CODEWITHSUNDEEP
phpmysql
Brendan Armstrong
Brendan Armstrong

Reputation: 43

Failing to create mySQL table

I'm trying to create a table if it does not already exist in my database. For this I'm running this test which is working as intended:

$conn = mysql_connect("localhost", "twa222", "twa222bg");
mysql_select_db("airline222", $conn) or die ("Database not found " . mysql_error() );
$val = mysql_query("SELECT 1 from '$FLIGHTID'");

However my problem comes when I try to create the table itself, which is giving me the following error: Problem with query: 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 ''passenger' SMALLINT NOT NULL, 'booking' CHAR(6), 'seat' VARCHAR(3))' at line 2

This is the code that is attempting to generate the table

if(!$val)
{
    $sql = "CREATE TABLE ".$FLIGHTID." (
    passenger SMALLINT NOT NULL PRIMARY KEY,
    booking CHAR(6), seat VARCHAR(3) )";
    $rs = mysql_query($sql) or die ("Problem with query" . mysql_error());
}
mysql_close($conn);

I originally thought it was the ".$FLIGHTID." that was causing the problem but when I changed that to simply be ABC I still got the same error.

Can anyone see where I am going wrong?

EDIT: My SQL output when using ABC is:

CREATE TABLE ABC ( passenger SMALLINT NOT NULL PRIMARY KEY, booking CHAR(6), seat VARCHAR(3) )

Without using ABC it is:

CREATE TABLE ( passenger SMALLINT NOT NULL PRIMARY KEY, booking CHAR(6), seat VARCHAR(3) )

Upvotes: 4

Views: 71

Answers (1)

Jens
Jens

Reputation: 69440

You use single quotes arround column names what is not allowed. Single qoutes indicates that the value inside is a litaral:

Change:

$val = mysql_query("SELECT 1 from '$FLIGHTID'");

to:

$val = mysql_query("SELECT 1 from $FLIGHTID");

Use mysqli_*or PDOinstead of deprecated mysql_* API.

Upvotes: 4

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