Count Distinct on Converted and Concatenated Columns and Grouping by Each Record
select
distinct convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height) as FormatName
from Creative
where CreativeFileDate > '1 SEP 18'
Query pulls the unique records as my per my concatenation. How do I most efficiently find the counts of each now?
Thank you
tsql
add a comment |
select
distinct convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height) as FormatName
from Creative
where CreativeFileDate > '1 SEP 18'
Query pulls the unique records as my per my concatenation. How do I most efficiently find the counts of each now?
Thank you
tsql
add a comment |
select
distinct convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height) as FormatName
from Creative
where CreativeFileDate > '1 SEP 18'
Query pulls the unique records as my per my concatenation. How do I most efficiently find the counts of each now?
Thank you
tsql
select
distinct convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height) as FormatName
from Creative
where CreativeFileDate > '1 SEP 18'
Query pulls the unique records as my per my concatenation. How do I most efficiently find the counts of each now?
Thank you
tsql
tsql
asked Nov 12 '18 at 22:24
joru100joru100
135
135
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You'll want to use group by and count
https://www.w3schools.com/sql/sql_groupby.asp
https://www.w3schools.com/sql/sql_count_avg_sum.asp
My SQL is a bit rusty, but your query should look something like:
select convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height) as FormatName, count(0)
from Creative
where CreativeFileDate > '1 SEP 18'
group by convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height)
Cheers
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
});
}
});
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%2f53270975%2fcount-distinct-on-converted-and-concatenated-columns-and-grouping-by-each-record%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
You'll want to use group by and count
https://www.w3schools.com/sql/sql_groupby.asp
https://www.w3schools.com/sql/sql_count_avg_sum.asp
My SQL is a bit rusty, but your query should look something like:
select convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height) as FormatName, count(0)
from Creative
where CreativeFileDate > '1 SEP 18'
group by convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height)
Cheers
add a comment |
You'll want to use group by and count
https://www.w3schools.com/sql/sql_groupby.asp
https://www.w3schools.com/sql/sql_count_avg_sum.asp
My SQL is a bit rusty, but your query should look something like:
select convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height) as FormatName, count(0)
from Creative
where CreativeFileDate > '1 SEP 18'
group by convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height)
Cheers
add a comment |
You'll want to use group by and count
https://www.w3schools.com/sql/sql_groupby.asp
https://www.w3schools.com/sql/sql_count_avg_sum.asp
My SQL is a bit rusty, but your query should look something like:
select convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height) as FormatName, count(0)
from Creative
where CreativeFileDate > '1 SEP 18'
group by convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height)
Cheers
You'll want to use group by and count
https://www.w3schools.com/sql/sql_groupby.asp
https://www.w3schools.com/sql/sql_count_avg_sum.asp
My SQL is a bit rusty, but your query should look something like:
select convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height) as FormatName, count(0)
from Creative
where CreativeFileDate > '1 SEP 18'
group by convert(varchar(8), Creative.Width) + 'x' + convert(varchar(8), Creative.Height)
Cheers
answered Nov 12 '18 at 23:04
JackJack
14219
14219
add a comment |
add a comment |
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.
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%2f53270975%2fcount-distinct-on-converted-and-concatenated-columns-and-grouping-by-each-record%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