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Home/ Questions/Q 8066253
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Editorial Team
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Editorial Team
Asked: June 5, 20262026-06-05T11:55:34+00:00 2026-06-05T11:55:34+00:00

I have a table ________________________________________________________________________________ | Message | |_______________________________________________________________________________| | ID(INT) | Text(TEXT) |read(TINYINT(0/1))

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I have a table

________________________________________________________________________________
|                         Message                                               |
|_______________________________________________________________________________|
| ID(INT) |  Text(TEXT) |read(TINYINT(0/1)) |deleted(TINYINT(0/1))|  User_id    |
| 1       | How Are You?|      0            |        0            |    6        |
| 2       | Fine        |      0            |        1            |    4        |
| 3       | Message 3   |      1            |        0            |    6        |
| 4       | Message 4   |      0            |        1            |    6        |
| 5       | Message 5   |      1            |        0            |    5        |
|_________|_____________|___________________|_____________________|_____________|

Now I want to select the message where user_id=6 and also select the column count for read=0 and read=1 seperately.I know this can be done by group by command and iam currently doing it with two sql queries .May somebody join them in one

select message,count(*) from message where User_id=6 and read=0 group by id;//for unread message
select message,count(*) from message where User_id=6 and read=1 group by id;//for read message
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  1. Editorial Team
    Editorial Team
    2026-06-05T11:55:35+00:00Added an answer on June 5, 2026 at 11:55 am

    Why don’t you

    select
      message,
      sum(read) as num_read,
      sum(case read when 0 then 1 else 0 end) as num_unread
    from message
    where user_id = 6
    group by user_id
    

    ? You can drop either the where or the group by line, depending on your real need and SQL implementation.

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