将变量作为变量传递给url时出错

Question

我试图将会话变量传递给变量'var‘中的url，这是一个新变量，请帮助我解决下面代码行7中的问题。

php

           if($user){
                 $response["error"] = false;
                 $_SESSION['vault_no'] = $user['vault_no'];

                $to=$email;
                $subject = "Reset profile password";
                $txt = "Click on link to reset your profile password-> www.miisky.com/appmiisky/reset_pro.php?var = $_SESSION['vault_no']" ;
                $headers =  'From: innovation@miisky.com' . "\r\n" ;
                $headers .= "BCC: prajwalkm7@gmail.com\r\n";
                $headers .=  'Reply-To: innovation@miisky.com' . "\r\n" .
                 'X-Mailer: PHP/' . phpversion(); 
                /*$headers = 'From: innovation@miisky.com' . "\r\n" .
                            'Cc: innovation@miisky.com' . "\r\n" .
               'Reply-To: innovation@miisky.com' . "\r\n" .
                'X-Mailer: PHP/' . phpversion();*/

                mail($to,$subject,$txt,$headers);
                 echo json_encode($response);
            }

Answer 1

双引号(")中的所有文本都用作字符串。

$txt = "Click on link to reset your profile password-> www.miisky.com/appmiisky/reset_pro.php?var = $_SESSION['vault_no']" ;

您需要使用字符串与您的会话值联系以获得url。

$txt = "Click on link to reset your profile password-> www.miisky.com/appmiisky/reset_pro.php?var=" . $_SESSION['vault_no'];