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我想检查用户输入的电子邮件 ID 是否唯一，因此最初我有我的变量Boolean valid = false;。单击按钮时，我正在输入输入的电子邮件 ID，并使用正则表达式检查它的有效电子邮件 ID 表达式，然后我使用 asyntask 来检查其唯一性。我的 onclicklistner 中的代码是

if (emailid.matches(regexp) && emailid.length() > 0) { new Validate().execute(); Toast.makeText(getApplicationContext(), valid.toString(), Toast.LENGTH_LONG).show(); if (valid) { data.putString("eid", eid); data.putString("firstname", firstname); data.putString("lastname", lastname); data.putString("emailid", emailid); Intent i = new Intent(getApplicationContext(), GamesFragment.class); startActivity(i); } else { Toast.makeText(getApplicationContext(), "Email Address Already Exist", Toast.LENGTH_LONG).show(); } } else { Toast.makeText(getApplicationContext(), "Check Your Email Address", Toast.LENGTH_LONG).show(); }

这里我面临的问题是，当我第一次输入唯一的电子邮件并单击按钮时，Validate()asynctask 检查并将valid变量设置为true，但它不会进入下一个活动GamesFragment，因为我valid = false最初已声明。现在，当我再次单击该按钮时，它会转到下一个活动，因为该valid变量由于先前的单击而设置为 true。现在我的Validate()异步任务是 

private class Validate extends AsyncTask<Void, Void, Void> {
        @Override
        protected Boolean doInBackground(Void... params) {
            ArrayList<NameValuePair> emailId = new ArrayList<NameValuePair>();
            emailId.add(new BasicNameValuePair("email", emailid));

            try {
                HttpClient httpclient = new DefaultHttpClient();
                HttpPost httppost = new HttpPost("url/validate.php");
                httppost.setEntity(new UrlEncodedFormEntity(emailId));
                HttpResponse response = httpclient.execute(httppost);
                HttpEntity entity = response.getEntity();
                iss = entity.getContent();

            } catch(Exception e) {
                Log.e("pass 1", "Connection Error");
                e.printStackTrace();
            }

            try {
                BufferedReader reader = new BufferedReader
                        (new InputStreamReader(iss,"iso-8859-1"),8);
                StringBuilder sb = new StringBuilder();
                while ((line = reader.readLine()) != null)
                    sb.append(line + "\n");
                iss.close();
                result = sb.toString();
            } catch(Exception e) {
                e.printStackTrace();
            }

            try {
                JSONObject json_data = new JSONObject(result);
                code=(json_data.getInt("code"));

                if(code == 1)
                    valid = true;
                else
                    valid = false;
                Log.e("pass 3", "valid "+valid);
            } catch(Exception e) {
                e.printStackTrace();
            }


            return null;
        }
    }

请帮助我不明白为什么会这样。

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1 回答 1

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创建函数来检查验证。

private boolean function validate(String emailid){

if (emailid.matches(regexp) && emailid.length() > 0) {
return true;
}


return false;

}

使用该功能来决定事件

if(validate(emailid)){ // if function return true then email is valid and good to go.

 new Validate().execute();


}

对于第二个条件，您必须在onPostExecute()Validate(); 的异步任务中检查它；

 @Override
    protected void onPostExecute(Object o) {

        super.onPostExecute(o);
        if(code == 1){

    // check if response is valid than 
Intent i = new Intent(getApplicationContext(), GamesFragment.class);
                        startActivity(i);

} }

于 2016-07-28T12:34:31.590 回答