小编Mus*_*hin的帖子

我正在开发一个wordpress插件,它将表单提交到另一个页面.但是当我尝试将表单提交到另一个页面时,该页面会返回一些php错误.我的表单代码如下

echo "<form action='".plugins_url()."/wp_voting_poll/frontend_poll_process.php'     method='post'>";
echo "<input type='hidden' name='hide' value='$ques' />";
        $total_vote_count = $wpdb->get_var( "SELECT COUNT(*) FROM $table_result WHERE question_uid='$ques'" );
        if($ques!=""){
        echo "<table>";

        foreach($ans_data as $ans_res){

         //   $ans=$ans_res->answer;
         $answer_id=$ans_res->id;
         $type=$ans_res->answer_type;


               $vote_count = $wpdb->get_var( "SELECT COUNT(*) FROM $table_result WHERE answer_id='$answer_id'" );
                if($vote_count==0){
                    error_reporting(0);
                }
                $vote_percent=($vote_count*100)/$total_vote_count;
             echo "<tr> <td>";  
           echo "<div class='answer_div'>";    
               if($type==1){
             echo "<div class='input'><input type='radio' name='ans_name[]' value='$answer_id'/>".$ans_res->answer."<br/></div>";
             }
             elseif($type==0){

             echo "<div class='input'><input type='checkbox' name='ans_name[]' value='$answer_id'/>".$ans_res->answer."<br/></div>";
             }
             if($backend==0){
             echo "</td> <td>";


             echo "<h4> total vote counted $vote_percent% </h4>";

            // echo …

我是android和java的新手.最近我遇到了使用baseadapter将获取的json数据显示到listview中的问题.起初我使用了这段代码

  @Override
  protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

   new TheTask().execute();
  }
     class TheTask extends AsyncTask<Void,Void,String>
  {

@Override
protected String doInBackground(Void... params) {
    String str = null;
    try
    {
      HttpClient httpclient = new DefaultHttpClient();
      HttpPost httppost = new HttpPost("http://10.0.2.2/BSDI/show.php");
      HttpResponse response = httpclient.execute(httppost);
      str =  EntityUtils.toString(response.getEntity());     
    } 
    catch (IOException e) {
           // TODO Auto-generated catch block
           e.printStackTrace();
        }

    return str;

  }



@Override
    protected void onPostExecute(String result) {

    super.onPostExecute(result);

    String response = result.toString();
    try {


        ArrayList<String> stringArray = new ArrayList<String>();

        JSONArray …

我很了解php.但我是一个非常新的php框架学习者.我正在学习codeigniter 2.0.3.我正在尝试开发一个简单的hello world应用程序.

我试过这段代码..

class Hello extends Controller{
  function Hello(){
    parent::Controller();
  }
  function you(){
    $this->load->view("you_view");
  }

}

但不幸的是我收到了这个错误......

致命错误:第7行的C:\ wamp\www\CodeIgniter\application\controllers\hello_world.php中找不到"控制器"类

我使用 AudioRecord 类录制了声音。但它无法通过音乐播放器播放。所以我在音频文件中添加了一些标题信息。虽然它现在可以通过音乐播放器播放，但它滞后且嘈杂，并且无法播放准确的声音。任何帮助，将不胜感激 。我的代码如下

private void rawToWave(final File rawFile, final File waveFile) throws IOException {

    byte[] rawData = new byte[(int) rawFile.length()];
    DataInputStream input = null;
    try {
        input = new DataInputStream(new FileInputStream(rawFile));
        input.read(rawData);
    } finally {
        if (input != null) {
            input.close();
        }
    }

    DataOutputStream output = null;
    try {
        int myBitsPerSample= 2;
        int myFormat = 1;
        long myChannels = 1;
        long mySampleRate = 8000 ;
        long myByteRate = mySampleRate * myChannels * myBitsPerSample/8;
        int myBlockAlign = (int) (myChannels * …