请帮助:)。我发现了这个错误:
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
关于这个php代码调用:
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是字符串并正确传递。
connect()方法是:
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。
答案 0 :(得分:2)
您不能将表名用作参数。
如果要点是创建几个结构相同但名称不同的表,我建议使用类似的东西:
$table_names = array('a', 'b', 'c');
foreach($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
}