CREATE TABLE $var help

[r00tk1LL]

Here's a simple syntax question, how would I use a variable for a table name in this query?

$table = 'CREATE TABLE `$var` //How should I use this variable??
          ('
        . ' `id` INT(15) NOT NULL AUTO_INCREMENT PRIMARY KEY, '
        . ' `pics` TEXT NOT NULL, '
        . ' `thumbs` TEXT NOT NULL'
        . ' )'
        . ' ENGINE = myisam;';

 

Thanks

[marcus]

Try putting the table variable inside double quotes.

[Fadion]

$var = 'tableName';
$table = "CREATE TABLE $var";

 

Inside single quotes it interprets $var as a string so it prings '$var'.

[r00tk1LL]

Im trying it like this:

$table = "CREATE TABLE $var ('
        . ' `id` INT(15) NOT NULL AUTO_INCREMENT PRIMARY KEY, '
        . ' `pics` TEXT NOT NULL, '
        . ' `thumbs` TEXT NOT NULL'
        . ' )'
        . ' ENGINE = myisam;";

and it doesn't seem to work??

[r00tk1LL]

I also tried to do this:

 

'CREATE TABLE "$var"';

but no success..

 

Any ideas?

[Fadion]

A mysql query is constructed this way:

 

$var = 'yourTable'
$query = "CREATE TABLE $var";
$results = @mysql_query($query) or die(mysql_error());

 

If it isnt working, then ure not using mysql_query() (???) or uve got something in assigning your $var.

[r00tk1LL]

Is there a way to see what error this is giving me?

 

My variable is

$uname

 

and print $uname; displays the right value, I just cant see what the problem is with sql.

[r00tk1LL]

OK here's what I discovered if I run this with single quotes:

$table = 'CREATE TABLE `$uname` ('
        . ' `id` INT(15) NOT NULL AUTO_INCREMENT PRIMARY KEY, '
        . ' `pics` TEXT NOT NULL, '
        . ' `thumbs` TEXT NOT NULL'
        . ' )'
        . ' ENGINE = myisam;';

It makes the table, but calls is $uname, the variables name instead of value.

 

If i run this with double quotes:

$table = "CREATE TABLE `$uname` ('
        . ' `id` INT(15) NOT NULL AUTO_INCREMENT PRIMARY KEY, '
        . ' `pics` TEXT NOT NULL, '
        . ' `thumbs` TEXT NOT NULL'
        . ' )'
        . ' ENGINE = myisam;";

It finds the variables value, but fails to make the table!

 

PLEASE HELP

 

[bruckerrlb]

$table = "CREATE TABLE `$uname` ('
        . ' `id` INT(15) NOT NULL AUTO_INCREMENT PRIMARY KEY, '
        . ' `pics` TEXT NOT NULL, '
        . ' `thumbs` TEXT NOT NULL'
        . ' )'
        . ' ENGINE = myisam;";

$results = @mysql_query($table) or die(mysql_error());

I think this would work

[r00tk1LL]

it says:

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(15) NOT NULL AUTO_INCREMENT PRIMARY KEY, ' . ' `p' at line 1

[r00tk1LL]

So the problem looks like MYSQL wants the create table in single quotes, but to get the variable value it has to be in double quotes, how can I get around this?

[bruckerrlb]

$table = "CREATE TABLE '$uname' (
       'id' INT(15) NOT NULL AUTO_INCREMENT PRIMARY KEY, 
       'pics' TEXT NOT NULL, 
       'thumbs' TEXT NOT NULL
       )'
       ENGINE = myisam;";

$results = @mysql_query($table) or die(mysql_error());

 

I think this may work better, I usually use phpmyadmin to create tables, but I know this option is not always available

[r00tk1LL]

It says:

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(15) NOT NULL AUTO_INCREMENT PRIMARY KEY, ' . ' `p' at line 1