PHP PDO Create Table using $variable value as name

Question

2.50/5 (2 votes)

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Scenario:

TO crate a table we use the query as shown below

PHP

$query = 'CREATE TABLE xyz ' .
			'(Id varchar(30), Val varchar(30))';

Here the table name is "xyz". I know the table name so i was able to create it. In the next case i do not know the name of the table and the name is in the variable $TableName. How do i create the table now

PHP

$TableName = "xyz"; -->Calculated in run time
$query = 'CREATE TABLE '.$TableName.' ' .
			'(Id varchar(30), Val varchar(30))';

When i execute the table is not being created. Is this format correct?

Posted 3-Apr-13 6:06am

amarasat

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Comments

enhzflep 3-Apr-13 13:05pm

See here: Can Php Pdo statements accept the table name as parameter?

Yes, the format is correct.
Summary: Can't bind either a column name or a table name. You have to use string concatenation.

Prasad Khandekar 3-Apr-13 17:02pm

I am sure he is using string concatenation. ($query = 'CREATE TABLE ' . $TableName . ' ' . '(Id varchar(30), Val varchar(30))';

enhzflep 3-Apr-13 17:44pm

Me too. It's an odd question - examples of string concatenation yet nothing that is specific to pdo (at least, so it seems).
Even the strings that are concatenated could simply be written in a single expression.
For instance, I still cannot work out why one would write this

$query = 'CREATE TABLE xyz ' .
'(Id varchar(30), Val varchar(30))';

instead of

$tblName = 'someName';
$query = "CREATE TABLE $tblName (Id varchar(30), Val varchar(30))";

or

$tblName = 'someName';
$query = 'CREATE TABLE ' . $tblName . ' (Id varchar(30), Val varchar(30))';

or even

$tblName = 'someName';
$query = sprintf('CREATE TABLE %s (Id varchar(30), Val varchar(30))', $tblName);

I'd naturally have expected to see something like the following code. That is until I realized that prepared statements are but one of the advantages of Pdo - the other being the fact it works with a bunch of different DBs, i.e sqLite.

$params = array(':tableName' => 'xyz');
$query = $pdo->prepare('select * from :tableName');
$query->execute();

Prasad Khandekar 3-Apr-13 17:11pm

Hello Amarsat,

The problem might be in the table name generation. Have you tried running the second snippet in which the $TableName is set to value "xyz". Please also have a look at (http://www.w3cyberlearnings.com/PHP_MySQL_PDO_Create_Table)

Member 12880505 1-Dec-16 2:55am

Sir,can you please tell me how to take variable as column name in create query.That to in foreach loop

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amarasat · Accepted Answer · 2013-04-03T11:05:00

Solution 1

PHP

$query = "CREATE TABLE $TableName".
                 '(Id varchar(30), Val varchar(30))';

This has worked fine. Thanks for all your support enhzflep and Prasad

Posted 3-Apr-13 11:05am

amarasat

Updated 3-Apr-13 11:06am

v2