Question

This query works fine for me:

$query = "
         SELECT 
           p.topnode_id,
           p.param_key,
           p.param_value 
         FROM
           tbl_params p
         INNER JOIN
           tbl_clients c
         ON
           c.client_id = p.client_id
         WHERE
           p.client_id = ?
         ";

However, if I put AS in the query it throws me an error:

$query = "
         SELECT 
           p.topnode_id AS topnode_id,
           p.param_key AS key,
           p.param_value AS value
         FROM
           tbl_params p
         INNER JOIN
           tbl_clients c
         ON
           c.client_id = p.client_id
         WHERE
           p.client_id = ?
         ";

What seems to be the problem?

Answer 1

You need to escape key with backticks since it is a reserved word in Mysql

as `key`