Remove the single quote around the WORD. It causes the column name to be converted as string.

SELECT word, SUM(amount) 
FROM Data 
Group By word

As Willem Renzema said, you've misunderstood how GROUP BY works. Since it doesn't seem like you've understood what he said, let me try saying it a little differently.

GROUP BY, logically enough, is used to group together rows from your result set. Normally you provide a list of the columns to use to group your rows together. GROUP BY sch.id, cv.member_id tells SQL to identify the unique sets of values for those two columns, and to group the rows in the result set by those values. In your case, there are two unique value pairs for those two values:

• cv.member_id = 82, sch.id = 17101
• cv.member_id = 82, sch.id = 17153

So, you'll get two groups of rows - three that have the first pair of values, and one that has the second pair.

Adding additional columns to a GROUP BY clause will never result in fewer groups - either the new column(s) are the same in all rows (in which case you have the same number of groups), or the new column(s) have different values form some rows in one or more of hour original groups (in which case, you'll now have more groups).

Also (as pointed out by Willem), you've got a syntax error. The columns in a GROUP BY list are separated by commas. In your GROUP BY sch.id AND cv.member_id, you're grouping by a calculation: sch.id AND cv.member_id, or the result of treating both sch.id and cv.member_id as if they were Boolean values. Since neither is 0, when converted to Booleans both evaluate to 1 (true), and the combination (true AND true) is true. So, you wind up with just one group, of 4 rows.

Let's step back, and consider what (it looks like) you're actually trying to do. For a given member_id, you want the total time they're involved in activities of the types "Booked" or "Present".

Note that the total time is calculated out of the schedule_event table. Also, note that a given member_id can be associated with the same schedule_event more than once. So, to get total time, we need to identify the distinct schedule_event rows that our member_id is tied to, and sum the time for those unique values.

That being the case, the simplest way to proceed is to use a sub-query to get the list of distinct schedule_events our member_id is tied to, and then sum the total times for those distinct events.

Here's a query that will do just that:

SELECT `member_id`
      ,SUM(`totalTime`) as `totalTime`
  FROM (
        SELECT DISTINCT
            cv.member_id AS `member_id`,
            sch.id AS `scheduleId`,
            TIMESTAMPDIFF(SECOND, sch.starts_at, sch.ends_at) AS `totalTime`
        FROM 
            `schedule_event` AS `sch`
            INNER JOIN `client_visit` AS `cv` ON cv.schedule_event_id = sch.id
            INNER JOIN `type_of_visit` AS `tov` ON tov.id = cv.status_type_id
        WHERE 
            (tov.type = 'TYPE_BOOKED' OR tov.type = 'TYPE_PRESENT') 
            AND cv.member_id = 82
       ) sq
 GROUP BY `member_id`;

The subquery (imaginatively labeled sq) is basically your original query. I changed your LEFT JOIN to an INNER JOIN, as we must have a client_visit record to identify both the member_id, and the type of visit. However, I removed the SUM on totalTime; at this point, we just want to know the time each schedule_event will take. I also added DISTINCT - we don't care how many time this schedule_event appears with this member_id; the total time will be the same whether it shows up once, three times, or 207 times.

Once we've identified the schedule_event data that our member_id is connected to, then we want the total time for all those schedule_event rows. So, we take the results of the sub-query, group them by member_id (in case it would ever be necessary to pull this back for multiple member_id values), and sum up the calculated times for each schedule_event row.

Since joanolo had goen to the trouble to set up a dbfiddle for your problem, I took his work and added this query at the end, so you could see the results were what you wanted; the updated dbfiddle link is here.

I hope this helps clarify how GROUP BY actually works for you.

Cannot you can simply add a sum() to the query?

SELECT user_id,
       count(*) total,
       sum(price) price
       FROM ad
       GROUP BY user_id;