PHP quotes with a like query

Question

I have gotten myself into a tangled mess with quotes.

I am trying to do a like query that looks something like this

$myQuery = 'SELECT col1 , col2 from my_table WHERE col_value LIKE '%$my_string_variable%';

But this query would obviously give an error. What is the right syntax for this? I tried messing with double-quotes, but that didn't work either.

Answer 1

Combine double and single quotes.

$myQuery = "SELECT col1 , col2
            FROM my_table
            WHERE col_value LIKE '%$my_string_variable%'";

Although I prefer to protect my code against SQL Injection..

$myQuery = "SELECT col1 , col2
            FROM my_table
            WHERE col_value LIKE '%" . mysql_real_escape_string($my_string_variable) . "%'";

Answer 2

Surround the whole thing in double quotes.

$query = "SELECT col1 , col2 from my_table WHERE col_value LIKE '%$my_string_variable%'";