Question

我有两个数据库，我想在userid密码条件下选择一个数据库，但是现在它不能处理这个条件。我怎样才能实现它？目前，只选择第一个数据库，而不是第二个数据库。

<?php
    if($_SESSION['s_activId'] == 'om' && isset($_SESSION['s_userType']) == 'om')
    {
      $dbName = "kal";
      $link = mysql_connect("localhost","root","");
      mysql_select_db($dbName) or die("colud not connect to database". mysql_error());
    }
    else if($_SESSION['s_activId'] == 'om' && isset($_SESSION['s_userType']) == 'om1')
    {
      $dbName = "kal1";
      $link = mysql_connect("localhost","root","");
      mysql_select_db($dbName) or die("colud not connect to database". mysql_error());
    }
?>

Answer 1

如果您在脚本中使用$_SESSION，则必须启动会话，否则该会话不可见。

If测试需要稍微修正，isset()返回一个布尔值，而不是数组中出现的数据。

<?php
// add this to gain access to the session
session_start();

if($_SESSION['s_activId'] == 'om' && 
   isset($_SESSION['s_userType']) && 
   $_SESSION['s_userType'] == 'om')
{
    $dbName = "kal";
    $link = mysql_connect("localhost","root","");
    mysql_select_db($dbName) or die("could not connect to database". mysql_error());
}
else if($_SESSION['s_activId'] == 'om' && 
        isset($_SESSION['s_userType']) && 
        $_SESSION['s_userType'] == 'om1')
{
    $dbName = "kal1";
    $link = mysql_connect("localhost","root","");
    mysql_select_db($dbName) or die("could not connect to database". mysql_error());
}

?>

Answer 2

试试这个：

<?php

session_start(); //starts the session, then you can access its members

if($_SESSION['s_activId'] == 'om' && $_SESSION['s_userType'] == 'om')
{
    $dbName = "kal";
    $link = mysql_connect("localhost","root","");
    mysql_select_db($dbName) or die("could not connect to database". mysql_error());
}
else if($_SESSION['s_activId'] == 'om' && $_SESSION['s_userType'] == 'om1')
{
    $dbName = "kal1";
    $link = mysql_connect("localhost","root","");
    mysql_select_db($dbName) or die("could not connect to database". mysql_error());
}

?>

Answer 3

您无法将isset与字符串进行比较，因为它返回true / false。此外，您必须在要使用会话的文件中使用session_start()。

修复你的代码：

   <?php
    session_start();
    if(isset($_SESSION['s_activId']) && isset($_SESSION['s_userType']
    && $_SESSION['s_activId'] == 'om' && $_SESSION['s_userType']=='om')
    {
      $dbName = "kal";
      $link = mysql_connect("localhost","root","");
      mysql_select_db($dbName) or die("colud not connect to database". mysql_error());
    }
    else if(isset($_SESSION['s_activId']) && isset($_SESSION['s_userType']
    && $_SESSION['s_activId'] == 'om1' && $_SESSION['s_userType']=='om1')
        {
      $dbName = "kal1";
      $link = mysql_connect("localhost","root","");
      mysql_select_db($dbName) or die("colud not connect to database". mysql_error());
    }

?>

但是这个代码的一个更好的方法是使用mysqli，因为不推荐使用mysql。

<?php
    session_start();
    if(isset($_SESSION['s_activId']) && isset($_SESSION['s_userType'] )
    {
      if ($_SESSION['s_activId'] == 'om' && $_SESSION['s_userType']=='om')
      {
        $dbName = "kal";
        $link = mysqli_connect("localhost","root","", $dbName);
      }
      else if($_SESSION['s_activId'] == 'om1' && $_SESSION['s_userType']=='om1') 
      {
        $dbName = "kal1";
        $link = mysqli_connect("localhost","root","",$dbName);
      }
    } 
?>

这应该有效。

如果您正在使用其他mysql函数，则第二个代码将无法正常工作，除非您将所有这些代码更改为mysqli。

例如mysql_fetch_array($result)将成为mysqli_fetch_array($link, $result)

Answer 4

不要使用 mysql _ 功能。它们已被弃用。使用 mysqli _ 或预备语句。

[注意：MySQLi扩展程序适用于MySQL 4.1.13或更高版本。]

session_start();

if($_SESSION['s_activId'] == 'om' && isset($_SESSION['s_userType']) && $_SESSION['s_userType'] == 'om')
{
    $dbName = "kal";
    $link = mysqli_connect("localhost","root","",$dbName) or die("could not connect to database". mysqli_connect_error();
}
else if($_SESSION['s_activId'] == 'om' && isset($_SESSION['s_userType']) && $_SESSION['s_userType'] == 'om1')
{
    $dbName = "kal1";
    $link = mysqli_connect("localhost","root","",$dbName) or die("could not connect to database". mysqli_connect_error();
}

?>

Answer 5

试试这段代码，可能会有所帮助。代码很简单为了调试您的问题，请使用

 echo print_r($_SESSION, true);

要查看您要使用的值。

<?php
session_start();

$sActivId = isset($_SESSION['s_activId']) ? $_SESSION['s_activId'] : "";
$sUserType = isset($_SESSION['s_userType']) ? $_SESSION['s_userType'] : ""; 
$dbName = "";

if($sActivId != 'om'){
    throw new Exception("The user is not valid or active");
}
switch($sUserType){
    case 'om': 
        $dbName = "kal";
        break;
    case 'om1':
        $dbName = "kal1";
        break;
    default:
        break;
}
if(empty($dbName)){
    throw new Exception("The db Name is undefined");
}

$link = mysql_connect("localhost","root","");
if(!$link){
    throw new Exception("'Could not connect: ' . mysql_error()");
}

mysql_select_db($dbName) or die("colud not connect to database". mysql_error());


?>

在userid和password条件上选择数据库

5 个答案: