Android 通过多协议发送HTTP Post时的进度对话和toast失败

Android 通过多协议发送HTTP Post时的进度对话和toast失败,android,http,dialog,handler,toast,Android,Http,Dialog,Handler,Toast,我还是一个相对业余的安卓编程爱好者。我有一个很烦人的问题。当我在我的类中单击“发送”时,它不会显示进度对话框,也会忽略我对我的2个编辑文本按钮的验证。我的代码可能非常混乱和无序。我事先道歉 我需要做的是显示正在运行的进度对话,并在处理程序的末尾使其退出。我还需要我的祝酒词来展示,全班同学停止工作 public class share extends Activity { ProgressDialog dialog; DefaultHttpClient client = n

我还是一个相对业余的安卓编程爱好者。我有一个很烦人的问题。当我在我的类中单击“发送”时,它不会显示进度对话框,也会忽略我对我的2个编辑文本按钮的验证。我的代码可能非常混乱和无序。我事先道歉

我需要做的是显示正在运行的进度对话,并在处理程序的末尾使其退出。我还需要我的祝酒词来展示,全班同学停止工作

    public class share extends Activity {
    ProgressDialog dialog;
    DefaultHttpClient client = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://api...");


    @Override

    protected void onCreate(Bundle savedInstanceState) {
    // TODO Auto-generated method stub
    super.onCreate(savedInstanceState);
    setContentView(R.layout.send);
    Button Deliver;

    Deliver = (Button) findViewById (R.id.Send);
    Deliver.setOnClickListener(new View.OnClickListener() {

      @Override
      public void onClick(View v) {
      final ProgressDialog dialog = ProgressDialog.show(share.this, "","Uploading...", true);
      dialog.show();
      Handler handler = new Handler();
      handler.post(new Runnable() {
      public void run() {

    EditText etxt_user = (EditText) findViewById(R.id.user_email);
      EditText etxt_pass = (EditText) findViewById(R.id.friend_email);

    if(file == null){

    Toast display = Toast.makeText(share.this, "There are no videos to send", Toast.LENGTH_SHORT);
 display.setGravity(Gravity.BOTTOM|Gravity.LEFT, 0, 0); 
 display.show();

    startActivity(new Intent("android.main.SHARE"));

                         }

    else{

     Pattern pattern= Pattern.compile(
     "[a-zA-Z0-9\\+\\.\\_\\%\\-\\+]{1,256}" +
    "\\@" +
    "[a-zA-Z0-9][a-zA-Z0-9\\-]{0,64}" +
     "(" +"\\." +  "[a-zA-Z0-9][a-zA-Z0-9\\-]{0,25}" +")+"
    );

     Matcher matcher=pattern.matcher(etxt_user.getText().toString());// match the contents of the first edit text
    Matcher matcher1=pattern.matcher(etxt_pass.getText().toString());

     if (!matcher.matches()&&(etxt_user.getText().toString()==null))
                          {
     Toast display1 = Toast.makeText(share.this, "Please enter correct email address", Toast.LENGTH_SHORT);
    display1.show();
    startActivity(new Intent("com..SHARE"));


     }
    else
        {
            //proceed with program                
        }

    if (!matcher1.matches()&&!(etxt_pass.getText().toString()==null)){

    Toast display2= Toast.makeText(share.this, "You entered wrong email Format in the second box", Toast.LENGTH_LONG);
 display2.setGravity(Gravity.TOP|Gravity.LEFT, 0, 0);                         
 startActivity(new Intent("com.apapa.vrsixty.SHARE"));                        
    }
    else
    {
          //proceed
     }

    try{

     MultipartEntity me = new MultipartEntity();

    me.addPart("file", new FileBody(new File("/sdcard/videocapture_example.H264")));

    me.addPart("userEmail", new StringBody(etxt_user.getText().toString())); 
     me.addPart("friendEmail", new StringBody(etxt_pass.getText().toString()));

     httppost.setEntity(me);

    HttpResponse responsePOST = client.execute(httppost);  
     HttpEntity resEntity = responsePOST.getEntity(); 
    InputStream inputstream = resEntity.getContent();
    BufferedReader buffered = new BufferedReader(new InputStreamReader(inputstream));
        StringBuilder stringbuilder = new StringBuilder();
        String currentline = null; 
        while ((currentline = buffered.readLine()) != null) { 
         stringbuilder.append(currentline + "\n"); 
        String result = stringbuilder.toString(); 
        Log.v("HTTP UPLOAD REQUEST",result); 
        inputstream.close();  } }
        catch (Exception e) {
        e.printStackTrace();
                }               
                         }dialog.dismiss();
                }
            });
                }
            });

当我完成此操作并按send时,它会给我一个没有进度对话的黑屏,即使我对编辑文本进行了验证,或者如果没有要发送的文件,它也会运行程序,而不会停止程序并显示祝酒词。提前感谢各位

看来您的httppost正在将其发送到“http://”,这不是一个完全有效的地址,这是您的问题的一部分

还有:那是逐字逐句的代码吗?它不可能运行。一开始你有这一行

new HttpPost("http://"");

结尾有两个引号,这实际上使第二个引号后面的所有内容都成为字符串,直到它变成结束引号。

感谢您的回复。请原谅我的错误,请忽略第二行,我实际上将其作为新的HttpPost(“http://...api"). 它实际上是有效的,因为当我转到phpmyadmin中的api调试器时,我看到了post的结果。它只是没有显示对话或祝酒词。抱歉,“共享。这”是此代码所在的当前类,出于某种原因,它不允许我添加“这”而不显示错误。太好了,您的代码看起来好多了:)我会看一看。谢谢,从现在起我将接受您的建议。连我都很难读懂lol