mysqli ::准备SQL错误

问题描述：

Help please :). I'm gettig this error:

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

on this php code call:

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 is string and is passed correctly.

connect() method is:

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
", mysqli_connect_error());
        exit();
    }
}

TIA.

请帮助:)。我发现这个错误： p>

 警告：mysqli :: prepare（）[mysqli.prepare] :( 42000/1064）：你的SQL语法有错误; 检查与MySQL服务器版本对应的手册，以便在*** / classes / db.mysql.class.php的第1行'？（id INT NOT NULL AUTO_INCREMENT，PRIMARY KEY（id））'附近使用正确的语法 在第69行
 
警告：mysqli :: prepare（）[mysqli.prepare] :( 42000/1064）：您的SQL语法有错误; 检查与您的MySQL服务器版本相对应的手册，以便在第75行的*** / classes / db.mysql.class.php第1行“？”附近使用正确的语法
  code>  pre  > 
 
 关于这个php代码调用： p> 
 
 
  public function createTable（$ tableName）{
 
 $ this-＆gt; connect（）; \  n 
 if（$ stmt = $ this-＆gt; dbSocket-＆gt; prepare（“CREATE TABLE？（id INT NOT NULL AUTO_INCREMENT，PRIMARY KEY（id））”））{
 $ stmt-＆gt; bind_param（“  s“，$ tableName）; 
 $ stmt-＆gt; execute（）; 
 $ stmt-＆gt; close（）; 
} 
 
 if（$ stmt = $ this-＆gt; dbSocket-＆gt;  prepare（“INSERT INTO sys_userTables（userTableName）VALUES（u_？）”））{
 $ stmt-＆gt; bind_param（“s”，$ tableName）; 
 $ stmt-＆gt; execute（）; 
 $ stmt  - ＆gt; close（）; 
} 
 
 $ this-＆gt; disonnect（）; 
} 
  code>  pre> 
 
  $ tableName是字符串并传递 正确。 p> 
 
 
 connect（）方法是： p> 
 
 
  private function connect（）{
 $ this-＆gt; dbSocket = new mysqli  （$ this-＆gt; dbHost，$ this-＆gt; dbUser，$ thi  s-＆gt; dbPassword，$ this-＆gt; dbDatabase）; 
 if（mysqli_connect_errno（））{
 printf（“Brakpołączeniazserwerem MySQL。  Kodbłędu：％s 
“，mysqli_connect_error（））; 
 exit（）; 
} 
} 
  code>  pre> 
 
  TIA。 p> \  n  div>

答

You cannot use a table name as a parameter.

If the point of this is to create several tables with the same structure but different name, I suggest using something like:

$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
}

mysqli ::准备SQL错误

相关推荐