我正在尝试在我的项目上创建特征的搜索和显示类型，但是，当我尝试在搜索栏中键入某些内容时，什么都没有显示，如果没有条目，我想显示一条错误消息。值，但显示具有所述值的条目。下面是我的代码，我将非常感谢您提出的任何建议和/或解决方案，正如我所说，我只是一个初学者。
<?php
$db = mysqli_connect('', '', '', '');
if ($db->connect_error){
die("Error: " . $db->connect_error);
}
$sql = "SELECT * FROM Quiz";
if(isset($_GET['search'])){
$name = mysqli_real_escape_string($db, htmlspecialchars($_GET['search']));
$sql = "SELECT * FROM Quiz WHERE name = '$name'";
}
?>
<!DOCTYPE html>
<html>
<head>
<title> Quizzes </title>
<link rel = "stylesheet" type = "text/css" href = "quiz.css">
</head>
<body>
<div class = "header">
<h2> Quizzes </h2>
</div>
<form method = "GET" action = "">
<div class = "input-group">
<label>Search for a quiz: </label>
<input type = "type" name = "name" placeholder = "search">
</div>
<div class="input-group">
<input type="submit" class="btn" name="search">
</div>
<?php
while ($row = $result -> fetch_assoc()){
echo $row['name'];
}
?>
</form>
</body>
</html>
这是一个如何连接到数据库的示例：
DEFINE('SERVER', '127.0.0.1');
DEFINE('PORT', '');
DEFINE('PSEUDO', 'root');
DEFINE('PWD', '');
DEFINE('DB_NAME', 'EEE');
function connectDB() {
static $db = null;
if ($db === null) {
try {
$pdo_options[PDO::ATTR_ERRMODE] = PDO::ERRMODE_EXCEPTION;
$db = new PDO("mysql:host=". SERVER .";port=". PORT .";dbname=". DB_NAME, PSEUDO, PWD, $pdo_options);
$db->exec('SET CHARACTER SET utf8');
} catch (Exception $exc) {
throw $exc;
}
}
return $db;
}