How to Concatenate table name with with a variable value in mySQL

Question

I'm trying to create a dynamic code that would ready from any table with the certain name but the difference between each table name is a number that is generated by a variable: for example :

//that's how I get my variable the value for example is = 3

$pid = $GLOBALS["localid"];

//the table name for example is tablename_3

$strTable = "tablename_" .$pid;

//here's how the query should look like

$query = "SELECT * FROM . $strTable . where .....;

I'm making a mistake somewhere but can't figure it out and would appreciate a little help please

That is not how you use SQL. Predefine your queries. Don't patch them together! — juergen d
– juergen d, Commented Nov 28, 2017 at 11:25
Change your query to this :- $query = "SELECT * FROM $strTable where ....."; — Yash Parekh
– Yash Parekh, Commented Nov 28, 2017 at 11:32
Validate the Global variable first and check the quotes, you don't need dots in the double quotes string for a variable. it takes variable as a value "SELECT * from $table_name where = $value"; — Cruzer
– Cruzer, Commented Nov 28, 2017 at 11:36

Sam Edward · Accepted Answer · 2017-12-08 15:28:05Z

3

Remove the dots and also make sure you have single quotes aroung where

$query = "SELECT * FROM $strTable where '.....';

answered Dec 8, 2017 at 15:28

Sam Edward

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DigiLive · Accepted Answer · 2017-11-28 11:32:11Z

1

Besides the comments about do or don't build your queries like this...

You're not closing the quotes properly.

$query = "SELECT * FROM . $strTable . where .....; //Double quote not closed.

should be:

$query = 'SELECT * FROM' . $strTable . 'where .....'; //Single quoted strings concatenated with variable.

or

$query = "SELECT * FROM $strTable where ....."; //Variable inside double quoted string.

answered Nov 28, 2017 at 11:32

DigiLive

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