我经历了多个尝试完成此任务的过程,但无法完全理解。我继续使用Stack Overflow找到重复的答案。我发现了一些,并尝试将它们调整为我的代码。仍然没有用。我试图说我是否已登录到个人资料页面,如果用户名在数据库中,我希望您显示他们的名字。如果不在数据库中,那么就不要显示任何内容。以下是我提出的建议:
if (isset($_SESSION['user_id'])){
$username = mysqli_real_escape_string($con, $_POST['username']);
$sql = "SELECT * FROM users WHERE username = '".$username."'";
$result = mysqli_query($con,$sql);
if(mysqli_num_rows($result)>=1){
echo "User was found in the database";
}
else{
echo "User was not found in the database.";
}
}
所以if isset user_id基本上说如果我已经登录,那么就这样做。之后的代码是尝试查找是否在该数据库中找到用户名。如果是,那就说出来了。如果没有,那就不要了。我希望这很清楚!谢谢!
编辑:这是HTML代码:
<?php
session_start();
ob_start();
include_once('dbconnect.php');
?>
<div class="banner_container">
<div class="jumbotron text-center">
<?php
if (isset($_SESSION['user_id']) && isset($_POST['username'])){
if(($_SESSION['user_id'] != "") && ($_POST['username'] != "")){
$user_id = $_SESSION['user_id'];
$username = mysqli_real_escape_string($con, $_POST['username']);
$sql = "SELECT * FROM users WHERE username = '".$username."' AND user_id !=".$user_id;
$result = mysqli_query($con, $sql);
if(mysqli_num_rows($result) > 0){
echo "User was found in the database";
}
else{
echo "User was not found in the database.";
}
}
else{
echo "Username or user Id is empty";
}
}
?>
<?php
echo "<h1>";
echo $_SESSION['first_name'];
echo " ";
echo $_SESSION['last_name'];
echo "</h1>";
echo "<p>";
echo '"';
echo $_SESSION['quote'];
echo '"';
echo "<br>";
echo $_SESSION['who'];
echo "</p>";
?>
</div>
</div>
<li><a href="http://www.quotin.co">Home</a></li>
<li><a href="quotin_about"> About</a></li>
<li class="qotd"><a href="quotin_qotd"> Quote of the Day</a></li>
<li class="all_categories"><a href="quotin_categories">All Categories</a></li>
<li><a href="http://www.quotin.co/quotin_authors"> Authors</a></li>
<?php
if(isset($_SESSION['user_id'])){
echo $_SESSION['user_id'];
echo '<li id="active" class="dropdown">';
echo '<a id="act_color" href="#" class="dropdown-toggle" data-toggle="dropdown">';
echo $_SESSION['first_name'];
echo "'s";
echo ' ';
echo "Profile";
echo '<b class ="caret"></b></a>';
echo '<ul class="dropdown-menu">';
echo '<li><a href="profile.php"> Profile</a></li>';
echo '<li><a href="logout.php">Log out</a></li>';
echo '</ul>';
echo '</li>';
echo '</a>';
echo '</li>';
} else {
}
?>
这是我所在的个人资料页面。
答案 0 :(得分:0)
使用此代码:
if (isset($_SESSION['user_id']) && isset($_POST['username'])){
if(($_SESSION['user_id'] != "") && ($_POST['username'] != "")){
$user_id = $_SESSION['user_id'];
$username = mysqli_real_escape_string($con, $_POST['username']);
$sql = "SELECT * FROM users WHERE username = '".$username."' AND user_id !=".$user_id;
$result = mysqli_query($con, $sql);
if(mysqli_num_rows($result) > 0){
echo "User was found in the database";
}
else{
echo "User was not found in the database.";
}
}
else{
echo "Username or user Id is empty";
}
}