Question

它显示＆＃34;登录失败！＆＃34;当我用android登录数据库时。请看一下。

代码点火器数据库代码

<?php

class api扩展了CI_Controller {

public function __construct() {
    parent::__construct();
    $this->load->model('m_activity', '', TRUE);
    $this->load->model('m_project', '', TRUE);
}

public function index() {
    $this->load->view('api/recentactivities');
}
public function stud_login($std_id, $std_pass){
        // $std_id = $this->input->post('std_id');
        // $std_pass = $this->input->post('std_pass');
        $this->db->select('*');
        $this->db->from('student');
        $this->db->where('std_id',$std_id);
        $this->db->where('std_pass',$std_pass);
        $query = $this->db->get();
        $result = $query->result_array();
        $data['android'] = $result;
        echo json_encode($data);
}

}

和我的java控制器页面

公共类登录页面扩展了活动{

    public static final String url = "http://ss.data.com/api/std_login";
    public static String result = "";

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.login);

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

            public void onClick(View v) {
                WebPageTask task = new WebPageTask();
                task.execute();
            }

            });
    }

    public String login(){

        EditText edittext_user = (EditText)findViewById(R.id.editText);
        EditText edittext_pass = (EditText)findViewById(R.id.editText2);
        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
        nameValuePairs.add(new BasicNameValuePair("std_id",
                edittext_user.getText().toString()));
        nameValuePairs.add(new BasicNameValuePair("std_pass",
                edittext_pass.getText().toString()));

        return postData(url,nameValuePairs);
    }

    public String postData(String url,List<NameValuePair> nameValuePairs){

        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost(url);
        BufferedReader in = null;
        try {
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpclient.execute(httppost);
            in = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
            StringBuffer sb = new StringBuffer("");
            String line = "";
            String NL = System.getProperty("line.separator");
            while ((line = in.readLine()) != null) {
                sb.append(line + NL);
            }
            in.close();
            String page = sb.toString();
            return page;
        }catch (ClientProtocolException e) {
            //TODO Auto-generated catch block
        }catch (IOException e){
            //TODO Auto-generated catch block
        }
        return null;
    }
    private class WebPageTask extends AsyncTask<Void, Void, String>{
        @Override
        protected String doInBackground(Void... param){
            result = login();
            return result;
        }

        @Override
        protected void onPostExecute(String result){
            if (result.trim().equals("OK")){
                Toast.makeText(Loginpage.this, "Login Success.",
                        Toast.LENGTH_LONG).show();
            }else {
                Toast.makeText(Loginpage.this, "Login Fail.",
                        Toast.LENGTH_LONG).show();
            }
        }
    }

} 请帮帮我。我试了整整一天。它不再起作用了。

Answer 1

这是因为你的代码的这一部分：

public String login(){

    EditText edittext_user = (EditText)findViewById(R.id.editText);
    EditText edittext_pass = (EditText)findViewById(R.id.editText2);
    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    nameValuePairs.add(new BasicNameValuePair("std_id",
            edittext_user.getText().toString()));
    nameValuePairs.add(new BasicNameValuePair("std_pass",
            edittext_pass.getText().toString()));

    return postData(url,nameValuePairs);

你已经宣布了你的名字.ValuePairs.add＆＃34;因此，如果你明白我的意思，你必须在＆＃34; pass.getText＆＃34;之前背诵你的editText变量。

换句话说，试试这个：

nameValuePairs.add(new BasicNameValuePair("std_id",
            edittext_user.getText().toString()));
    nameValuePairs.add(new BasicNameValuePair("std_pass",
//recite text from intent
var<openTextEditor.class> {
insert.editText.java;
//close snippet
return true
}
edittext_pass.getText().toString()));

从android页面到数据库的登录错误（Codeigniter）

1 个答案: