使用数据库连接的 PHPUnit 模拟函数 [英] PHPUnit Mocking functions which use Database Connection

问题描述

我是 PHPUnit 的新手，我正在尝试测试需要从数据库中获取所有任务的函数 getAllTask​​s().我尝试了一切，但我只是让我的代码更糟.所以请帮我解决这个问题.TaskTest.php 是我试图进行测试但它不起作用的东西.并且确定是否有更好的方法来做某事，我也喜欢学习新东西.这是我的代码:

I am new to using PHPUnit and i am trying to test function getAllTasks() which need to fetch all tasks from database. I tried everything but i am just making my code worse. So please help me to solve the problem. TaskTest.php is something i tried to make test but it dont works. And sure if there are better ways to do something, i like to learn new stuff too. Here is my code:

我更改了 TaskTest.php 的代码，并设法通过了测试.有人可以告诉我这是否是测试此功能的好方法，或者有更好的方法吗?谢谢！

I changed code for TaskTest.php and i managed to get test pass. Can someone please tell me if this is good way to test this function, or there are better ways? Thanks!

任务.php

<?php

  require_once 'Database.php';

  class Task {
    private $db;

    public function __construct() {
      $this->db = new Database;
    }

    public function getAllTasks() {
      $this->db->query('SELECT * FROM tasks');
      $results = $this->db->resultSet();

      return $results;
    }
} 

数据库.php

<?php
  class Database {

    private $host = 'localhost';
    private $user = 'root';
    private $pass = '123456';
    private $dbname = 'todolist';

    private $dbh;
    private $stmt;
    private $error;

    public function __construct(){
      // Set DSN
      $dsn = 'mysql:host=' . $this->host . ';dbname=' . $this->dbname;
      $options = array(
        PDO::ATTR_PERSISTENT => true,
        PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION
      );

        // Create PDO instance
      try {
        $this->dbh = new PDO($dsn, $this->user, $this->pass, $options);
      } catch(PDOException $e){
        $this->error = $e->getMessage();
        echo $this->error;
      }
    }

public function query($sql){
      $this->stmt = $this->dbh->prepare($sql);
      $this->execute();
    }

public function execute(){
      return $this->stmt->execute();
    }

    public function resultSet(){
      $this->execute();
      return $this->stmt->fetchAll(PDO::FETCH_ASSOC);

    }
}

TaskTest.php

TaskTest.php

<?php
require_once './src/Task.php';
require_once './src/Database.php';
use PHPUnit\Framework\TestCase;

  class TaskTest extends TestCase {

    public function testGetAllTasks() {

      $table = array(
        array(
          'task_id' => '1',
          'task_desc' => 'Task One Test'
        ),
        array(
          'task_id' => '2',
          'task_desc' => 'Task Two Test'
        )
      );

      $dbase = $this->getMockBuilder('Database')
        ->getMock();

      $dbase->method('resultSet')
          ->will($this->returnValue($table));

      $expectedResult = [
                          'task_id' => '1',
                          'task_desc' => 'Task One Test',
                        ];

      $task = new Task();
      $actualResult =  $task->getAllTasks();

      $this->assertEquals($expectedResult, $actualResult[0]);

    }
}

推荐答案

您将模拟传递给 Task 类构造函数，但它不会对它做任何事情.

You pass the mock to the Task class constructor, but it doesn't do anything with it.

$task = new Task($resultSetMock);

更新了代码以便使用:

class Task {
    private $db;

    public function __construct( ?Database $db = null ) {
        // set the db if none is provided
        if( is_null($db) )
        {
            $db = new Database;
        }

        $this->db = $db;
    }

    // ...
}

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