Question

我建立一个社交网络，我使用php会话，当用户转到另一个页面时允许信息留在页面上，但是当mysql脚本更新一个值时。它确实反映了所做的更改，除非用户注销并重新登录。任何想法？

谢谢。。

<?php

session_start();

$login_email = $_SESSION['email'] ;
$login_pass = $_SESSION['pass'] ;

$target_path = "pictures/";

$target_path = $target_path . basename( $_FILES['file']['name']); 

if (move_uploaded_file($_FILES['file']['tmp_name'], $target_path) and

   $con = mysql_connect("localhost","root","naruto") and 

    mysql_select_db("users", $con) and 

    mysql_query (" UPDATE user_info SET profile_pic = ' $target_path ' WHERE email = '$login_email' AND password1 = '$login_pass' " ) ) {

        session_destroy ();

        include 'login.php';

        session_start ();

 if ( $login_email == $_SESSION['page_email'] && $login_pass == $_SESSION['page_pass ']){       


        header ('location:home.php');

} else{

    echo "There was an error uploading the file, please try again!";
}

    }



?>

$ _ SESSION [＆＃39; PAGE_EMAIL＆＃39;] $ _SESSION [＆＃39; page_pass＆＃39;]

Answer 1

这取决于您如何在用户页面上显示个人资料照片。如果它是通过存储的会话，我建议你创建一个函数来返回用户信息存储在$ _SESSION中，并在每个用户配置文件更新时调用它

mysql_query (" UPDATE user_info SET profile_pic = ' $target_path ' WHERE email = '$login_email' AND password1 = '$login_pass' " ) ) {

    session_destroy();

    include 'login.php';

    session_start ();
    $_SESSION['user'] = get_user_info(); // your new function to return user info

php会话和mysql更新

1 个答案: