Question

下面这段代码显示了从我的服务器到我的android工具栏或操作栏的内容，但我想将代码转换为Activity的正文，如用户全名，电子邮件用户名等。字符串{{1用于显示来自服务器的其他点我可以操作

URL = Config.Base_Url+"get/gtuspo.php";

PHP

txtName = (TextView) findViewById(R.id.name);
txtEmail = (TextView) findViewById(R.id.email);
txtUser = (TextView) findViewById(R.id.user);

// Showing Points on ToolBar Menu 
@Override
public boolean onPrepareOptionsMenu(final Menu menu) {
    // executing retreival task here
    String URL = Config.Base_Url+"get/gtuspo.php";

    StringRequest stringRequest = new StringRequest(Request.Method.POST, URL, new Response.Listener<String>() {
                @Override
                public void onResponse(String response) {
                    //System.out.println(response);
                    Toast.makeText(MainActivity.this,response,Toast.LENGTH_SHORT).show();

                    menu.findItem(R.id.points).setTitle("Points :" + response);
                }
            },
            new Response.ErrorListener() {
                @Override
                public void onErrorResponse(VolleyError error) {
                    Toast.makeText(MainActivity.this,error.toString(),Toast.LENGTH_SHORT).show();
                }
            }){
        @Override
        protected Map<String,String> getParams(){
            Map<String,String> params = new HashMap<String, String>();

            params.put("username",App.getInstance().getUsername());
            return params;
        }
    };
    RequestQueue requestQueue = Volley.newRequestQueue(this);
    requestQueue.add(stringRequest);
    return super.onPrepareOptionsMenu(menu);
}

String URL = Config.Base_Url+"get/gtuspo.php";

并在<?php include '../connect_database.php'; if($_SERVER['REQUEST_METHOD']=='POST') { $username = $_POST['username']; $sql = "SELECT points FROM users WHERE login = '$username'"; $result = $connect->query($sql); $row = $result->fetch_assoc(); echo $row["points"]; } else { echo '404 - Not Found <br/>'; echo 'the Requested URL is not found on this server '; } ?>上显示姓名。

Actionbar

PHP

String `URL = Config.Base_Url+"get/name.php";

这是数据库结构

<?php
include '../connect_database.php'; 
if($_SERVER['REQUEST_METHOD']=='POST') {
    $username = $_POST['username'];
    $sql = "SELECT fullname FROM users WHERE login = '$username'";
    $result = $connect->query($sql);
    $row = $result->fetch_assoc();
    echo $row["fullname"];
} else {
    echo '404 - Not Found <br/>';
    echo 'the Requested URL is not found on this server ';
}
?>

Answer 1

将其另存为 user_info.php

onReponse

现在，在您的Android代码中，在您的@Override public void onResponse(String response) { try { JSONObject jsonResponse= new JSONObject(response); txtName.setText(jsonResponse.getString("fullname")); txtEmail.setText(jsonResponse.getString("email")); txtPoints.setText(jsonResponse.getString("points"));// you need to create this textView txtPoints. } catch (Throwable t) { Log.e("onResponse", "The response: " + response + " seems to not be json formatted."); } }中执行以下操作：

注意： Volley 提供了一种获取json响应的方法，但为了不改变您的代码，我们将这样做。

String URL = Config.Base_Url + "get/user_info.php";

不要忘记：

    Pattern p = Pattern.compile("^(?:.*?\\sof\\s*)?(.*)");
    Matcher m = p.matcher("94km SSE of Taron, Papua New Guinea");
    if (m.find()) {
        System.out.println(m.group(1));
    }

在android上显示服务器的详细信息

1 个答案: