Question

我知道问题在于mysqli_query的第一个参数，但我不知道我的生活中找出了确切的参数应该是什么。我可以使用mysql_query，但我正在尝试学习OOP并转换为mysqli。我希望能够在需要时调用connect（并最终断开连接），然后在连接后执行查询。任何帮助都会很棒。

database.php中

class Database{

private $db_host = "localhost"; 
private $db_user = "root"; 
private $db_pass = "root"; 
private $db_name = "db";

public function __construct(){

    $this->connect();

}

public function connect()
{
    //If database is not connected, look to connect
    if(!$this->con)
    {
        $myconn = mysqli_connect($this->db_host,$this->db_user,$this->db_pass);
        //If database is connected, select table
        if($myconn)
            {
            $seldb = mysqli_select_db($this->db_name,$myconn);
            //If database is selected, return conn
            if($seldb)
            {
                $this->con = true; 
                return true; 
            } else
            {
                return false; 
            }
        } else
        {
            return false; 
        }
    } else
    {
        return true; 
    }
}

public function disconnect()
{
if($this->con)
{
    if(mysqli_close())
    {
        $this->con = false; 
        return true; 
    }
    else
    {
        return false; 
    }
}
}

}

school.php

require_once('database.php');

class Schools{

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

public function getAllSchools($connection){

$schoolsql = "SELECT SID, schoolName, schoolCity, schoolCountry, schoolRegion FROM schools ORDER BY SID";

$schoolquery = mysqli_query($connection->con, $schoolsql);
$results = array();

while($row = mysqli_fetch_array($schoolquery))
        {
           $results[] = array(
              'id' => $row['SID'],
              'name' => $row['schoolName'],
              'city' => $row['schoolCity'],
              'country' => $row['schoolCountry'],
              'region' => $row['schoolRegion']
           );
        }

        $json = json_encode($results);
        echo $json; 

}
}

$school = new Schools;

我知道这就是失败的地方。 $ schoolquery = mysqli_query（$ connection-＆gt; con，$ schoolsql）;

我只是不知道mysqli连接变量对数据库类的期望是什么？

这组代码“有效”。我知道它们是安全问题，但是现在我正在学习我只是想从数据库中获得我期望的结果 - 但最终代码结构并不是我想要的。我更喜欢让database.php文件包含一个我可以在schools.php文件中实例化的类。现在它只是创建连接，我可以将其加载到Schools类中。我已经在SO（以及Google）上查看了几个mysqli类连接问题，并尝试了几种排列，但我无法使Database类选项工作。

database.php中

$db_host = "localhost"; 
$db_user = "root"; 
$db_pass = "root"; 
$db_name = "db";

$mysqli = new mysqli($db_host, $db_user, $db_pass, $db_name);
if (!$mysqli) {
    die("Connection failed: " . mysqli_connect_error());
}

schools.php

class Schools{ 

function __construct($mysqli){
    $this->mysqli = $mysqli;
}

//Method to return all schools
function getAllSchools () {
    $query = "SELECT SID, schoolName, schoolCity, schoolCountry, schoolRegion FROM schools ORDER BY SID ASC";
    $results = array();
        if ($result = $this->mysqli->query($query) ) 
        {
            while ($row = $result->fetch_assoc()) {
                $results[] = array(
                  'id' => $row['SID'],
                  'name' => $row['schoolName'],
                  'city' => $row['schoolCity'],
                  'country' => $row['schoolCountry'],
                  'region' => $row['schoolRegion']
               );
            }
        $json = json_encode($results);
        echo $json;
        $result->close();
        } else {
            echo "No results";
        }
    }
 //Method to return one school  
 function getASchool ($id) {

    $query = "SELECT SID, schoolName, schoolCity, schoolCountry, schoolRegion FROM schools WHERE SID=$id";
    $results = array();
        if ($result = $this->mysqli->query($query) ) 
        {
            while ($row = $result->fetch_assoc()) {
                $results[] = array(
                  'id' => $row['SID'],
                  'name' => $row['schoolName'],
                  'city' => $row['schoolCity'],
                  'country' => $row['schoolCountry'],
                  'region' => $row['schoolRegion']
               );
            }
        $json = json_encode($results);
        echo $json;
        $result->close();
        } else {
            echo "No results";
        }
}   
}

$school = new Schools($mysqli);
$school->getASchool("120");

Answer 1

我更喜欢让database.php文件包含一个类

你拥有它。请注意，Mysqli已经是一流的课程了。

我可以在schools.php文件中实例化。

你不能拥有它。实例化mysqli类意味着创建一个到mysql的新连接，你肯定只希望每个脚本一次。因此，您只需要将其实例化一次，然后将此唯一实例传递给其他类，就像您一样。

Answer 2

注意：关于代码结构有几点可以/应该改进，但是这个答案只是为了回答你的直接问题。

有两部分：

在class Database，而不是return true;，您需要return $this->con，如下所示：

class Database{

    private $db_host = "localhost"; 
    private $db_user = "root"; 
    private $db_pass = "root"; 
    private $db_name = "db";
    // Add this class variable to store con
    private $con = FALSE;

    public function __construct(){
        // Return the connection
        return $this->connect();

    }

    public function connect() {
        //If database is not connected, look to connect
        if( ! $this->con) {
            $myconn = mysqli_connect($this->db_host,$this->db_user,$this->db_pass);
            //If database is connected, select table
            if($myconn) {
                $seldb = mysqli_select_db($this->db_name,$myconn);
                //If database is selected, return conn
                if($seldb) {
                    // Assign the class variable to the connection
                    $this->con = $myconn; 
                    // Return the connection
                    return $this->con; 
                } else {
                    return false; 
                }
            } else {
                return false; 
            }
        } else {
            // Return the connection
            return $this->con; 
        }
    }
// ... the rest of your class ...

}

然后，在class Schools中，您应该为$connection使用类变量，如下所示：

class Schools{
    // Class variable to store your database connection
    private $connection;

    public function __construct() {   
        // Assign the connection to your class variable
        $this->connection = new Database();
        $this->getAllSchools();
    }

    public function getAllSchools($connection){
        $schoolsql = "SELECT SID, schoolName, schoolCity, schoolCountry, schoolRegion FROM schools ORDER BY SID";
        // Use the class variable to connect
        $schoolquery = mysqli_query($this->connection, $schoolsql);
        $results = array();
    // ... rest of your code
    }
}

根据扩展答案

修改：

首先，你的路线在这里： $school = new Schools($mysqli); - 这是dependency injection的一种形式。这是一个很好的学习/熟悉的东西，因为它在编写那些依赖＆＃34;依赖的课程时非常有价值。在其他课程上。

为了理清下面代码为什么没有按照您的期望进行操作，我需要查看您的代码版本。请注意，我提供的Database类修订版有三行新return行。如果缺少任何一个，代码可能无法正常工作。

代码实际上应该失败并发生致命错误 - 函数getAllSchools需要一个参数，但对它的调用不包括该参数。

最后，最好的办法是学习如何调试它。通过修改Schools类功能，进行一些故障排除，如下所示：

public function getAllSchools(){
    $schoolsql = "SELECT SID, schoolName, schoolCity, schoolCountry, schoolRegion FROM schools ORDER BY SID";
    // Use the class variable to connect
    $schoolquery = mysqli_query($this->connection, $schoolsql);
    // Is there an issue? Let's output it:
    var_dump(mysqli_error($this->connection));
    $results = array();
// ... rest of your code
}

向我们提供错误的详细信息，并确保使用您生成错误的特定代码更新我们。

php类mysqli_query问题

2 个答案: