Throwback in dates without Weekends
I often use this query1
to go back 6 days on a variable date:
query1 = SELECT
DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ`
AS S WHERE S.`I` <= 6;
With the help of this auxiliary table with a sequence inserted there.
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES (1),(2),(3),(4),(5),(6);
Question:
How can I avoid output weekends days using query1?
In a such way that the output go back in 6 useful days, in this example starting on '2018-11-21' the output would be ->
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
| 2018-11-15 |
| 2018-11-14 |
mysql mysqli
add a comment |
I often use this query1
to go back 6 days on a variable date:
query1 = SELECT
DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ`
AS S WHERE S.`I` <= 6;
With the help of this auxiliary table with a sequence inserted there.
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES (1),(2),(3),(4),(5),(6);
Question:
How can I avoid output weekends days using query1?
In a such way that the output go back in 6 useful days, in this example starting on '2018-11-21' the output would be ->
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
| 2018-11-15 |
| 2018-11-14 |
mysql mysqli
add a comment |
I often use this query1
to go back 6 days on a variable date:
query1 = SELECT
DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ`
AS S WHERE S.`I` <= 6;
With the help of this auxiliary table with a sequence inserted there.
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES (1),(2),(3),(4),(5),(6);
Question:
How can I avoid output weekends days using query1?
In a such way that the output go back in 6 useful days, in this example starting on '2018-11-21' the output would be ->
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
| 2018-11-15 |
| 2018-11-14 |
mysql mysqli
I often use this query1
to go back 6 days on a variable date:
query1 = SELECT
DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ`
AS S WHERE S.`I` <= 6;
With the help of this auxiliary table with a sequence inserted there.
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES (1),(2),(3),(4),(5),(6);
Question:
How can I avoid output weekends days using query1?
In a such way that the output go back in 6 useful days, in this example starting on '2018-11-21' the output would be ->
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
| 2018-11-15 |
| 2018-11-14 |
mysql mysqli
mysql mysqli
edited Nov 22 '18 at 10:13
Madhur Bhaiya
19.5k62236
19.5k62236
asked Nov 21 '18 at 11:25
Nelson Bustier
387
387
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
We can utilize DayName()
function to get the name of the weekday corresponding to a Date. We will utilize this function's result to restrict weekends by NOT IN ('Saturday', 'Sunday')
.
Also, we will need to increase the number generator range upto 10. Because there is a possibility that we can come across 2 weekends (total 4 days) on either side of 5 weekdays.
So, we need 2 (first pair of weekend days) + 5 (weekdays) + 2 (second pair of weekend days) + 1 (6th weekday) = 10 dates to consider. An example of this edge case would be when an input date is Sunday.
We will need to use LIMIT 6
to restrict the result upto 6 days only, in the non-edge cases.
Schema (MySQL v5.7)
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES
(1),(2),(3),(4),(5),(6),(7),(8),(9),(10);
Query
SELECT
DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ` AS S
WHERE S.`I` <= 10
AND DAYNAME(DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')) NOT IN ('SATURDAY', 'SUNDAY')
ORDER BY VAR DESC
LIMIT 6;
Result
| VAR |
| ---------- |
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
| 2018-11-15 |
| 2018-11-14 |
View on DB Fiddle
Edge Case Demo - Input date: 25 Nov 2018 (Sunday)
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES
(1),(2),(3),(4),(5),(6),(7),(8),(9),(10);
Query #2
SELECT
DATE_FORMAT((DATE('2018-11-25') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ` AS S
WHERE S.`I` <= 10
AND DAYNAME(DATE_FORMAT((DATE('2018-11-25') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')) NOT IN ('SATURDAY', 'SUNDAY')
ORDER BY VAR DESC
LIMIT 6;
Result
| VAR |
| ---------- |
| 2018-11-23 |
| 2018-11-22 |
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
View on DB Fiddle
Madhur Bhaiya, But my idea is that it would continue for six days. In this case starting on 2018-11-21 the intended output would be: | 2018-11-21 | 2018-11-20 | 2018-11-19 | 2018-11-16 | 2018-11-15 | 2018-11- 14 |; How is this possible ?
– Nelson Bustier
Nov 22 '18 at 10:02
@NelsonBustier please specify this in the question. I will update answer accordingly.
– Madhur Bhaiya
Nov 22 '18 at 10:03
Madhur Bhaiya, Already done . TY
– Nelson Bustier
Nov 22 '18 at 10:12
@NelsonBustier you will need 6 days only ? not more than that ?
– Madhur Bhaiya
Nov 22 '18 at 10:12
1
Madhur Bhaiya, Done! Perfect answer. It works. TY
– Nelson Bustier
Nov 22 '18 at 10:20
|
show 2 more comments
Your Answer
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1 Answer
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oldest
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oldest
votes
We can utilize DayName()
function to get the name of the weekday corresponding to a Date. We will utilize this function's result to restrict weekends by NOT IN ('Saturday', 'Sunday')
.
Also, we will need to increase the number generator range upto 10. Because there is a possibility that we can come across 2 weekends (total 4 days) on either side of 5 weekdays.
So, we need 2 (first pair of weekend days) + 5 (weekdays) + 2 (second pair of weekend days) + 1 (6th weekday) = 10 dates to consider. An example of this edge case would be when an input date is Sunday.
We will need to use LIMIT 6
to restrict the result upto 6 days only, in the non-edge cases.
Schema (MySQL v5.7)
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES
(1),(2),(3),(4),(5),(6),(7),(8),(9),(10);
Query
SELECT
DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ` AS S
WHERE S.`I` <= 10
AND DAYNAME(DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')) NOT IN ('SATURDAY', 'SUNDAY')
ORDER BY VAR DESC
LIMIT 6;
Result
| VAR |
| ---------- |
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
| 2018-11-15 |
| 2018-11-14 |
View on DB Fiddle
Edge Case Demo - Input date: 25 Nov 2018 (Sunday)
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES
(1),(2),(3),(4),(5),(6),(7),(8),(9),(10);
Query #2
SELECT
DATE_FORMAT((DATE('2018-11-25') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ` AS S
WHERE S.`I` <= 10
AND DAYNAME(DATE_FORMAT((DATE('2018-11-25') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')) NOT IN ('SATURDAY', 'SUNDAY')
ORDER BY VAR DESC
LIMIT 6;
Result
| VAR |
| ---------- |
| 2018-11-23 |
| 2018-11-22 |
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
View on DB Fiddle
Madhur Bhaiya, But my idea is that it would continue for six days. In this case starting on 2018-11-21 the intended output would be: | 2018-11-21 | 2018-11-20 | 2018-11-19 | 2018-11-16 | 2018-11-15 | 2018-11- 14 |; How is this possible ?
– Nelson Bustier
Nov 22 '18 at 10:02
@NelsonBustier please specify this in the question. I will update answer accordingly.
– Madhur Bhaiya
Nov 22 '18 at 10:03
Madhur Bhaiya, Already done . TY
– Nelson Bustier
Nov 22 '18 at 10:12
@NelsonBustier you will need 6 days only ? not more than that ?
– Madhur Bhaiya
Nov 22 '18 at 10:12
1
Madhur Bhaiya, Done! Perfect answer. It works. TY
– Nelson Bustier
Nov 22 '18 at 10:20
|
show 2 more comments
We can utilize DayName()
function to get the name of the weekday corresponding to a Date. We will utilize this function's result to restrict weekends by NOT IN ('Saturday', 'Sunday')
.
Also, we will need to increase the number generator range upto 10. Because there is a possibility that we can come across 2 weekends (total 4 days) on either side of 5 weekdays.
So, we need 2 (first pair of weekend days) + 5 (weekdays) + 2 (second pair of weekend days) + 1 (6th weekday) = 10 dates to consider. An example of this edge case would be when an input date is Sunday.
We will need to use LIMIT 6
to restrict the result upto 6 days only, in the non-edge cases.
Schema (MySQL v5.7)
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES
(1),(2),(3),(4),(5),(6),(7),(8),(9),(10);
Query
SELECT
DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ` AS S
WHERE S.`I` <= 10
AND DAYNAME(DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')) NOT IN ('SATURDAY', 'SUNDAY')
ORDER BY VAR DESC
LIMIT 6;
Result
| VAR |
| ---------- |
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
| 2018-11-15 |
| 2018-11-14 |
View on DB Fiddle
Edge Case Demo - Input date: 25 Nov 2018 (Sunday)
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES
(1),(2),(3),(4),(5),(6),(7),(8),(9),(10);
Query #2
SELECT
DATE_FORMAT((DATE('2018-11-25') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ` AS S
WHERE S.`I` <= 10
AND DAYNAME(DATE_FORMAT((DATE('2018-11-25') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')) NOT IN ('SATURDAY', 'SUNDAY')
ORDER BY VAR DESC
LIMIT 6;
Result
| VAR |
| ---------- |
| 2018-11-23 |
| 2018-11-22 |
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
View on DB Fiddle
Madhur Bhaiya, But my idea is that it would continue for six days. In this case starting on 2018-11-21 the intended output would be: | 2018-11-21 | 2018-11-20 | 2018-11-19 | 2018-11-16 | 2018-11-15 | 2018-11- 14 |; How is this possible ?
– Nelson Bustier
Nov 22 '18 at 10:02
@NelsonBustier please specify this in the question. I will update answer accordingly.
– Madhur Bhaiya
Nov 22 '18 at 10:03
Madhur Bhaiya, Already done . TY
– Nelson Bustier
Nov 22 '18 at 10:12
@NelsonBustier you will need 6 days only ? not more than that ?
– Madhur Bhaiya
Nov 22 '18 at 10:12
1
Madhur Bhaiya, Done! Perfect answer. It works. TY
– Nelson Bustier
Nov 22 '18 at 10:20
|
show 2 more comments
We can utilize DayName()
function to get the name of the weekday corresponding to a Date. We will utilize this function's result to restrict weekends by NOT IN ('Saturday', 'Sunday')
.
Also, we will need to increase the number generator range upto 10. Because there is a possibility that we can come across 2 weekends (total 4 days) on either side of 5 weekdays.
So, we need 2 (first pair of weekend days) + 5 (weekdays) + 2 (second pair of weekend days) + 1 (6th weekday) = 10 dates to consider. An example of this edge case would be when an input date is Sunday.
We will need to use LIMIT 6
to restrict the result upto 6 days only, in the non-edge cases.
Schema (MySQL v5.7)
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES
(1),(2),(3),(4),(5),(6),(7),(8),(9),(10);
Query
SELECT
DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ` AS S
WHERE S.`I` <= 10
AND DAYNAME(DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')) NOT IN ('SATURDAY', 'SUNDAY')
ORDER BY VAR DESC
LIMIT 6;
Result
| VAR |
| ---------- |
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
| 2018-11-15 |
| 2018-11-14 |
View on DB Fiddle
Edge Case Demo - Input date: 25 Nov 2018 (Sunday)
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES
(1),(2),(3),(4),(5),(6),(7),(8),(9),(10);
Query #2
SELECT
DATE_FORMAT((DATE('2018-11-25') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ` AS S
WHERE S.`I` <= 10
AND DAYNAME(DATE_FORMAT((DATE('2018-11-25') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')) NOT IN ('SATURDAY', 'SUNDAY')
ORDER BY VAR DESC
LIMIT 6;
Result
| VAR |
| ---------- |
| 2018-11-23 |
| 2018-11-22 |
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
View on DB Fiddle
We can utilize DayName()
function to get the name of the weekday corresponding to a Date. We will utilize this function's result to restrict weekends by NOT IN ('Saturday', 'Sunday')
.
