Question

我是Android开发的新手，我在这里有一些我建立的代码，可以在我的移动应用程序中使用登录功能。我试图将我输入文本中输入数据的值与mysql数据库进行比较，但似乎我没有明白这一点，我真的不知道我的代码是否引导我进入某事，或者它只是废话，可以你们帮帮我吗？顺便说一句，我没有得到任何错误，单击按钮后没有任何结果。

这是我的代码：

**

protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        // Edit Text
        inputEmail = (EditText) findViewById(R.id.inputEmail);
        inputPassword = (EditText) findViewById(R.id.inputPassword);

        // Create button
        Button btnSubmit = (Button) findViewById(R.id.btnLogin);

        // button click event
        btnSubmit.setOnClickListener(new View.OnClickListener() {

            @Override
            public void onClick(View view) {
                // creating new product in background thread
                new CheckLogin().execute();
            }
        });     
    }

    class CheckLogin extends AsyncTask<String, String, String> {

        /**
         * Before starting background thread Show Progress Dialog
         * */
        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            pDialog = new ProgressDialog(Main.this);
            pDialog.setMessage("Logging in..");
            pDialog.setIndeterminate(false);
            pDialog.setCancelable(true);
            pDialog.show();
        }

        /**
         * Creating product
         * */
        protected String doInBackground(String... args) {
            String eadd = inputEmail.getText().toString();
            String password = inputPassword.getText().toString();

            httpclient = new DefaultHttpClient();
            httppost = new HttpPost("http://10.0.2.2/TheCalling/log_in.php");


            try {
                nameValuePairs = new ArrayList<NameValuePair>();

                nameValuePairs.add(new BasicNameValuePair("eadd", eadd));
                nameValuePairs.add(new BasicNameValuePair("password", password));

                httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                response = httpclient.execute(httppost);

                if(response.getStatusLine().getStatusCode()==200){
                    entity = response.getEntity();

                    if(entity != null){

                        InputStream instream = entity.getContent();
                        JSONObject jsonResponse = new JSONObject(convertStreamToString(instream));

                        String inputEmail = jsonResponse.getString("eadd");
                        String inputPassword = jsonResponse.getString("password");

                        if(eadd.equals(inputEmail) && password.equals(inputPassword)){

                            SharedPreferences sp = getSharedPreferences("logindetails", 0);
                            SharedPreferences.Editor spedit = sp.edit();

                            spedit.putString("eadd", eadd);
                            spedit.putString("password", password);

                            spedit.commit();
                            Toast.makeText(getBaseContext(), "SUCCESS!", Toast.LENGTH_SHORT).show();


                        }else{
                            Toast.makeText(getBaseContext(), "Invalid login details", Toast.LENGTH_SHORT).show();                   
                        }

                    }

                }
            } catch (Exception e) {
                e.printStackTrace();
            }

            return null;
        }

**

这是我的php文件：

**

<?php

$db_host = "localhost";
$db_user = "root";
$db_password = "";
$db_name = "sample_login";
$response = array();

    $db_connect = mysql_connect($db_host, $db_user, $db_password) or die(mysql_error());
    $db = mysql_select_db($db_name);
    $eadd = $_POST['eadd'];
    $password = $_POST['password'];

    // mysql inserting a new row
    $result = mysql_query("SELECT * FROM users WHERE eadd = '$eadd' and password = '$password'");

    // check if row inserted or not
    if ($result) {
        // successfully inserted into database
        $response["success"] = 1;
        $response["message"] = "Login succes";

        // echoing JSON response
        echo json_encode($response);
    } else {
        // failed to insert row
        $response["success"] = 0;
        $response["message"] = "Oops! An error occurred.";

        // echoing JSON response
        echo json_encode($response);   
}
?>

**

Answer 1

我不是Android开发人员，但这就是我要说的。

php文件很好。唯一的问题是，你没有返回你似乎在android文件中访问的电子邮件和密码。这里：

InputStream instream = entity.getContent（）; JSONObject jsonResponse = new JSONObject（convertStreamToString（instream））; String inputEmail = jsonResponse.getString（“eadd”）; String inputPassword = jsonResponse.getString（“password”）;
如果查询正确，mysql_query（）会在成功时返回资源，如果出错则返回FALSE。你需要检查mysql_num_rows（$ result）＆gt; 0表示有效匹配。参考http://php.net/manual/en/function.mysql-num-rows.php

记住

if ($result) {
}

将始终为真。

Answer 2

1）你可以发布方法'convertStreamToString'吗？

2）您是否检查了

  echo json_encode($response);

返回正确的值并且不返回null？

Android登录与php / mysql无法正常工作

2 个答案: