请帮助：）。我得到这个错误：
Warning: mysqli::prepare() [mysqli.prepare]: (42000/1064): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '?(id INT NOT NULL AUTO_INCREMENT, PRIMARY KEY(id))' at line 1 in ***/classes/db.mysql.class.php on line 69
Warning: mysqli::prepare() [mysqli.prepare]: (42000/1064): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '?)' at line 1 in ***/classes/db.mysql.class.php on line 75
在这个php代码调用上：
public function createTable($tableName) {
$this->connect();
if ($stmt = $this->dbSocket->prepare("CREATE TABLE ?(id INT NOT NULL AUTO_INCREMENT, PRIMARY KEY(id))")) {
$stmt->bind_param("s", $tableName);
$stmt->execute();
$stmt->close();
}
if ($stmt = $this->dbSocket->prepare("INSERT INTO sys_userTables(userTableName) VALUES (u_?)")) {
$stmt->bind_param("s", $tableName);
$stmt->execute();
$stmt->close();
}
$this->disonnect();
}
$tableName是字符串，并且传递正确。
connect（）方法为：
private function connect() {
$this->dbSocket = new mysqli($this->dbHost, $this->dbUser, $this->dbPassword, $this->dbDatabase);
if (mysqli_connect_errno()) {
printf("Brak połączenia z serwerem MySQL. Kod błędu: %s'n", mysqli_connect_error());
exit();
}
}
TIA。
不能将表名用作参数。
如果这样做的目的是创建几个具有相同结构但不同名称的表，我建议使用以下内容：
$table_names = array('a', 'b', 'c');
foreach($table_names as $name) {
$query = "CREATE TABLE `$name` (id INT NOT NULL AUTO_INCREMENT, PRIMARY KEY(id))";
// run query or add it to a collection to run later
// or append a ';' to the end of the string and do it with a multi_query
}