我正在为我的网站创建博客输入系统，如果已经有输入URL，我希望它创建相同的URL加一个数字。
因此，以URL“ test”为例，如果URL测试已经存在，我希望它为“ test1”，如果URL“ test1”已经存在，我希望它为“ test2”。
这是我现在（不是真的）正在做的事情：
如您所见，它并不是按照我想要的方式工作。它只能从“ test”变为“ test1”-如何使其动态地无限重命名？
哦，这仅供我个人使用，不需要PDO。
编辑：好的，眉毛跳动后，我收到了将所有内容切换到PDO的信息：
$title = $_POST['title'];
$author = $_POST['author'];
$content = $_POST['content'];
$date = new DateTime('now');
$date = $date->format('Y-m-d');
$domainurl = 'https://test.com/entry.php?';
$maxlength = (75 - strlen($domainurl));
if (strlen($_POST['title']) > $maxlength) {
$url = wordwrap($_POST['title'], $maxlength);
$url = substr($url, 0, strpos($url, "\n"));
$url = strtr(strtolower($url), ' ', '-');
}else{
$url = strtr(strtolower($_POST['title']), ' ', '-');
}
$url = preg_replace('/[^A-Za-z0-9\-]/', '', $url);
try {
$options = [
\PDO::ATTR_ERRMODE => \PDO::ERRMODE_EXCEPTION,
\PDO::ATTR_DEFAULT_FETCH_MODE => \PDO::FETCH_ASSOC,
\PDO::ATTR_EMULATE_PREPARES => false,
];
$conn = new PDO("mysql:charset=utf8mb4;host=$servername;dbname=$dbname", $username, $password);
$sql = "SELECT count(*) FROM blog WHERE url = ?";
$sql = $conn->prepare($sql);
$sql->execute([$url]);
$rowCount = $sql->fetchColumn();
if ($rowCount >= 1) {
$url = $url . $rowCount;
}
try {
$sql = $conn->prepare("INSERT INTO blog (date, title, url, author, content)
VALUES (:date, :title, :url, :author, :content)");
$sql->bindParam(':date', $date);
$sql->bindParam(':title', $title);
$sql->bindParam(':url', $url);
$sql->bindParam(':author', $author);
$sql->bindParam(':content', $content);
$sql->execute();
$rowCount = $sql->rowCount();
$id = $conn->lastInsertId();
if ($rowCount > 0) {
echo json_encode(array ('type' => 'success', 'message' =>"Blog entry created successfully"));
}else{
echo json_encode(array ('type' => 'danger', 'message' =>"Blog entry <strong>not</strong> created successfully"));
}
}catch(PDOException $e){
echo json_encode(array ('type' => 'danger', 'message'=>$e->getMessage()));
}
}catch(PDOException $e){
echo json_encode(array ('response'=>'error','errorMessage'=>$e->getMessage()));
}
$conn = null;
尝试一下：