Selecting MYSQL rows with same field names and adding a prefix

Question

I\'m trying to make a mysql query to select several tables and LEFT join them, however they all have same columns names \'user\' etc. I want to rename all the fields in this man

Answer 1

You cannot supply a shorthand to alias columns you must do it explicitly for each column name. In general anyway, it is typically recommended to name all columns explicitly in the SELECT list rather than using SELECT *, since it allows you to deterministically specify the column order, and protects you against accidentally pulling in a large BLOB later on if one ever gets added to the table ( or any other schema changes ).

SELECT
  mod_backup_accounts.user AS account_user,
  mod_backup_subscriptions.user AS subscription_user,
  ...
  ...
FROM
  mod_backup_accounts
  LEFT JOIN `mod_backup_subscriptions` ON `mod_backup_accounts`.subscription_id = `mod_backup_subscriptions`.package_id

Answer 2

I totally understand your problem about duplicated field names.

I needed that too until I coded my own function to solve it. If you are using PHP you can use it, or code yours in the language you are using for if you have this following facilities.

The trick here is that mysql_field_table() returns the table name and mysql_field_name() the field for each row in the result if it's got with mysql_num_fields() so you can mix them in a new array.

You can also modify the function to only add the "column." prefix when the field name is duplicated.

Regards,

function mysql_rows_with_columns($query) {
    $result = mysql_query($query);
    if (!$result) return false; // mysql_error() could be used outside
    $fields = mysql_num_fields($result);
    $rows = array();
    while ($row = mysql_fetch_row($result)) { 
        $newRow = array();
        for ($i=0; $i<$fields; $i++) {
            $table = mysql_field_table($result, $i);
            $name = mysql_field_name($result, $i);
            $newRow[$table . "." . $name] = $row[$i];
        }
        $rows[] = $newRow;
    }
    mysql_free_result($result);
    return $rows;
}