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Create table where tablename is a variable

Posted on 2006-06-26
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Last Modified: 2008-03-10
I am trying to create a table where the tablename is a variable. When I do, it creates a table with the variable (it will look like $tablename) instead of the value that $tablename represents.

Example code:
$mysql = 'CREATE TABLE $tablename (id INT NOT NULL PRIMARY KEY AUTO_INCREMENT, col1 VARCHAR( 150 ) NOT NULL, col2 VARCHAR( 150 ) NOT NULL)';

I am sure the variable is working correctly.

I need this ASAP... thanks.
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Question by:cbonnock
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20 Comments
 
LVL 6

Accepted Solution

by:
soapergem earned 250 total points
ID: 16988685
Use double quotes instead of single quotes.

"CREATE TABLE $tablename...
0
 
LVL 37

Assisted Solution

by:Harisha M G
Harisha M G earned 250 total points
ID: 16988687
Hi, try changing single quote to double quote  !

$mysql = "CREATE TABLE $tablename (id INT NOT NULL PRIMARY KEY AUTO_INCREMENT, col1 VARCHAR( 150 ) NOT NULL, col2 VARCHAR( 150 ) NOT NULL)";


---
Harish
0
 
LVL 6

Expert Comment

by:soapergem
ID: 16988690
Alternatively, you can separate the variable from the rest of the string using the concatenation operator (aka the period):

'CREATE TABLE ' . $tablename . ' (id INT...
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LVL 37

Expert Comment

by:Harisha M G
ID: 16988694
cbonnock, the main difference between single and double quotes is that in the double quotes, variable substitution takes place, while in single quotes, it doesn't.

So, if you have

$str = "ABC";

$text1 = '$str';
$text2 = "$str";

text1 will have $str, and text2 will have ABC.
0
 

Author Comment

by:cbonnock
ID: 16988727
I have tried all of the following combinations with no improvement:

$mysql = 'CREATE TABLE $tablename (id INT NOT NULL PRIMARY KEY AUTO_INCREMENT, col1 VARCHAR( 150 ) NOT NULL, col2 VARCHAR( 150 ) NOT NULL)';

$mysql = "CREATE TABLE $tablename (id INT NOT NULL PRIMARY KEY AUTO_INCREMENT, col1 VARCHAR( 150 ) NOT NULL, col2 VARCHAR( 150 ) NOT NULL)";

$mysql = 'CREATE TABLE '.$tablename.' (id INT NOT NULL PRIMARY KEY AUTO_INCREMENT, col1 VARCHAR( 150 ) NOT NULL, col2 VARCHAR( 150 ) NOT NULL)';

$mysql = "CREATE TABLE ".$tablename." (id INT NOT NULL PRIMARY KEY AUTO_INCREMENT, col1 VARCHAR( 150 ) NOT NULL, col2 VARCHAR( 150 ) NOT NULL)";

$mysql = 'CREATE TABLE ".$tablename." (id INT NOT NULL PRIMARY KEY AUTO_INCREMENT, col1 VARCHAR( 150 ) NOT NULL, col2 VARCHAR( 150 ) NOT NULL)';

What else might work?
0
 
LVL 6

Expert Comment

by:soapergem
ID: 16988740
What errors are you getting? And did you remember to actually assign a value to $tablename first?
0
 
LVL 37

Expert Comment

by:Harisha M G
ID: 16988743
cbonnock, before executing the query, echo it..

$tablename = "something";

$mysql = "CREATE TABLE $tablename (id INT NOT NULL PRIMARY KEY AUTO_INCREMENT, col1 VARCHAR( 150 ) NOT NULL, col2 VARCHAR( 150 ) NOT NULL)";

echo $mysql;

mysql_query($mysql) or die(mysql_error());
0
 

Author Comment

by:cbonnock
ID: 16988775
OK... now I see the problem... the result of the variable is an email address... the @ symbol is giving the sql query a problem... is there a way I can still use the email with the @ as a table name? Otherwise it works fine with just double quotes around the whole thing.
0
 
LVL 37

Expert Comment

by:Harisha M G
ID: 16988781
cbonnock, try \@
0
 
LVL 37

Expert Comment

by:Harisha M G
ID: 16988787
cbonnock, or a better and prefered way is to use...

$tablename = "something";

$mysql = "CREATE TABLE $tablename (id INT NOT NULL PRIMARY KEY AUTO_INCREMENT, col1 VARCHAR( 150 ) NOT NULL, col2 VARCHAR( 150 ) NOT NULL)";

mysql_query(mysql_real_escape_string($mysql)) or die(mysql_error());
0
 
LVL 37

Expert Comment

by:Harisha M G
ID: 16988795
Or you may need to put the optional quotes for the table and field names...


$mysql = "CREATE TABLE `$tablename` (`id` INT NOT NULL PRIMARY KEY AUTO_INCREMENT, `col1` VARCHAR( 150 ) NOT NULL, `col2` VARCHAR( 150 ) NOT NULL)";
0
 

Author Comment

by:cbonnock
ID: 16988804
no... same problem.
0
 
LVL 7

Expert Comment

by:maUru
ID: 16988869
you cant make a table with an @ inside it
0
 
LVL 7

Expert Comment

by:maUru
ID: 16988886
try converting the @ symbol to AT

eg:

$tablename = str_replace('@', 'AT', $tablename);
$query = "CREATE $tablename etcetc.............."
0
 

Author Comment

by:cbonnock
ID: 16988887
I worked around it... thanks to all for the quick responses.
0
 
LVL 7

Expert Comment

by:maUru
ID: 16988910
cbonnock, why the HELL did you award the points to mgh_mgharish and not soapergem?!
0
 
LVL 6

Expert Comment

by:soapergem
ID: 16988916
I was just wondering the same thing.
0
 

Author Comment

by:cbonnock
ID: 16989229
My apologies... with all of the other questions I had, I didn't even realize you said the same thing... i would have split the points had I noticed... is there anything i can do about that now?
0
 

Author Comment

by:cbonnock
ID: 16991426
Thanks again.
0

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