mysqli ::准备SQL错误

小编典典

mysqli ::准备SQL错误

sql

请帮忙：）。我gettig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

在此php代码调用中：

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，可以正确传递。

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 po艂膮czenia z serwerem MySQL. Kod b艂臋du: %s\n", mysqli_connect_error());
        exit();
    }
}

TIA。

阅读 164

2021-05-16

共1个答案

小编典典

您不能将表名称用作参数。

如果这样做的目的是创建具有相同结构但名称不同的多个表，我建议使用类似以下内容的表：

$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
}

2021-05-16