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Home/ Questions/Q 9200323
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T22:44:27+00:00 2026-06-17T22:44:27+00:00

For example this works fine: $dropTable = $dbConnection->prepare(‘DROP TABLE IF EXISTS announcements’); $dropTable->execute(); $createTable

  • 0

For example this works fine:

$dropTable = $dbConnection->prepare('DROP TABLE IF EXISTS announcements');
$dropTable->execute();

$createTable = $dbConnection->prepare('CREATE TABLE announcements(
            id MEDIUMINT NOT NULL AUTO_INCREMENT,
            announcements TEXT NOT NULL,
            PRIMARY KEY (id))');
$createTable->execute();

But this fails:

$dropTable = $dbConnection->prepare('DROP TABLE IF EXISTS :tableToDrop');
$dropTable->bindParam(':tableToDrop', $_GET['table']);
$dropTable->execute();

$createTable = $dbConnection->prepare('CREATE TABLE announcements(
        id MEDIUMINT NOT NULL AUTO_INCREMENT,
            announcements TEXT NOT NULL,
            PRIMARY KEY (id))');
$createTable->execute();

With Error:

Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42000]:   
  Syntax error or access violation: 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 xxxx/createTables.php:9  
Stack trace: #0 xxxxx/createTables.php(9):   
  PDO->prepare('DROP TABLE IF E...') #1 {main} thrown in xxxx/createTables.php on line 9

I’m sure it’s something trivial, but I’ve been going at it for hours. Cheers.

Edit: Turns out you can’t bindParam with a table name.Is there anyway to do a secure prepared statement with a dynamic table name?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T22:44:28+00:00Added an answer on June 17, 2026 at 10:44 pm

    Safe way (and u should use that, event if what u wrote could work):

    $t=array('t1'=>'t1','t2'=>'t2'....'tn'=>'tn');
    
    $sql = "drop table {$t[$_GET['table']]} ..."
    
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