Question

尝试在index.php上显示数据，用户名运行良好，电子邮件未显示。

server.php

$query = "SELECT * FROM users WHERE username='$username' AND password='$password'";
$results = mysqli_query($db, $query);
if (mysqli_num_rows($results) == 1) {


  $_SESSION['email'] = $email;
  $_SESSION['username'] = $username;
  $_SESSION['success'] = "You are now logged in";
  header('location: index.php');

index.php

<?php
// Echo session variables that were set on previous page
echo "Your email is: " . $_SESSION['email'] . "<br>";
echo "Your username is: " . $_SESSION['username'] . "";
?>

Answer 1

查询数据后未获取结果。使用mysqli_fetch_assoc

请注意，我已将mysqli_real_escape_string应用于您的WHERE条件参数，以解决与SQL注入相关的问题。 Although it is not a fool-proof solution!

更改为以下内容：

$query = "SELECT * 
          FROM users 
          WHERE username = '" . mysqli_real_escape_string($db, $username) . 
          " AND password = '" . mysqli_real_escape_string($db, $password) . "'";
$results = mysqli_query($db, $query);
if (mysqli_num_rows($results) == 1) {

  $row = mysqli_fetch_assoc($db, $results);
  $_SESSION['email'] = $row['email'];
  $_SESSION['username'] =  $row['username'];
  $_SESSION['success'] = "You are now logged in";

请阅读how to prevent SQL injection related issues using Prepared Statements上的。

Answer 2

这是完整的代码，$ username出现（用不同的用户测试）

if (isset($_POST['login_user'])) {
  $username = mysqli_real_escape_string($db, $_POST['username']);
  $password = mysqli_real_escape_string($db, $_POST['password']);

  if (empty($username)) {
    array_push($errors, "Username is required");
  }
  if (empty($password)) {
    array_push($errors, "Password is required");
  }

  if (count($errors) == 0) {
    $password = md5($password);
    $query = "SELECT * FROM users WHERE username='$username' AND password='$password'";
    $results = mysqli_query($db, $query);
    if (mysqli_num_rows($results) == 1) {


      $_SESSION['email'] = $email;
      $_SESSION['username'] = $username;
      $_SESSION['success'] = "You are now logged in";
      header('location: index.php');
    }else {
        array_push($errors, "Wrong username or password combination");
    }
  }
}

Answer 3

您没有从结果中获取电子邮件，因此$ email变量为null。如果该电子邮件位于您的用户表中，则需要从结果中分配该电子邮件。

为什么数据不显示php mysql

3 个答案: