我是JSON使用的初学者。
所以我尝试从JSON回复中提取图像的URL。
这是允许我获取数组的代码：
// Instantiate the RequestQueue.
RequestQueue queue = Volley.newRequestQueue(getActivity());
//String url ="http://www.google.com";
String url = "http://ws.audioscrobbler.com/2.0/?method=album.search&album="+albumName+"&api_key=c51f8eb36bad&format=json";
// Request a string response from the provided URL.
StringRequest stringRequest = new StringRequest(Request.Method.GET, url,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
// Display the first 500 characters of the response string.
//mTextView.setText("Response is: "+ response.substring(0,500));
Log.i("RESPONSE","Response is: "+ response);
JSONObject jsono = new JSONObject();
try {
jsono = new JSONObject(response);
//String url = jsono.getString("results");
//Log.i("RESPONSE",url);
} catch (JSONException e) {
e.printStackTrace();
Log.d ("RESPONSE",e.getMessage());
}
JSONArray jsonArray = new JSONArray();
try {
jsonArray = jsono.getJSONObject("results").getJSONObject("albummatches").getJSONArray("album");
} catch (JSONException e) {
e.printStackTrace();
Log.d ("RESPONSE",e.getMessage());
}
for (int i = 0; i < jsonArray.length(); i++) {
JSONObject object = new JSONObject();
try {
object = jsonArray.getJSONObject(i);
Log.i("RESPONSE",object.getString("image"));
} catch (JSONException e) {
e.printStackTrace();
}
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Log.i("RESPONSE","That didn't work!");
}
});
queue.add(stringRequest);
以下是JSON答案中此部分的结构：
{
"album": [
{
"name": "DD Y Ponle Play",
"artist": "Jumbo",
"id": "2528039",
"url": "http://www.last.fm/music/Jumbo/DD+Y+Ponle+Play",
"image": [
{
"#text": "http://images.amazon.com/images/P/B00005LN6S.01._SCMZZZZZZZ_.jpg",
"size": "small"
},
{
"#text": "http://images.amazon.com/images/P/B00005LN6S.01._SCMZZZZZZZ_.jpg",
"size": "medium"
},
{
"#text": "http://images.amazon.com/images/P/B00005LN6S.01._SCMZZZZZZZ_.jpg",
"size": "large"
},
{
"#text": "http://images.amazon.com/images/P/B00005LN6S.01._SCMZZZZZZZ_.jpg",
"size": "extralarge"
}
]
}
]
}
如何获取给定尺寸的图像的网址？
非常感谢你的建议。
答案 0 :(得分：1)
您可以使用谷歌GSON。将其作为依赖项导入
首先创建一个专辑类。
public class Albums {
private List<Album> album;
public List<Album> getAlbum() {
return album;
}
public class Album{
private String name;
private String artist;
private String id;
private String url;
public String getName() {
return name;
}
public String getArtist() {
return artist;
}
public String getId() {
return id;
}
public String getUrl() {
return url;
}
public List<Image> getImage() {
return image;
}
public class Image {
@SerializedName("#text")
private String text;
private String size;
public String getText() {
return text;
}
public String getSize() {
return size;
}
}
private List<Image> image;
}
}
现在，在您获得上述JSON对象的代码中，请尝试以下代码
Gson gson = new Gson();
// Im assuming "response" as the above JSON object
Albums albums = gson.fromJson(response.optString("album"),Albums.class);
这会将您的json映射到java对象。（注意：您可以根据需要从POJO中删除不需要的对象）
您可以使用getter函数
获取图像答案 1 :(得分：0)
正如Raghunandan所提到的，当你需要提取JSONArray时，你正在提取一个JSONObject，然后从那个数组中提取一个JSONObject。
答案 2 :(得分：0)
解析JSON数组实际上非常简单：
JSONArray jsonarray = new JSONArray("album");
for (int i = 0; i < jsonarray.length(); i++) {
JSONObject jsonobject = jsonarray.getJSONObject(i);
String url = jsonobject.getString("url");
}
希望它有所帮助!!!
答案 3 :(得分：0)
JSON只不过是键值表示。抓住它并不难。你的代码应该是这样的，
更新：这只会打印包含size = medium
String response = "{\"album\":[{\"name\":\"DD Y Ponle Play\",\"artist\":\"Jumbo\",\"id\":\"2528039\",\"url\":\"http://www.last.fm/music/Jumbo/DD+Y+Ponle+Play\",\"image\":[{\"#text\":\"http://images.amazon.com/images/P/B00005LN6S.01._SCMZZZZZZZ_.jpg\",\"size\":\"small\"},{\"#text\":\"http://images.amazon.com/images/P/B00005LN6S.01._SCMZZZZZZZ_.jpg\",\"size\":\"medium\"},{\"#text\":\"http://images.amazon.com/images/P/B00005LN6S.01._SCMZZZZZZZ_.jpg\",\"size\":\"large\"},{\"#text\":\"http://images.amazon.com/images/P/B00005LN6S.01._SCMZZZZZZZ_.jpg\",\"size\":\"extralarge\"}]}]}";
JSONObject myObject = new JSONObject(response);
JSONArray myArray = myObject.getJSONArray( "album" );
for(int i=0; i<myArray.length(); i++)
{
JSONObject myIterator = myArray.getJSONObject( i );
JSONArray arrayOne = myIterator.getJSONArray( "image" );
for(int j=0; j<arrayOne.length(); j++)
{
JSONObject myInnerIterator = arrayOne.getJSONObject( j );
if(myInnerIterator.has( "size" ))//check if 'size' key is present
if(myInnerIterator.getString( "size" ).equalsIgnoreCase( "medium" ))
System.out.println( myInnerIterator.getString( "#text" ) );
}
}
答案 4 :(得分：0)
找到图像的网址＆＃34; medium＆＃34;我确实喜欢这个：
ArrayList<String> listUrl = new ArrayList<String>();
for(int i=0; i<jsonArray.length(); i++)
{
JSONObject myIterator = null;
try {
myIterator = jsonArray.getJSONObject( i );
JSONArray arrayOne = myIterator.getJSONArray( "image" );
for(int j=0; j<arrayOne.length(); j++)
{
JSONObject myInnerIterator = arrayOne.getJSONObject( j );
String s = myInnerIterator.getString( "size" )+myInnerIterator.getString("#text");
if (s.contains("medium") && s.contains("https")){
listUrl.add (s.replace("medium",""));
Log.i("RESPONSE",s.replace("medium",""));
}
}
} catch (JSONException e) {
e.printStackTrace();
}
我认为必须有更好的......但是它可以做到这一点！