【问题标题】:Extending a database class扩展数据库类
【发布时间】:2011-04-18 05:57:51
【问题描述】:

我有一个数据库类:

<?php

class dbConnect
{
    var $strHost = "";
    var $strDatabase = "";
    var $strUser = "";
    var $strPassword = "";

    var $intLinkID = 0;
    var $intQueryID = 0;
    var $arrRecord = array();
    var $intRow;

    var $intErrno = 0;
    var $strError = "";

    function dbConnect()
    {
        $this->Connect();
    }

    function dbHalt($strError)
    {
        $objError = new errorHandle();

        $objError->reportError("Database error: $strError", false);
        $this->intErrno = mysql_errno($this->intLinkID);
        $this->strError = mysql_error($this->intLinkID);

        $strError = sprintf("MySQL Error: %s (%s)", $this->intErrno, $this->strError);

        $objError->reportError($strError, true);
    }

    function Connect()
    {
        if (0 == $this->intLinkID) {

            $this->intLinkID = mysql_connect($this->strHost, $this->strUser, $this->strPassword);
            if (!mysql_select_db($this->strDatabase, $this->intLinkID)) {
                $this->dbHalt("cannot use database " . $this->strDatabase);
            }

        }

        if (!$this->intLinkID) {
            $this->dbHalt("Database connection failed");
        }


    }
}

class dbMain extends dbConnect
{
// Database connection settings.

// Dev server.
    var $strHost = 'db_host_here';
    var $strUser = 'db_User_here';
    var $strPassword = 'db_password_here';
    var $strDatabase = 'db_name_here';
}

调用$objDB = new dbMain(); 时可以正常工作

但我想要的是在另一个类(对于主 CMS)中动态设置数据库详细信息,例如:

$objDB = new dbMain($user, $pass, $dbname, $dbhost);

【问题讨论】:

  • 为了清楚起见,您应该将访问修饰符声明为 private 而不是 var。不再需要 var 关键字。它可以在 PHP5 中工作,但会在 PHP5 直到 5.3 版本中引发 E_STRICT 警告,从该版本开始,它已被弃用。

标签: php database class extend


【解决方案1】:

使用类构造函数。传入值。

class dbMain 
{
   function __construct($strHost='db_host_here',$strUser='db_User_here',$strPassword='db_password_here',$strDatabase='db_name_here'){
        $this->strHost = $strHost;
        $this->strUser = $strUser;
        $this->strPassword = $strPassword;
        $this->strDatabase = $strDatabase;
    }
}

【讨论】:

  • 类 dbDynamic 扩展 dbConnect { function __construct($strHost='localhost',$strUser='user',$strPassword='password',$strDatabase='database'){ $this->strHost = $str主机; $this->strUser = $strUser; $this->strPassword = $strPassword; $this->strDatabase = $strDatabase; } }
  • 以上带来mysql_connect not valid link resource的错误?
  • 检查您的参数。 mysql_connect() 期间检查错误;使用 mysql_error()
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