Mysql query multiple select statements output on multiple columns









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2
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Table



+------+-------+------------------+
|CLIENT| VALUE | DATETIME |
+------+-------+------------------+
| A | 1 | 2018-11-10 09:00 |
| B | 1 | 2018-11-10 09:00 |
| C | 1 | 2018-11-10 09:00 |
| D | 2 | 2018-11-10 08:00 |
| E | 2 | 2018-11-10 08:00 |
| F | 3 | 2018-11-10 08:00 |
| A | 1 | 2018-11-10 07:00 |
| B | 2 | 2018-11-10 07:00 |
| C | 2 | 2018-11-10 07:00 |
| D | 3 | 2018-11-10 06:00 |
| E | 1 | 2018-11-10 06:00 |
| F | 2 | 2018-11-10 06:00 |
| A | 1 | 2018-11-08 08:00 |
| B | 2 | 2018-11-08 08:00 |
| C | 2 | 2018-11-08 08:00 |
| D | 1 | 2018-11-08 08:00 |
| E | 1 | 2018-11-08 07:00 |
| F | 2 | 2018-11-08 07:00 |


I'm newbie to mysql, and I'm in trouble with this query.



I have only one table named "table" with three columns.



This table records many data every day at different time from a specific set of client A,B,C,D,E,F



With one query I need to create new table with one row for each client and with the following 4 columns:



  1. first column should contain the newest value recordered in the table for each client

  2. second column should contain the percentage of time the value is equal to 1 for each client during the last 24 hours

  3. third column should contain the percentage of time the value is equal to 1 for each client during the last 7 days

  4. as previous column but during the last 30 days

I hope someone can help me.



What I would like to receive:



+------+-------------+-----------+--------------+--------------+
|CLIENT| NEWEST VALUE| LAST 24 H | LAST 7 DAYS | LAST 30 DAYS |
+------+-------------+-----------+--------------+--------------+
| A | 1 | 100% | 100% | ... |
| B | 1 | 50% | 66% | ... |
| C | 1 | 50% | 33% | ... |
| D | 2 | 0% | 33% | ... |
| E | 2 | 50% | 66% | ... |
| F | 3 | 0% | 0% | ... |


This piece of code works fine to create the "NEWST VALUE" column



SELECT
client,
value,
max(datetime)
FROM
table
GROUP BY
client;


and this one create the "LAST 24 H" column



SELECT
client,
count(if(value = 1,1, null))/count(value),
FROM
table
WHERE
date(datetime) < CURRENT_DATE() - interval 1 day
GROUP BY
repository_name;


but I'm not able to put all the output together in one new table










share|improve this question























  • If the query is still not resolved, can you please provide a db-fiddle.com OR, Create Table and Insert Into statements would be helpful, so that we can reproduce the case
    – Madhur Bhaiya
    Nov 10 at 17:57














up vote
2
down vote

favorite












Table



+------+-------+------------------+
|CLIENT| VALUE | DATETIME |
+------+-------+------------------+
| A | 1 | 2018-11-10 09:00 |
| B | 1 | 2018-11-10 09:00 |
| C | 1 | 2018-11-10 09:00 |
| D | 2 | 2018-11-10 08:00 |
| E | 2 | 2018-11-10 08:00 |
| F | 3 | 2018-11-10 08:00 |
| A | 1 | 2018-11-10 07:00 |
| B | 2 | 2018-11-10 07:00 |
| C | 2 | 2018-11-10 07:00 |
| D | 3 | 2018-11-10 06:00 |
| E | 1 | 2018-11-10 06:00 |
| F | 2 | 2018-11-10 06:00 |
| A | 1 | 2018-11-08 08:00 |
| B | 2 | 2018-11-08 08:00 |
| C | 2 | 2018-11-08 08:00 |
| D | 1 | 2018-11-08 08:00 |
| E | 1 | 2018-11-08 07:00 |
| F | 2 | 2018-11-08 07:00 |


I'm newbie to mysql, and I'm in trouble with this query.



I have only one table named "table" with three columns.



This table records many data every day at different time from a specific set of client A,B,C,D,E,F



With one query I need to create new table with one row for each client and with the following 4 columns:



  1. first column should contain the newest value recordered in the table for each client

  2. second column should contain the percentage of time the value is equal to 1 for each client during the last 24 hours

  3. third column should contain the percentage of time the value is equal to 1 for each client during the last 7 days

  4. as previous column but during the last 30 days

I hope someone can help me.



What I would like to receive:



+------+-------------+-----------+--------------+--------------+
|CLIENT| NEWEST VALUE| LAST 24 H | LAST 7 DAYS | LAST 30 DAYS |
+------+-------------+-----------+--------------+--------------+
| A | 1 | 100% | 100% | ... |
| B | 1 | 50% | 66% | ... |
| C | 1 | 50% | 33% | ... |
| D | 2 | 0% | 33% | ... |
| E | 2 | 50% | 66% | ... |
| F | 3 | 0% | 0% | ... |


This piece of code works fine to create the "NEWST VALUE" column



SELECT
client,
value,
max(datetime)
FROM
table
GROUP BY
client;


and this one create the "LAST 24 H" column



SELECT
client,
count(if(value = 1,1, null))/count(value),
FROM
table
WHERE
date(datetime) < CURRENT_DATE() - interval 1 day
GROUP BY
repository_name;


but I'm not able to put all the output together in one new table










share|improve this question























  • If the query is still not resolved, can you please provide a db-fiddle.com OR, Create Table and Insert Into statements would be helpful, so that we can reproduce the case
    – Madhur Bhaiya
    Nov 10 at 17:57












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Table



+------+-------+------------------+
|CLIENT| VALUE | DATETIME |
+------+-------+------------------+
| A | 1 | 2018-11-10 09:00 |
| B | 1 | 2018-11-10 09:00 |
| C | 1 | 2018-11-10 09:00 |
| D | 2 | 2018-11-10 08:00 |
| E | 2 | 2018-11-10 08:00 |
| F | 3 | 2018-11-10 08:00 |
| A | 1 | 2018-11-10 07:00 |
| B | 2 | 2018-11-10 07:00 |
| C | 2 | 2018-11-10 07:00 |
| D | 3 | 2018-11-10 06:00 |
| E | 1 | 2018-11-10 06:00 |
| F | 2 | 2018-11-10 06:00 |
| A | 1 | 2018-11-08 08:00 |
| B | 2 | 2018-11-08 08:00 |
| C | 2 | 2018-11-08 08:00 |
| D | 1 | 2018-11-08 08:00 |
| E | 1 | 2018-11-08 07:00 |
| F | 2 | 2018-11-08 07:00 |


I'm newbie to mysql, and I'm in trouble with this query.



I have only one table named "table" with three columns.



This table records many data every day at different time from a specific set of client A,B,C,D,E,F



With one query I need to create new table with one row for each client and with the following 4 columns:



  1. first column should contain the newest value recordered in the table for each client

  2. second column should contain the percentage of time the value is equal to 1 for each client during the last 24 hours

  3. third column should contain the percentage of time the value is equal to 1 for each client during the last 7 days

  4. as previous column but during the last 30 days

I hope someone can help me.



What I would like to receive:



+------+-------------+-----------+--------------+--------------+
|CLIENT| NEWEST VALUE| LAST 24 H | LAST 7 DAYS | LAST 30 DAYS |
+------+-------------+-----------+--------------+--------------+
| A | 1 | 100% | 100% | ... |
| B | 1 | 50% | 66% | ... |
| C | 1 | 50% | 33% | ... |
| D | 2 | 0% | 33% | ... |
| E | 2 | 50% | 66% | ... |
| F | 3 | 0% | 0% | ... |


This piece of code works fine to create the "NEWST VALUE" column



SELECT
client,
value,
max(datetime)
FROM
table
GROUP BY
client;


and this one create the "LAST 24 H" column



SELECT
client,
count(if(value = 1,1, null))/count(value),
FROM
table
WHERE
date(datetime) < CURRENT_DATE() - interval 1 day
GROUP BY
repository_name;


but I'm not able to put all the output together in one new table










share|improve this question















Table



+------+-------+------------------+
|CLIENT| VALUE | DATETIME |
+------+-------+------------------+
| A | 1 | 2018-11-10 09:00 |
| B | 1 | 2018-11-10 09:00 |
| C | 1 | 2018-11-10 09:00 |
| D | 2 | 2018-11-10 08:00 |
| E | 2 | 2018-11-10 08:00 |
| F | 3 | 2018-11-10 08:00 |
| A | 1 | 2018-11-10 07:00 |
| B | 2 | 2018-11-10 07:00 |
| C | 2 | 2018-11-10 07:00 |
| D | 3 | 2018-11-10 06:00 |
| E | 1 | 2018-11-10 06:00 |
| F | 2 | 2018-11-10 06:00 |
| A | 1 | 2018-11-08 08:00 |
| B | 2 | 2018-11-08 08:00 |
| C | 2 | 2018-11-08 08:00 |
| D | 1 | 2018-11-08 08:00 |
| E | 1 | 2018-11-08 07:00 |
| F | 2 | 2018-11-08 07:00 |


I'm newbie to mysql, and I'm in trouble with this query.



I have only one table named "table" with three columns.



This table records many data every day at different time from a specific set of client A,B,C,D,E,F



With one query I need to create new table with one row for each client and with the following 4 columns:



  1. first column should contain the newest value recordered in the table for each client

  2. second column should contain the percentage of time the value is equal to 1 for each client during the last 24 hours

  3. third column should contain the percentage of time the value is equal to 1 for each client during the last 7 days

  4. as previous column but during the last 30 days

I hope someone can help me.



What I would like to receive:



+------+-------------+-----------+--------------+--------------+
|CLIENT| NEWEST VALUE| LAST 24 H | LAST 7 DAYS | LAST 30 DAYS |
+------+-------------+-----------+--------------+--------------+
| A | 1 | 100% | 100% | ... |
| B | 1 | 50% | 66% | ... |
| C | 1 | 50% | 33% | ... |
| D | 2 | 0% | 33% | ... |
| E | 2 | 50% | 66% | ... |
| F | 3 | 0% | 0% | ... |


This piece of code works fine to create the "NEWST VALUE" column



SELECT
client,
value,
max(datetime)
FROM
table
GROUP BY
client;


and this one create the "LAST 24 H" column



SELECT
client,
count(if(value = 1,1, null))/count(value),
FROM
table
WHERE
date(datetime) < CURRENT_DATE() - interval 1 day
GROUP BY
repository_name;


but I'm not able to put all the output together in one new table







mysql sql mariadb multiple-columns multiple-select






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 days ago









marc_s

565k12610911243




565k12610911243










asked Nov 10 at 16:06









Luca Folin

132




132











  • If the query is still not resolved, can you please provide a db-fiddle.com OR, Create Table and Insert Into statements would be helpful, so that we can reproduce the case
    – Madhur Bhaiya
    Nov 10 at 17:57
















  • If the query is still not resolved, can you please provide a db-fiddle.com OR, Create Table and Insert Into statements would be helpful, so that we can reproduce the case
    – Madhur Bhaiya
    Nov 10 at 17:57















If the query is still not resolved, can you please provide a db-fiddle.com OR, Create Table and Insert Into statements would be helpful, so that we can reproduce the case
– Madhur Bhaiya
Nov 10 at 17:57




If the query is still not resolved, can you please provide a db-fiddle.com OR, Create Table and Insert Into statements would be helpful, so that we can reproduce the case
– Madhur Bhaiya
Nov 10 at 17:57












1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










You can use conditional aggregation. Assuming pre-8.0 MySQL, only the most recent value is really tricky. Here is one approach:



select t.client,
max(case when t.datetime = c.maxdt then t.value end) as most_recent_value,
avg(case when t.datetime >= now() - interval 1 day
then (t.value = 1)
end) as last_day_percentage,
avg(case when t.datetime >= now() - interval 7 day
then (t.value = 1)
end) as last_7day_percentage,
avg(case when t.datetime >= now() - interval 30 day
then (value = 1)
end) as last_30day_percentage
from t join
(select t.client, max(t.datetime) as maxdt
from t
group by t.client
) c
on c.client = t.client
group by t.client;


Note that this logic uses a MySQL extension where boolean values are treated as numbers in a numeric context, with 1 for true and 0 for false.



The average produces "0" or "1" for the time period in question, with NULL values for any other record. The avg() function ignores NULL values.






share|improve this answer




















  • thk you so much this solve my problem
    – Luca Folin
    Nov 11 at 9:21










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










You can use conditional aggregation. Assuming pre-8.0 MySQL, only the most recent value is really tricky. Here is one approach:



select t.client,
max(case when t.datetime = c.maxdt then t.value end) as most_recent_value,
avg(case when t.datetime >= now() - interval 1 day
then (t.value = 1)
end) as last_day_percentage,
avg(case when t.datetime >= now() - interval 7 day
then (t.value = 1)
end) as last_7day_percentage,
avg(case when t.datetime >= now() - interval 30 day
then (value = 1)
end) as last_30day_percentage
from t join
(select t.client, max(t.datetime) as maxdt
from t
group by t.client
) c
on c.client = t.client
group by t.client;


Note that this logic uses a MySQL extension where boolean values are treated as numbers in a numeric context, with 1 for true and 0 for false.



The average produces "0" or "1" for the time period in question, with NULL values for any other record. The avg() function ignores NULL values.






share|improve this answer




















  • thk you so much this solve my problem
    – Luca Folin
    Nov 11 at 9:21














up vote
0
down vote



accepted










You can use conditional aggregation. Assuming pre-8.0 MySQL, only the most recent value is really tricky. Here is one approach:



select t.client,
max(case when t.datetime = c.maxdt then t.value end) as most_recent_value,
avg(case when t.datetime >= now() - interval 1 day
then (t.value = 1)
end) as last_day_percentage,
avg(case when t.datetime >= now() - interval 7 day
then (t.value = 1)
end) as last_7day_percentage,
avg(case when t.datetime >= now() - interval 30 day
then (value = 1)
end) as last_30day_percentage
from t join
(select t.client, max(t.datetime) as maxdt
from t
group by t.client
) c
on c.client = t.client
group by t.client;


Note that this logic uses a MySQL extension where boolean values are treated as numbers in a numeric context, with 1 for true and 0 for false.



The average produces "0" or "1" for the time period in question, with NULL values for any other record. The avg() function ignores NULL values.






share|improve this answer




















  • thk you so much this solve my problem
    – Luca Folin
    Nov 11 at 9:21












up vote
0
down vote



accepted







up vote
0
down vote



accepted






You can use conditional aggregation. Assuming pre-8.0 MySQL, only the most recent value is really tricky. Here is one approach:



select t.client,
max(case when t.datetime = c.maxdt then t.value end) as most_recent_value,
avg(case when t.datetime >= now() - interval 1 day
then (t.value = 1)
end) as last_day_percentage,
avg(case when t.datetime >= now() - interval 7 day
then (t.value = 1)
end) as last_7day_percentage,
avg(case when t.datetime >= now() - interval 30 day
then (value = 1)
end) as last_30day_percentage
from t join
(select t.client, max(t.datetime) as maxdt
from t
group by t.client
) c
on c.client = t.client
group by t.client;


Note that this logic uses a MySQL extension where boolean values are treated as numbers in a numeric context, with 1 for true and 0 for false.



The average produces "0" or "1" for the time period in question, with NULL values for any other record. The avg() function ignores NULL values.






share|improve this answer












You can use conditional aggregation. Assuming pre-8.0 MySQL, only the most recent value is really tricky. Here is one approach:



select t.client,
max(case when t.datetime = c.maxdt then t.value end) as most_recent_value,
avg(case when t.datetime >= now() - interval 1 day
then (t.value = 1)
end) as last_day_percentage,
avg(case when t.datetime >= now() - interval 7 day
then (t.value = 1)
end) as last_7day_percentage,
avg(case when t.datetime >= now() - interval 30 day
then (value = 1)
end) as last_30day_percentage
from t join
(select t.client, max(t.datetime) as maxdt
from t
group by t.client
) c
on c.client = t.client
group by t.client;


Note that this logic uses a MySQL extension where boolean values are treated as numbers in a numeric context, with 1 for true and 0 for false.



The average produces "0" or "1" for the time period in question, with NULL values for any other record. The avg() function ignores NULL values.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 10 at 16:11









Gordon Linoff

743k32285390




743k32285390











  • thk you so much this solve my problem
    – Luca Folin
    Nov 11 at 9:21
















  • thk you so much this solve my problem
    – Luca Folin
    Nov 11 at 9:21















thk you so much this solve my problem
– Luca Folin
Nov 11 at 9:21




thk you so much this solve my problem
– Luca Folin
Nov 11 at 9:21

















 

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