小编Rog*_*vis的帖子

我正在尝试解析像这样的JSON字符串

[
   {
      "updated_at":"2012-03-02 21:06:01",
      "fetched_at":"2012-03-02 21:28:37.728840",
      "description":null,
      "language":null,
      "title":"JOHN",
      "url":"http://rus.JOHN.JOHN/rss.php",
      "icon_url":null,
      "logo_url":null,
      "id":"4f4791da203d0c2d76000035",
      "modified":"2012-03-02 23:28:58.840076"
   },
   {
      "updated_at":"2012-03-02 14:07:44",
      "fetched_at":"2012-03-02 21:28:37.033108",
      "description":null,
      "language":null,
      "title":"PETER",
      "url":"http://PETER.PETER.lv/rss.php",
      "icon_url":null,
      "logo_url":null,
      "id":"4f476f61203d0c2d89000253",
      "modified":"2012-03-02 23:28:57.928001"
   }
]

进入一个对象列表.

List<ChannelSearchEnum> lcs = (List<ChannelSearchEnum>) new Gson().fromJson( jstring , ChannelSearchEnum.class);

这是我正在使用的对象类.

import com.google.gson.annotations.SerializedName;

public class ChannelSearchEnum {



@SerializedName("updated_at")
private String updated_at;

@SerializedName("fetched_at")
private String fetched_at;

@SerializedName("description")
private String description;

@SerializedName("language")
private String language;

@SerializedName("title")
private String title;

@SerializedName("url")
private String url;

@SerializedName("icon_url")
private String icon_url;

@SerializedName("logo_url")
private …

到目前为止,这是我的代码:

String path = Environment.getExternalStorageDirectory().toString()+"/Pictures";

        AssetManager mgr = getAssets();

        try {

            String list[] = mgr.list(path);
            Log.e("FILES", String.valueOf(list.length));

            if (list != null)
                for (int i=0; i<list.length; ++i)
                    {
                        Log.e("FILE:", path +"/"+ list[i]);
                    }

        } catch (IOException e) {
            Log.v("List error:", "can't list" + path);
        }

然而,虽然我在那个目录中有文件,但它返回list.length = 0 ...任何想法？

想象一下,你在2d空间中有两个点,你需要将这些点中的一个旋转X度,另一个点作为中心.

float distX = Math.abs( centerX -point2X );
float distY = Math.abs( centerY -point2Y );

float dist = FloatMath.sqrt( distX*distX + distY*distY );

到目前为止,我只是找到了两点之间的距离......任何想法我应该从哪里开始？

我在Python 3中编写了一个程序,但现在想将其转换为Python 2代码.是否有任何实用程序自动执行此操作？

在java/android中复制/克隆对象的最佳方法是什么？

rlBodyDataObj rlbo = bdoTable.get(name);

现在代码从hashTable中分配一个对象,但我需要得到它的一个克隆,这样我就能多次使用它.

我只是在学习AsyncTask并希望将它用作一个单独的类,而不是一个子类.

例如,

class inetloader extends AsyncTask<String, Void, String> {


    @Override
    protected String doInBackground(String... urls) {

        String response = "";

            DefaultHttpClient client = new DefaultHttpClient();
            HttpGet httpGet = new HttpGet(urls[0]);
            try {
                HttpResponse execute = client.execute(httpGet);
                InputStream content = execute.getEntity().getContent();

                BufferedReader buffer = new BufferedReader(
                        new InputStreamReader(content));
                String s = "";
                while ((s = buffer.readLine()) != null) {
                    response += s;
                }

            } catch (Exception e) {
                e.printStackTrace();
            }

        return response;
    }


    @Override
    protected void onPostExecute(String result) {
        Log.e("xx",result);
        // how …

如何在javascript中获取光标X和Y？

var $curX=(ns6)?e.pageX : event.clientX+ietruebody().scrollLeft;
var $curY=(ns6)?e.pageY : event.clientY+ietruebody().scrollTop;

我发现了这两个,但它们看起来像"未定义".有任何想法吗？

谢谢!

感谢Torious,得到了:

    private void smudge() {


        for (int i = 0; i < COUNT*2; i += 2) {

            float xOriginal = matrixOriganal[i+0];
            float yOriginal = matrixOriganal[i+1];

            float distX = Math.abs(pointX-xOriginal);
            float distY = Math.abs(pointY-yOriginal);

            float dist = FloatMath.sqrt( distX*distX + distY*distY );

            float coof = ( bubbleSize - dist ) / bubbleSize;

        float oc = (float) -Math.sin(coof * 2*Math.PI) * 0.15f ;

            if ( dist < bubbleSize )
            {
            matrixVertsMoved[i+0] = xOriginal + smudgeAmount * (coof+oc);
            matrixVertsMoved[i+1] = yOriginal;
            }
            else …

假设我有像这样的心电图数据阵列

http://joachim.behar.perso.neuf.fr/Joachim/ECG_tuto_1/ECG_filters_basics/ecg_sample.txt

我需要使用这样的算法过滤50Hz的噪音

http://joachim.behar.perso.neuf.fr/Joachim/ECG_tuto_1/ECG_filters_basics/notch_filter.m

结果应该是如图2所示(红色过滤,蓝色 - 未过滤):

在Java/android中执行此操作的最佳方式是什么？这个图书馆会帮助https://sites.google.com/site/piotrwendykier/software/jtransforms吗？

谢谢!)

例如,我有一个PHP数组,例如这个

<?php $s= array('a','b','c','d','e','f') ; ?>

我需要在JavaScript中循环它,任何想法我该怎么做？

for ( i=0 ; i < <?php echo sizeof($s) ?> ; i++) {
document.write('<?php echo $s [somehow need to get the 'i' value into here] ?>');
}

有什么建议？谢谢!