mysqli->prepare issue with creating tables

Question

I'm trying to create a table with mysqli->prepare, but get an error while preparing the statement. My statement is as follows:

$mysqli->prepare("CREATE TABLE ? (id INT AUTO_INCREMENT NOT NULL, PRIMARY KEY (id))");

Then I just bind the table name to it

$mysqli->bind_param("s", $some_name);

And execute it.

$mysqli->execute();

The error I get is:

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 AUTO_INCREMENT NOT NULL, PRIMARY KE' at line 1

Can you help me?

what is the value of $some_name ? – Jean-François Savard Feb 01 '15 at 18:27 — Jean-François Savard, Feb 01 '15 at 18:27
just a string like "table1" – lukas293 Feb 01 '15 at 18:34 — lukas293, Feb 01 '15 at 18:34

score 2 · Answer 1 · answered Feb 01 '15 at 18:29

2

See http://php.net/manual/en/mysqli-stmt.prepare.php

In general, parameters are legal only in Data Manipulation Language (DML) statements, and not in Data Definition Language (DDL) statements.

answered Feb 01 '15 at 18:29

nlu

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Could i do this with pdo prepared statements? – lukas293 Feb 01 '15 at 18:35
http://stackoverflow.com/questions/3665784/php-pdo-prepared-statements-dont-work-when-creating-a-table – nlu Feb 01 '15 at 18:52

score 0 · Answer 2 · edited May 23 '17 at 11:43

0

This seems to have been answered before. Can I parameterize the table name in a prepared statement? You can't bind the table name

edited May 23 '17 at 11:43

Community

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answered Feb 01 '15 at 18:34

Zoltar

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mysqli->prepare issue with creating tables

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