从价值观和textviews比较JSONArray [英] Comparing values from textviews and JSONArray

问题描述

很抱歉的（貌似）懒惰的问题，但我一直在寻找，没有运气的解决方案（换句话说，我还没有找到一个解决方案，我明白了）。

我希望让用户通过输入用户名和密码的方式登录到一个应用程序，该用户名和密码必须匹配，从我已经从phpmyadmin的数据库中检索到jsonarray用户名和密码。的用户名和密码都必须在同一行中

下面是我用它来找回我的jsonarray功能：

 私人无效的getData（）{JSONArray json的;
尝试{
    。用户= editText1.getText（）的toString（）;
    。密码= editText2.getText（）的toString（）;    JSON = readJsonFromUrl（HTTP：//localhost/indextest.php功能=的getData）;       视频下载（1000）;
  }赶上（例外五）{
    Log.e（BACKGROUND_PROC，e.getMessage（））;
  }
 

}

我只需要知道如何搜索jsonarray因我从我的textviews检索值。

如果可能，我想找回我可以轻松地分配到一个值，如果像这样的语句：

 公共无效的onClick（视图v）{        开关（v.getId（））{
            案例R.id.button1：                如果（editText1 =空＆放大器;！＆放大器;！editText1.length（）= 0＆放大器;＆放大器; editText2 =空＆放大器;！＆放大器;！editText2.length（）= 0）{
                    的getData（）;
                    m_ProgressDialog = ProgressDialog.show（HomeScreen.this，
                              请稍候...，检查详细信息...，真正的）;
                        m_ProgressDialog.setCancelable（真）;                      如果（/ *的用户名和密码匹配* /）{
                      意图I =新意图（这一点，Afterlog.class）;
                      startActivity（ⅰ）;
                      }                    其他{
                        Toast.makeText（HomeScreen.this的用户名和密码不匹配任何在我们的数据库中...，Toast.LENGTH_SHORT）.show（）;
                    }
                }
                其他{
                    Toast.makeText（HomeScreen.this，请输入用户名和密码......，Toast.LENGTH_SHORT）.show（）;
                }
                打破;}
}
 

解决方案

找到了一个非常简单的一点教程这里：

<一个href=\"http://www.$c$crzheaven.com/2012/04/22/create-simple-login-form-php-android-connect-php-android/\" rel=\"nofollow\">http://www.$c$crzheaven.com/2012/04/22/create-simple-login-form-php-android-connect-php-android/

感谢所有帮助@meda：）

Sorry for the (seemingly) lazy question, but i've been looking for a solution with no luck (in other words, I haven't found a solution that i understand).

I want to have users log in to an app by way of entering a username and password, this username and password has to match a username and password from the jsonarray which i've retrieved from a phpmyadmin database. The username and password have to be in the same row.

Here's the function I use to retrieve my jsonarray:

private void getData(){

JSONArray json;
try{
    user = editText1.getText().toString();
    password = editText2.getText().toString();

    json = readJsonFromUrl("http://localhost/indextest.php?function=getdata");

       Thread.sleep(1000);
  } catch (Exception e) {
    Log.e("BACKGROUND_PROC", e.getMessage());
  }

}

I just need to know how to search the jsonarray for the values that i retrieve from my textviews.

If possible I would like to retrieve a value that I can easily assign to an if statement like such:

public void onClick(View v) {

        switch (v.getId()) {
            case R.id.button1:

                if  ( editText1 != null && editText1.length() != 0 && editText2 != null && editText2.length() != 0){    
                    getData();
                    m_ProgressDialog = ProgressDialog.show(HomeScreen.this,    
                              "Please wait...", "Checking Details...", true);
                        m_ProgressDialog.setCancelable(true);

                      if ( /*username and password match*/){
                      Intent i = new Intent(this, Afterlog.class);
                      startActivity(i);     
                      }

                    else{
                        Toast.makeText(HomeScreen.this, "The username and password did not match any in our database...", Toast.LENGTH_SHORT).show();   
                    }
                }
                else {
                    Toast.makeText(HomeScreen.this, "Please enter a user name AND a password...", Toast.LENGTH_SHORT).show();
                }
                break;

}
}

解决方案

Found a very simple and to the point tutorial here:

http://www.coderzheaven.com/2012/04/22/create-simple-login-form-php-android-connect-php-android/

Thanks for all the help @meda :)

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