I have a variable in php called as id. I want to use it as a table name.This code is not creating the table in php myadmin

Question

I have to use the $id variable as table name but it is not working.

       $sql1="CREATE TABLE $id(
              amt_to_be_paid INT(6),
              no_of_days_req INT(2),
              proposal TEXT NOT NULL,
              channel_link VARCHAR(100) NOT NULL,
              )";

spare comma at end of last column definition. voting to close as typo. — P.Salmon, May 01 '20 at 08:58

score 1 · Answer 1 · answered May 01 '20 at 23:27

Be sure that you have the full control on the $id variable and is not coming from user input.

You need to concatenate your $id variable to the query string, as following:

   $sql1="CREATE TABLE " . $id . "(
          amt_to_be_paid INT(6),
          no_of_days_req INT(2),
          proposal TEXT NOT NULL,
          channel_link VARCHAR(100) NOT NULL,
          )";

score 0 · Answer 2 · answered May 01 '20 at 14:11

0

Take a look at this answer for a detailed review of how to achieve this : How to include a PHP variable inside a MySQL statement

answered May 01 '20 at 14:11

mojo

988
6
17

I have a variable in php called as id. I want to use it as a table name.This code is not creating the table in php myadmin

2 Answers2