运行 PHP 静态属性中的时间错误和访问 Class 中的全局实例

Question

我设计了一个 PHP Class 和 Static 属性，但我得到一个没有任何 error log 的空白页面，请帮助我 [=22] =]代码?

<?php

ini_set("display_startup_errors", 1);
ini_set("display_errors", 1);
error_reporting(-1);

class Response
{
    public $Status;
    public $Message;

    function __construct() {
        $this->Status = FALSE;
        $this->Message = "";
    }        
}

$response = new Response();

class Connection
{
    static private $server = "localhost"
    static private $database = "Student";
    static private $user = "ram";
    static private $password = "ram12345!";

    static public $link;

    public static function Trigger() {
        try
        {
            if (!isset(self::$link)) {
                self::$link = mysql_connect("localhost", "bbminfoc_bbm", "Princess4BBM.>") or die("Couldn't make connection.");
                if(!self::$link)
                {
                    $response->Status = FALSE;
                    $response->Message = "Database Connection Failed !";
                }
                else
                {
                    if(!mysql_select_db(self::$database, self::$link))
                    {
                        $response->Status = FALSE;
                        $response->Message = "Couldn't select database Main !";
                    }
                }
            }
        }
        catch(Exception $e) {
            $response->Status = FALSE;
            $response->Message = "Exception: " . $e->getMessage();
        }

        if(!$response->Status)
        {
            unset(self::$link);
        }
    }
}

if(!isset(Connection::link))
{
    Connection::Trigger();
}

$outp = json_encode($response);

if(isset($outp))
{
    echo($outp);
}
else
{
    echo("No Data");
}

?>

请在这段代码中帮助我，我不知道我在上面粘贴的代码中做错了什么，因为我无法在 error_log 文件中看到日志信息。

Answer 1

正如@Yolo 在评论中已经提到的，Connection::Trigger(); 和 static private $server = "localhost" 中缺少一个分号。因为这会导致解析错误，脚本将在执行前失败，您不会看到任何错误。

要查看这些错误，您必须找到 php.ini 配置文件并设置 display_errors = on。此外，您还应该考虑使用带有自动代码检查功能的 IDE，它会在您键入代码时向您显示潜在的解析错误。

摘自 this 回答。

完全固定的代码如下所示：

<?php

ini_set("display_startup_errors", 1);
ini_set("display_errors", 1);
error_reporting(-1);

class Response
{
    public $Status;
    public $Message;

    function __construct() {
        $this->Status = FALSE;
        $this->Message = "";
    }        
}

$response = new Response();

class Connection
{
    static private $server = "localhost";
    static private $database = "Student";
    static private $user = "ram";
    static private $password = "ram12345!";

    static public $link = null;

    public static function Trigger() {
        global $response;

        try
        {
            if (self::$link !== null) {
                self::$link = mysql_connect("localhost", "bbminfoc_bbm", "Princess4BBM.>") or die("Couldn't make connection.");
                if(!self::$link)
                {
                    $response->Status = FALSE;
                    $response->Message = "Database Connection Failed !";
                }
                else
                {
                    if(!mysql_select_db(self::$database, self::$link))
                    {
                        $response->Status = FALSE;
                        $response->Message = "Couldn't select database Main !";
                    }
                }
            }
        }
        catch(Exception $e) {
            $response->Status = FALSE;
            $response->Message = "Exception: " . $e->getMessage();
        }

        if(!$response->Status)
        {
            self::$link = null;
        }
    }
}

if(Connection::$link !== null)
{
    Connection::Trigger();
}

$outp = json_encode($response);

if(isset($outp))
{
    echo($outp);
}
else
{
    echo("No Data");
}

?>

Answer 2

在 static private $server = "localhost" 你错过了最后一个 ;

此外，我收到此错误：
Fatal error: Cannot use isset() on the result of an expression (you can use "null !== expression" instead) in ... on line 61

因此，将 if(!isset(Connection::link)) 更改为 if (Connection::$link !== null)

这应该可以解决您的问题。

注意
请不要使用 mysql_* 扩展名。它已被弃用。如果您想继续使用 mysql，请更改为 PDO or mysqli

Answer 3

将其更改为：

if(!isset(Connection::link))
{
    Connection::Trigger();
}

这个：

if(!isset(Connection::$link))
{
    Connection::Trigger();
}

Answer 4

你不需要取消连接它会自动取消..在class,

之后

你在 localhost 后错过了分号，你没有在函数触发器中调用响应...

<?php


    ini_set("display_startup_errors", 1);
    ini_set("display_errors", 1);
    error_reporting(-1);

    class Response
    {
        public $Status;
        public $Message;

        function __construct() {
            $this->Status = FALSE;
            $this->Message = "";
        }        
    }

    $response = new Response();

    class Connection
    {
        static private $server = "localhost";
        static private $database = "listings";
        static private $user = "root";
        static private $password = "";

        static public $link;

        public static function Trigger() {

            global $response;
            try
            {
                if (!isset(self::$link)) {
                    self::$link = mysql_connect("localhost", "root", "") or die("Couldn't make connection.");
                    if(!self::$link)
                    {
                        $response->Status = FALSE;
                        $response->Message = "Database Connection Failed !";
                    }
                    else
                    {
                        if(!mysql_select_db(self::$database, self::$link))
                        {
                            $response->Status = FALSE;
                            $response->Message = "Couldn't select database Main !";
                        }
                    }
                }
            }
            catch(Exception $e) {
                $response->Status = FALSE;
                $response->Message = "Exception: " . $e->getMessage();
            }

        }
    }

    if(!isset(Connection::$link))
    {
        Connection::Trigger();
    }

    $outp = json_encode($response);

    if(isset($outp))
    {
        echo "hello";
        echo($outp);
    }
    else
    {

        echo "hello";
        echo("No Data");
    }

运行 PHP 静态属性中的时间错误和访问 Class 中的全局实例

Run time error in PHP Static Properties and Accessing Global Instance within a Class

php

static

static-methods

reference

self