Android 在同一ListView/GridView中调用多个JSON数组

更新：感谢Hariharan先生，现在我可以将两个JSON数组添加到一个ListView中（我已经更新了我的代码）

现在我需要将未收集的邮票更改为黑白颜色，我有一个代码可以更改黑白颜色，但是如果我实现了这一点，它会将所有邮票图像更改为黑白颜色

我应该怎么做才能只对未收集的邮票实施B/W：

这是更改图像颜色的功能：

public Bitmap toGrayscale(Bitmap bmpOriginal)
    {        
        int width, height;
        height = bmpOriginal.getHeight();
        width = bmpOriginal.getWidth();    

        Bitmap bmpGrayscale = Bitmap.createBitmap(width, height, Bitmap.Config.RGB_565);
        Canvas c = new Canvas(bmpGrayscale);
        Paint paint = new Paint();
        ColorMatrix cm = new ColorMatrix();
        cm.setSaturation(0);
        ColorMatrixColorFilter f = new ColorMatrixColorFilter(cm);
        paint.setColorFilter(f);
        c.drawBitmap(bmpOriginal, 0, 0, paint);
        return bmpGrayscale;
    }

这是ListViewAdapter

public class ListViewAdapter extends BaseAdapter {

    // Declare Variables
    Context context;
    LayoutInflater inflater;
    ArrayList<HashMap<String, String>> data;
    ImageLoader imageLoader;
    HashMap<String, String> resultp = new HashMap<String, String>();

    public ListViewAdapter(Context context,
            ArrayList<HashMap<String, String>> arraylist) {
        this.context = context;
        data = arraylist;
        imageLoader = new ImageLoader(context);
    }

    @Override
    public int getCount() {
        return data.size();
    }

    @Override
    public Object getItem(int position) {
        return null;
    }

    @Override
    public long getItemId(int position) {
        return 0;
    }

    public View getView(final int position, View convertView, ViewGroup parent) {
        // Declare Variables

        TextView tvStampName;
        ImageView stampImage;

        inflater = (LayoutInflater) context
                .getSystemService(Context.LAYOUT_INFLATER_SERVICE);

        View itemView = inflater.inflate(R.layout.single_stamp_layout, parent, false);
        // Get the position
        resultp = data.get(position);

        tvStampName = (TextView) itemView.findViewById(R.id.tvStampName);
        stampImage = (ImageView) itemView.findViewById(R.id.stampImage);

        tvStampName.setText(resultp.get(StampsActivity.TAG_STAMP_NAME));

        // Passes stampImage images URL into ImageLoader.class
        imageLoader.DisplayImage(TAG_STAMP_IMAGE_URL+resultp.get(StampsActivity.TAG_STAMP_IMAGE), stampImage);
        Animation myFadeInAnimation = AnimationUtils.loadAnimation(context, R.anim.fadein);
        stampImage.startAnimation(myFadeInAnimation);
        return itemView;
    }
}

我想将JSONArray的“collectedStamp”和“unCollectedStamp”合并到同一个列表中

这是更新的代码感谢Hariharan先生

当我试图调用分配给

'collectedStamp'

JSONArray的

'arraylistCollected'

时，我得到了这个错误

No value for collectedStamp 

当我成功地从“unCollectedStamp”JSONArray获取数据时

但是如果我将

'collectedStamp'

的for循环代码移到

unCollectedStamp

之前，我会得到一个错误

No value for unCollectedStamp

试试这个

您对这两个for循环使用相同的

jsonobject

。你应该分开给

for (int i = 0; i < jsonArrayUncollected.length(); i++) {
        HashMap<String, String> map = new HashMap<String, String>();
        JSONObject jsject = jsonArrayUncollected.getJSONObject(i);

        // Retrive JSON Objects
        map.put(TAG_STAMPS_ID, jsject .getString(TAG_STAMPS_ID));
        map.put(TAG_STAMP_IMAGE, jsject .getString(TAG_STAMP_IMAGE));
        map.put(TAG_STAMP_NAME, jsject .getString(TAG_STAMP_NAME));
        map.put(TAG_MESSAGE, jsject .getString(TAG_MESSAGE));
        // Set the JSON Objects into the array
        arraylistUncollected.add(map);
    }

    // Locate COLLECTED STAMPS array name in JSON
    jsonArrayCollected = jsonobject.getJSONArray(TAG_COLLECTED_STAMPS);

    for (int i = 0; i < jsonArrayCollected.length(); i++) {
        HashMap<String, String> map = new HashMap<String, String>();
        JSONObject jsject= jsonArrayCollected.getJSONObject(i);

        // Retrive JSON Objects
        map.put(TAG_STAMPS_ID, jsject.getString(TAG_STAMPS_ID));
        map.put(TAG_STAMP_IMAGE, jsject.getString(TAG_STAMP_IMAGE));
        map.put(TAG_STAMP_NAME, jsject.getString(TAG_STAMP_NAME));
        map.put(TAG_MESSAGE, jsject.getString(TAG_MESSAGE));

        // Set the JSON Objects into the array
        arraylistCollected.add(map);
    }

for（int i=0；i

编辑

HashMap<String,ArrayList<HashMap<String, String>>> result = new  HashMap<String,ArrayList<HashMap<String, String>>>();
result.put("Uncollected",arraylistUncollected);
result.put("Collected",arraylistCollected);

HashMap result=newhashmap（）；
结果。put（“未收集”，已收集）；
结果。放置（“已收集”，arraylistCollected）；

谢谢你，它很管用。现在我不想把两个数组都保存到一个列表中，我该怎么做呢？适配器需要一个arrayList，目前我只能传递一个我有“JSONArrayOnCollected”或“jsonArrayCollected”的arrayList。有没有办法将两个ArrayList组合成一个数组list@ahmadssb或者只使用一个ArrayList在一个数组中添加所有元素不要使用两个。最后一件事（对不起），如果我想通过“id”（在我的帖子中检查JSON结果）使用“result”数组列表，我想根据其属性对图像进行排序ID@ahmadssb告诉我，朋友。

No value for unCollectedStamp

for (int i = 0; i < jsonArrayUncollected.length(); i++) {
        HashMap<String, String> map = new HashMap<String, String>();
        JSONObject jsject = jsonArrayUncollected.getJSONObject(i);

        // Retrive JSON Objects
        map.put(TAG_STAMPS_ID, jsject .getString(TAG_STAMPS_ID));
        map.put(TAG_STAMP_IMAGE, jsject .getString(TAG_STAMP_IMAGE));
        map.put(TAG_STAMP_NAME, jsject .getString(TAG_STAMP_NAME));
        map.put(TAG_MESSAGE, jsject .getString(TAG_MESSAGE));
        // Set the JSON Objects into the array
        arraylistUncollected.add(map);
    }

    // Locate COLLECTED STAMPS array name in JSON
    jsonArrayCollected = jsonobject.getJSONArray(TAG_COLLECTED_STAMPS);

    for (int i = 0; i < jsonArrayCollected.length(); i++) {
        HashMap<String, String> map = new HashMap<String, String>();
        JSONObject jsject= jsonArrayCollected.getJSONObject(i);

        // Retrive JSON Objects
        map.put(TAG_STAMPS_ID, jsject.getString(TAG_STAMPS_ID));
        map.put(TAG_STAMP_IMAGE, jsject.getString(TAG_STAMP_IMAGE));
        map.put(TAG_STAMP_NAME, jsject.getString(TAG_STAMP_NAME));
        map.put(TAG_MESSAGE, jsject.getString(TAG_MESSAGE));

        // Set the JSON Objects into the array
        arraylistCollected.add(map);
    }

HashMap<String,ArrayList<HashMap<String, String>>> result = new  HashMap<String,ArrayList<HashMap<String, String>>>();
result.put("Uncollected",arraylistUncollected);
result.put("Collected",arraylistCollected);