i have trouble to load somedata, i already database and has query name "tipe" (type), in query tipe only has 2 type of string ("1" and "2") So, how to load data if im only want load sepecific data tipe="1"....
我有加载somedata的麻烦,我已经数据库并且有查询名称“tipe”(类型),在查询tipe中只有2种类型的字符串(“1”和“2”)那么,如果我只想加载如何加载数据特定数据tipe =“1”....
this my code in android :
这是我在android中的代码:
MovieFragment.java
// Parsing json
for (int i = 0; i < response.length(); i++) {
try {
JSONObject obj = response.getJSONObject(i);
Movie movie = new Movie();
movie.setTitle(obj.getString("name"));
movie.setThumbnailUrl(obj.getString("images1"));
//movie.setDescribe(obj.getString("describe"));
//movie.setRating(((Number) obj.get("rating"))
// .doubleValue());
movie.setYear(obj.getInt("id"));
movie.setTipe(obj.getString("tipe"));
/*// Genre is json array
JSONArray genreArry = obj.getJSONArray("genre");
ArrayList<String> genre = new ArrayList<String>();
for (int j = 0; j < genreArry.length(); j++) {
genre.add((String) genreArry.get(j));
}
movie.setGenre(genre);*/
// adding movie to movies array
movieList.add(movie);
} catch (JSONException e) {
e.printStackTrace();
}
}
1 个解决方案
#1
1
i would try somethig like this:
我会尝试这样的事情:
// Parsing json
for (int i = 0; i < response.length(); i++) {
try {
JSONObject obj = response.getJSONObject(i);
if (obj.getString("tipe").equals("tipe1){
Movie movie = new Movie();
movie.setTitle(obj.getString("name")); movie.setThumbnailUrl(obj.getString("images1"));
movie.setYear(obj.getInt("id"));
movie.setTipe(obj.getString("tipe"));
// adding movie to movies array
movieList.add(movie);
}
} catch (JSONException e) {
e.printStackTrace();
}
}
#1
1
i would try somethig like this:
我会尝试这样的事情:
// Parsing json
for (int i = 0; i < response.length(); i++) {
try {
JSONObject obj = response.getJSONObject(i);
if (obj.getString("tipe").equals("tipe1){
Movie movie = new Movie();
movie.setTitle(obj.getString("name")); movie.setThumbnailUrl(obj.getString("images1"));
movie.setYear(obj.getInt("id"));
movie.setTipe(obj.getString("tipe"));
// adding movie to movies array
movieList.add(movie);
}
} catch (JSONException e) {
e.printStackTrace();
}
}