datatables format column from multiple sources?(Server side processing)

Question

Im trying to get server side processing working with concatenated columns.

I came across this this post: Datatables - Server-side processing - DB column merging

But when I use that format I get SQL errors. But I also want to insert say... a space between the fields... is this possible?

Edit:

Example:

Table init:

var customer_Table = $('#customer_Table').DataTable( {
    "processing": true,
    "serverSide": true,
    "ajax": "php/server_processing.php",
    stateSave: true,
    "lengthMenu": [[10, 25, 50, -1], [10, 25, 50, "All"]],
    'order': 2, 'asc' ]],
    "columns": [{"title":"id","visible":false,"type":"html-string"},{"title":"name","visible":true,"type":"html-string"},[{"title":"address","visible":true,"type":"html-string"}
} );

Column scheme:

$columns = array(
    array( 'db' => 'id', 'dt' => "id" ),
    array( 'db' => 'name',  'dt' => "Name" ),
    array( 'db' => "`street` . ' ' . `city` . '<br>' . `postal` . ' ' . `country`" 'dt' => "address"
    )
);

Error:

{"error":"An SQL error occurred: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '`id`,`name`, `JCP`, ``street` . ' ' . `city` . '<br>' . `postal` . ' ' . `country``...' at line 1"}

Answer 1

Judging from the error message, the processing engine is expecting a valid column reference for 'db'. Each 'db' should refer to a valid column reference in the WHERE clause, and each 'dt' contains the label to be displayed for that column.

Your reference is, according to your snippet:

"`street` . ' ' . `city` . '<br>' . `postal` . ' ' . `country`"

Which doesn't mean much to most database engines I am familiar with. Try something like this:

$columns = array(
    array( 'db' => 'id', 'dt' => "id" ),
    array( 'db' => 'name',  'dt' => "Name" ),
    array( 'db' => "CONCAT(`street`, ' ', `city`, '<br>', `postal`, ' ', `country`)", 
            'dt' => "address")
);

Answer 2

$columns = array(
    array( 'db' => 'id', 'dt' => "id" ),
    array( 'db' => 'name',  'dt' => "Name" ),
    array( 'db' => "CONCAT(`street`, ' ', `city`, '<br>', `postal`, ' ', `country`)", 
            'as' => "address",
            'dt' => "address")
);