Question

我是PHP脚本的新手，来自Java背景。这是一个微不足道的事情，现在变成了我的脑力激荡。所以这就是问题所在，我为变量分配了一些值，当尝试在if / else语句中使用该值时，变量实际上不具有先前赋值的值。这是代码： -

<?php
    session_start();

    $email = $_POST["Email"];   
    $password = $_POST["Password"];

    $db_username="root";
    $db_password="root";
    $database="mydb";
    $localhost = "mysql";

    $con = mysql_connect($localhost,$db_username,$db_password);
    mysql_select_db($database,$con) or die( "Unable to select database");

    $query = "select * from photobook.users where email ='$email' and password ='$password';" ;

    $result = mysql_query($query);

    $num=mysql_num_rows($result);

    if($num == 1){
        while($row = mysql_fetch_array($result))
             {
                $_SESSION['email'] = $row['email'];

                $_SESSION['username'] = $row['username'];   
             }

        header("location: home.php");

    }

    else{

        include "photoBookProtocol.php";

                print "<br>email value after photobookprotocol file include is $email";
                print "<br>password value after photobookprotocol file include is $password";

        $obj=new Protocol();

        $var = $obj->loginCheck($email,$password);

                print "value of var received is $var";

        if($var == 0){
            session_destroy();
                        print "<br>user does not exist";
            //header("location: login.php");
        }
        else{
            $_SESSION['email'] = $var[0];
            $_SESSION['username'] = $var[1];
            print "<br>user exists";
            header("location: home.php");

        }   
    }

    mysql_close($con);
?>

因此，当我在“else”子句中传递loginCheck（$ email，$ password）中的$ email和$密码时，没有任何内容传递。知道为什么会这样吗？

Answer 1

您的变量范围没有任何问题，因此：

POST数据中没有任何内容
include会覆盖两个变量
loginCheck()正在接收正确的变量，但函数

作为旁注，由于您的脚本依赖于POST数据，因此您应该有条件在继续之前检查所需数据是否存在：

if(isset($_POST['Email'], $_POST['Password']))
{
    // something posted
}

在PHP中的if / else语句中访问时变量超出范围

1 个答案: