Question

我想显示表格中的用户数据，因此一旦用户登录网站，页面上会显示文字，上面写着＆＃34;欢迎（名字）（姓氏）。＆＃34;

名字的列是＆＃34;名字＆＃34;姓氏是＆＃34;姓氏＆＃34;。我尝试过很多不同的方法，但是所有这些方法似乎都会返回错误而我无法使用它。如果有人可以提供帮助，那就太棒了。

这是我的代码：

<html>
<head>
<title>Landing page</title>
<link rel="stylesheet" type="text/css" href="css.css">
</head>
<body>

<?php
/* ESTABLISH FIRST YOUR CONNECTION TO YOUR DATABASE */
$con = new mysqli("localhost", "root", "", "user_login"); /* REPLACE NECESSARY DATA */

if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}

if($stmt = $con->prepare("SELECT Username, Password FROM users WHERE Username = ? AND Password = ?")){
$stmt->bind_param("ss",$_POST["username"],$_POST["password"]);
$stmt->execute();
$stmt->store_result();
$checklog = $stmt->num_rows;

if($checklog > 0){
  /* HERE IS YOUR CODE WITH SUCCESSFUL LOGIN */  
  session_start();
  $_SESSION["logged_in"] = "YES";
  echo "<h1>You are now logged in</h1>";
  echo "<p><a href='secure1.php'>Link to protected file</a></p>";
  echo "<p><a href='secure2.php'>Link to protected file #2</a></p>";

}
else {
  /* HERE IS YOUR CODE WITH UNSUCCESSFUL LOGIN */
  session_start();
  $_SESSION["logged_in"] = "NO";
  echo "<h1>You are NOT logged in </h1>";
  echo "<p><a href='secure1.php'>Link to protected file</a></p>";
  echo "<p><a href='secure2.php'>Link to protected file #2</a></p>";
}

$stmt->close();
}

?>
<p><a href="public.html">Public Page</a></p>
<p><a href="logout.html">Logout</a></p>
</body>
</html>

我在做什么错，我该如何解决？

Answer 1

您可以执行$stmt->fetch(PDO::FETCH_ASSOC)

并将结果存储在$result

中

$result = $stmt->fetch(PDO::FETCH_ASSOC)  
$Firstname = $result["Firstname"];

由于您可能只有一个匹配的行，我建议您的条件必须为$checklog==1

所以，你应该有这样的东西

$checklog = $stmt->num_rows;
if($checklog == 1)  // Allows only for one match
{
session_start();
$_SESSION["logged_in"] = "YES";
$result = $stmt->fetch(PDO::FETCH_ASSOC);  // Fetch the row into $result
$Firstname = $result["Firstname"]; // This is your First Name
$Lastname = $result["Lastname"]; // This is your Last Name
}

您可以在此会话中拥有First Name和Last Name

$_SESSION["FirstName"] = $result["Firstname"];
$_SESSION["LastName"] = $result["Lastname"];

更新：

以下是专为您精心制作的酷炫且完全更新的版本：）

我花了10分钟来构建它：）

<?php
$conn = new PDO("mysql:host=localhost;dbname=user_login",'root','');
$name = $_POST["username"];
$pass = $_POST["password"];
$sql = "SELECT * FROM users WHERE Username = ? AND Password  = ?";
$q = $conn->prepare($sql);
$q->execute(array($name,$pass));
$count = $q->rowCount();
if ($count==1) 
{
$q->setFetchMode(PDO::FETCH_BOTH);
while($row = $q->fetch())
{
echo $row['FirstName']; //Just to show you output
session_start();
$_SESSION["FirstName"]  = $row['FirstName'];
}   
}
?>

Answer 2

试试这个..

<?php   if($stmt = $con->prepare("SELECT firstname,lastname,Username, Password FROM users WHERE Username = ? AND Password = ?")){
$stmt->bind_param("ss",$_POST["username"],$_POST["password"]);
$stmt->execute();
//$stmt->store_result();
//$checklog = $stmt->num_rows;
$results = $stmt->fetch(PDO::FETCH_ASSOC);
 $_SESSION['firstname'] = $results['firstname'];
 $_SESSION['lastname'] = $results['lastname'];

print_r($_SESSION);

显示登录数据的用户

2 个答案: