I want a query to get LIST of averages of users in postgres
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
0
I want to get the list of averages from a table who has the list of users
Not single average but a list
select avg(no_of_interactions) from table name;
where I want to get the list of avg interactions of user1,user2,user3 so on
it is giving me only one average
sql postgresql
edited Nov 16 '18 at 11:28
Kaushik Nayak
21.5k41332
asked Nov 16 '18 at 11:26
abdulabdul
176
add a comment |
0
I want to get the list of averages from a table who has the list of users
Not single average but a list
select avg(no_of_interactions) from table name;
where I want to get the list of avg interactions of user1,user2,user3 so on
it is giving me only one average
sql postgresql
edited Nov 16 '18 at 11:28
Kaushik Nayak
21.5k41332
asked Nov 16 '18 at 11:26
abdulabdul
176
SELECT user, avg(no_of_interactions) FROM table GROUP BY user
– 404
Nov 16 '18 at 11:29
Please provide sample data and desired results.
– Gordon Linoff
Nov 16 '18 at 11:39
add a comment |
0
0
0
I want to get the list of averages from a table who has the list of users
Not single average but a list
select avg(no_of_interactions) from table name;
where I want to get the list of avg interactions of user1,user2,user3 so on
it is giving me only one average
sql postgresql
edited Nov 16 '18 at 11:28
Kaushik Nayak
21.5k41332
asked Nov 16 '18 at 11:26
abdulabdul
176
I want to get the list of averages from a table who has the list of users
Not single average but a list
select avg(no_of_interactions) from table name;
where I want to get the list of avg interactions of user1,user2,user3 so on
it is giving me only one average
sql postgresql
sql postgresql
edited Nov 16 '18 at 11:28
Kaushik Nayak
21.5k41332
asked Nov 16 '18 at 11:26
abdulabdul
176
edited Nov 16 '18 at 11:28
Kaushik Nayak
21.5k41332
asked Nov 16 '18 at 11:26
abdulabdul
176
edited Nov 16 '18 at 11:28
Kaushik Nayak
21.5k41332
edited Nov 16 '18 at 11:28
Kaushik Nayak
21.5k41332
edited Nov 16 '18 at 11:28
Kaushik Nayak
21.5k41332
21.5k41332
asked Nov 16 '18 at 11:26
abdulabdul
176
asked Nov 16 '18 at 11:26
abdulabdul
176
asked Nov 16 '18 at 11:26
abdulabdul
176
176
SELECT user, avg(no_of_interactions) FROM table GROUP BY user
– 404
Nov 16 '18 at 11:29
Please provide sample data and desired results.
– Gordon Linoff
Nov 16 '18 at 11:39
add a comment |
SELECT user, avg(no_of_interactions) FROM table GROUP BY user
– 404
Nov 16 '18 at 11:29
Please provide sample data and desired results.
– Gordon Linoff
Nov 16 '18 at 11:39
SELECT user, avg(no_of_interactions) FROM table GROUP BY user– 404
Nov 16 '18 at 11:29
SELECT user, avg(no_of_interactions) FROM table GROUP BY user– 404
Nov 16 '18 at 11:29
Please provide sample data and desired results.
– Gordon Linoff
Nov 16 '18 at 11:39
Please provide sample data and desired results.
– Gordon Linoff
Nov 16 '18 at 11:39
add a comment |
1 Answer
1
active
oldest
votes
2
It should be
select <column>, avg(no_of_interactions)
from table <name>
group by <column>;
where <column> is userName or userId or anything else you want to use as grouping condition
answered Nov 16 '18 at 11:30
eXcalibureXcalibur
745
101 101 101 101 101 102 102 102 102 102 102 102 102 102 102 102
– abdul
Nov 16 '18 at 11:48
user_id:101 and user_id:102
– abdul
Nov 16 '18 at 11:49
user_id:101 and user_id:102,103,etc interaction average should be found and list of avg should be displayed
– abdul
Nov 16 '18 at 11:50
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53336933%2fi-want-a-query-to-get-list-of-averages-of-users-in-postgres%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
It should be
select <column>, avg(no_of_interactions)
from table <name>
group by <column>;
where <column> is userName or userId or anything else you want to use as grouping condition
answered Nov 16 '18 at 11:30
eXcalibureXcalibur
745
101 101 101 101 101 102 102 102 102 102 102 102 102 102 102 102
– abdul
Nov 16 '18 at 11:48
user_id:101 and user_id:102
– abdul
Nov 16 '18 at 11:49
user_id:101 and user_id:102,103,etc interaction average should be found and list of avg should be displayed
– abdul
Nov 16 '18 at 11:50
add a comment |
2
It should be
select <column>, avg(no_of_interactions)
from table <name>
group by <column>;
where <column> is userName or userId or anything else you want to use as grouping condition
answered Nov 16 '18 at 11:30
eXcalibureXcalibur
745
101 101 101 101 101 102 102 102 102 102 102 102 102 102 102 102
– abdul
Nov 16 '18 at 11:48
user_id:101 and user_id:102
– abdul
Nov 16 '18 at 11:49
user_id:101 and user_id:102,103,etc interaction average should be found and list of avg should be displayed
– abdul
Nov 16 '18 at 11:50
add a comment |
2
2
2
It should be
select <column>, avg(no_of_interactions)
from table <name>
group by <column>;
where <column> is userName or userId or anything else you want to use as grouping condition
answered Nov 16 '18 at 11:30
eXcalibureXcalibur
745
It should be
select <column>, avg(no_of_interactions)
from table <name>
group by <column>;
where <column> is userName or userId or anything else you want to use as grouping condition
answered Nov 16 '18 at 11:30
eXcalibureXcalibur
745
answered Nov 16 '18 at 11:30
eXcalibureXcalibur
745
answered Nov 16 '18 at 11:30
eXcalibureXcalibur
745
answered Nov 16 '18 at 11:30
eXcalibureXcalibur
745
745
101 101 101 101 101 102 102 102 102 102 102 102 102 102 102 102
– abdul
Nov 16 '18 at 11:48
user_id:101 and user_id:102
– abdul
Nov 16 '18 at 11:49
user_id:101 and user_id:102,103,etc interaction average should be found and list of avg should be displayed
– abdul
Nov 16 '18 at 11:50
add a comment |
101 101 101 101 101 102 102 102 102 102 102 102 102 102 102 102
– abdul
Nov 16 '18 at 11:48
user_id:101 and user_id:102
– abdul
Nov 16 '18 at 11:49
user_id:101 and user_id:102,103,etc interaction average should be found and list of avg should be displayed
– abdul
Nov 16 '18 at 11:50
101 101 101 101 101 102 102 102 102 102 102 102 102 102 102 102
– abdul
Nov 16 '18 at 11:48
101 101 101 101 101 102 102 102 102 102 102 102 102 102 102 102
– abdul
Nov 16 '18 at 11:48
user_id:101 and user_id:102
– abdul
Nov 16 '18 at 11:49
user_id:101 and user_id:102
– abdul
Nov 16 '18 at 11:49
user_id:101 and user_id:102,103,etc interaction average should be found and list of avg should be displayed
– abdul
Nov 16 '18 at 11:50
user_id:101 and user_id:102,103,etc interaction average should be found and list of avg should be displayed
– abdul
Nov 16 '18 at 11:50
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53336933%2fi-want-a-query-to-get-list-of-averages-of-users-in-postgres%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
SELECT user, avg(no_of_interactions) FROM table GROUP BY user– 404
Nov 16 '18 at 11:29
Please provide sample data and desired results.
– Gordon Linoff
Nov 16 '18 at 11:39