Question

我有asynctask和json的代码：

class loginn extends AsyncTask<Void, Void, Void>{
     String u;
     String usuario = user.getText().toString();
     String contraseña = pass.getText().toString();
        @Override
        protected Void doInBackground(Void... params) {


                HttpClient httpclient2 = new DefaultHttpClient();
                HttpPost httppost2 = new HttpPost("****");

                     try
                     {
                     nameValuePairs = new ArrayList<NameValuePair>();
                     nameValuePairs.add(new BasicNameValuePair("user", usuario));
                     nameValuePairs.add(new BasicNameValuePair("pass", contraseña));                     
                     httppost2.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                     HttpResponse response2 = httpclient2.execute(httppost2);
                     HttpEntity entity = response2.getEntity();
                     u = EntityUtils.toString(entity);
                     if(u.equals("1")){

                     runOnUiThread(new Runnable() {
                        public void run() {
                            Toast toast = Toast.makeText(login.this, "yes", Toast.LENGTH_SHORT);
                             toast.show();
                        }
                        });
                             }else
                             {
                                 runOnUiThread(new Runnable() {
                                    public void run() {
                                        Toast toast = Toast.makeText(login.this, "no", Toast.LENGTH_SHORT);
                                         toast.show();
                                    }
                                    });
                             }
                 }
                 catch(Exception e)
                 {

                 }
                    return null;


        }

        protected void onPostExecute(Void result) {

            }

php的查询很好，但是＆＃34; if＆＃34;总是返回＆＃34; no＆＃34;，但如果我把变量＆＃34; u＆＃34;在祝酒词上，它返回＆＃34; 1＆＃34;当存在用户并传递时，＆＃34; 0＆＃34;什么时候不存在。

php是：

<?php  

$con = mysql_connect('***', '****', '*****');  
mysql_query("SET CHARACTER SET utf8");  
mysql_query("SET NAMES utf8");  
$user = $_POST['user'];
$pass = $_POST['pass'];
if( $con )  
{  
mysql_select_db('*****');  
$query = "select * from usuarios where username='$user' and passw='$pass'";    
$res = mysql_query($query);
$count = 0;

    while($row = mysql_fetch_object($res))
      {
         $count++;
      }
if($count==1)
{
   echo json_encode(1);
}else
  {
    echo json_encode(0);
  }
}
php?>

Answer 1

尝试这样的事情

if(Integer.valueOf(u) == 1)

希望这会对你有所帮助。

Answer 2

class loginn extends AsyncTask<Void, Void, Boolean>{
  String usuario = user.getText().toString();
  String contraseña = pass.getText().toString();

  @Override
  protected Boolean doInBackground(Void... params) {
    HttpClient httpclient2 = new DefaultHttpClient();
    HttpPost httppost2 = new HttpPost("****");
    try {
      nameValuePairs = new ArrayList<NameValuePair>();
      nameValuePairs.add(new BasicNameValuePair("user", usuario));
      nameValuePairs.add(new BasicNameValuePair("pass", contraseña));                     
      httppost2.setEntity(new UrlEncodedFormEntity(nameValuePairs));
      HttpResponse response2 = httpclient2.execute(httppost2);
      HttpEntity entity = response2.getEntity();
      String u = EntityUtils.toString(entity);
      System.out.println("length " + u.length()); //should be 1
      System.out.println("'" + u + "'"); //should be '1'
      for (char c : u.toCharArray())
        System.out.println("current = " + ((int)c)); //should be 49, if more than one => look up in ASCII table
      return u.equals("1");
    }
    catch(Exception e) { 
      return false;
    }
  }

  protected void onPostExecute(Boolean result) {
    Toast.makeText(login.this, result ? "yes" : "no", Toast.LENGTH_SHORT).show();
  }
}

上面的代码是你的代码，但是稍微进行了优化，并且更易于阅读。

显然，u 不等于"1"。这可能是因为u中您没有看到的字符，例如空格。这意味着你的if条件完美无瑕。你应该关注u是什么。您可以通过调试我放入代码中的语句来实现。

更新

12-12 11:38:52.858: I/System.out(4445): length 5 
12-12 11:38:52.858: I/System.out(4445): '1 ' 
12-12 11:38:52.862: I/System.out(4445): current = 49 
12-12 11:38:52.862: I/System.out(4445): current = 9 
12-12 11:38:52.862: I/System.out(4445): current = 9 
12-12 11:38:52.862: I/System.out(4445): current = 9 
12-12 11:38:52.862: I/System.out(4445): current = 9 
and the toast shows "no" –

查看日志语句，会在响应中添加一些水平制表符。

由于第一个条目是49，您可以使用

return u.length() > 0 && u.charAt(0) == '1';

这应该可以解决您的问题，但您仍然需要找出发生这种情况的原因

修改

您可以尝试将EntityUtils.toString与字符集EntityUtils.toString(entity, "UTF-8)"

一起使用

Answer 3

响应中也必须出现空格或某个不可见的字符。

试试这个：

if(Pattern.compile("1").matcher(u).find())

如果（条件）在asynctask中不起作用

3 个答案: