如何让2个SESSIONS同时进行?如果有可能PHP

时间:2014-02-18 15:52:11

标签: php session phpmyadmin

我创建了一个用户登录页面,用户使用他们的名称和密码登录数据库(DBname = signin,table = information)我想保持会话继续进行,无论用户继续访问哪个页面显示“HELLO和用户名”)。到目前为止,这完美无缺。

当进入AddRental页面时(只有地址信息)我必须创建一个新的数据库连接和会话到DB = project和table = address。但由于某种原因,它会在第2行附近出现错误SESSION_START();

我想我需要同时保持2次会议,但我不知道该怎么做。有没有人有什么建议?

非常感谢

<?php
session_start();

    $hostname="localhost"; // Host name 
    $username="root"; // phpmyadmin username 
    $password=""; // phpmyadmin password 
    $dbname="project"; // Database name 
    $tblname="address"; // Table name

    // Connect to server and select databse.
mysql_connect("$hostname", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$dbname")or die("cannot select DB");


    // If form not yet submitted display form
    // To check if the Submit button was clicked
    if (isset($_POST['submit']))
    {
        // Get data entered into into form fields
        $PostCode = $_POST['PostCode'];
        $AddL1 = $_POST['AddL1'];
        $AddL2 = $_POST['AddL2'];
        $AddL3 = $_POST['AddL3'];
        $County = $_POST['County'];
        $Country = $_POST['Country'];   


        // Validation of the data entered into into form fields
        //if ((preg_match('/^([A-Za-z]+ ?)*$/', $name)) 
        //&& (preg_match('/^[A-Za-z0-9_]+([\.\-\+]{0,1}[A-Za-z0-9_])*@[A-Za-z0-9_]+([\.-]{0,1}[A-Za-z0-9_]+)*(\.[A-Za-z0-9]{2,4})+$/', $email))
        //&& (preg_match('/^[A-Z][A-Z]/', $country)) 
        //&& (preg_match('/^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[^a-zA-Z0-9])(?!.*\s)/', $password)))
        {           
            // Insert data taken form fields into table
            mysql_query("INSERT INTO `address` (`PostCode`, `AddL1`, `AddL2`, `AddL3`,`County`,`Country`)
            VALUES('$PostCode', '$AddL1', '$AddL2', '$AddL3', 'County','Country'") or die(mysql_error());
        }   
    }   
    // Close connection to database server
//  mysql_close($mysql_connect);
?> 

<!DOCTYPE html>
<html><head><title> Menu Form </title>
<link rel= "stylesheet" type="text/css" href="AddRental.css"/>
</head><body>

<div id="logo">
<img src="C:/Users/Daisy/Desktop/Website/Forms/FC.JPG"  width="30%" height="5%"/>
<div id="addr"> Add Rental </div>
</div>

<div id="content">
    <div id ="rform"> Insert Address
    <br>    
<FORM action = "AddRental.php" method="post">
            <label for="PostCode">PostCode</label>
            <input type="text" name="PostCode" id="PostCode" value="" maxlength="8" />
            <br>
            <label for="AddL1">AddL1</label>
            <input type="text" name="AddL1" id="AddL1" value="" maxlength="40" />
            <br>
            <label for="AddL2">AddL2</label>
            <input type="text" name="AddL2" id="AddL2" value="" maxlength="25" />
            <br>
            <label for="AddL3">AddL3</label>
            <input type="text" name="AddL3" id="AddL3" value="" maxlength="25" />
            <br>
            <label for="County">County</label>
            <input type="text" name="County" id="County" value="" maxlength="20" />
            <br>
            <label for="Country">Country</label>
            <input type="text" name="Country" id="Country" value="" maxlength="20" />
            <br>
            <button type="submit" name="submit" >SUBMIT</button></div>

</FORM>

    <div id ="lilmenu"> <?php echo "Hello ".$_SESSION['myusername']; ?>
    <br>Options <br>

        <FORM METHOD="LINK" ACTION="Welcome.php">
            <INPUT TYPE="submit" VALUE="Back"></FORM><br>
        <FORM name="LogOut" method="post" action="logout.php">
            <INPUT name="LogOut" TYPE="submit" VALUE="Log-Out"></FORM>
    </div>
</div>



<div id="footer"> &copy; Copyright 2014 </div>
</body></html>

1 个答案:

答案 0 :(得分:0)

从你的评论中,你得到"You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 2",这是由你的SQL查询引起的,而不是会话。

我不支持mysql_*功能,但在您使用时,请确认您在使用mysql_real_escape_string()之前清理输入。否则输入中的'"等特殊字符可能会导致SQL查询出错。