problem.
protected void parseJson()
{
JSONObject object=null;
try {
object=new JSONObject(json);
myArray=object.getJSONArray(MY_ARRAY);
Log.e("Array Length",""+myArray.length());
key_id=new String[myArray.length()];
key_name=new String[myArray.length()];
for (int i=0;i<=myArray.length();i++)
{
JSONObject fetchObject=myArray.optJSONObject(i);
if(fetchObject==null)
{
//do nothing
}
else
{
key_id[i] = fetchObject.getString(KEY_ID);
key_name[i] = fetchObject.getString(KEY_NAME);
}
}
} catch (JSONException e) {
e.printStackTrace();
}
}