Also, we will need to increase the number generator range upto 10. Because there is a possibility that we can come across 2 weekends (total 4 days) on either side of 5 weekdays.
So, we need 2 (first pair of weekend days) + 5 (weekdays) + 2 (second pair of weekend days) + 1 (6th weekday) = 10 dates to consider. An example of this edge case would be when an input date is Sunday.
We will need to use LIMIT 6
to restrict the result upto 6 days only, in the non-edge cases.
Schema (MySQL v5.7)
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES
(1),(2),(3),(4),(5),(6),(7),(8),(9),(10);
Query
SELECT
DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ` AS S
WHERE S.`I` <= 10
AND DAYNAME(DATE_FORMAT((DATE('2018-11-21') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')) NOT IN ('SATURDAY', 'SUNDAY')
ORDER BY VAR DESC
LIMIT 6;
Result
| VAR |
| ---------- |
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
| 2018-11-15 |
| 2018-11-14 |
View on DB Fiddle
Edge Case Demo - Input date: 25 Nov 2018 (Sunday)
CREATE TABLE `HELPER_SEQ` (`I` tinyint(3) UNSIGNED NOT NULL);
INSERT INTO `HELPER_SEQ` (`I`) VALUES
(1),(2),(3),(4),(5),(6),(7),(8),(9),(10);
Query #2
SELECT
DATE_FORMAT((DATE('2018-11-25') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')
AS VAR
FROM `HELPER_SEQ` AS S
WHERE S.`I` <= 10
AND DAYNAME(DATE_FORMAT((DATE('2018-11-25') - INTERVAL(S.`I` - 1) DAY), '%Y-%m-%d')) NOT IN ('SATURDAY', 'SUNDAY')
ORDER BY VAR DESC
LIMIT 6;
Result
| VAR |
| ---------- |
| 2018-11-23 |
| 2018-11-22 |
| 2018-11-21 |
| 2018-11-20 |
| 2018-11-19 |
| 2018-11-16 |
View on DB Fiddle
edited Nov 22 '18 at 10:38
answered Nov 21 '18 at 18:03
Madhur Bhaiya
19.5k62236
19.5k62236
Madhur Bhaiya, But my idea is that it would continue for six days. In this case starting on 2018-11-21 the intended output would be: | 2018-11-21 | 2018-11-20 | 2018-11-19 | 2018-11-16 | 2018-11-15 | 2018-11- 14 |; How is this possible ?
– Nelson Bustier
Nov 22 '18 at 10:02
@NelsonBustier please specify this in the question. I will update answer accordingly.
– Madhur Bhaiya
Nov 22 '18 at 10:03
Madhur Bhaiya, Already done . TY
– Nelson Bustier
Nov 22 '18 at 10:12
@NelsonBustier you will need 6 days only ? not more than that ?
– Madhur Bhaiya
Nov 22 '18 at 10:12
1
Madhur Bhaiya, Done! Perfect answer. It works. TY
– Nelson Bustier
Nov 22 '18 at 10:20
|
show 2 more comments
Madhur Bhaiya, But my idea is that it would continue for six days. In this case starting on 2018-11-21 the intended output would be: | 2018-11-21 | 2018-11-20 | 2018-11-19 | 2018-11-16 | 2018-11-15 | 2018-11- 14 |; How is this possible ?
– Nelson Bustier
Nov 22 '18 at 10:02
@NelsonBustier please specify this in the question. I will update answer accordingly.
– Madhur Bhaiya
Nov 22 '18 at 10:03
Madhur Bhaiya, Already done . TY
– Nelson Bustier
Nov 22 '18 at 10:12
@NelsonBustier you will need 6 days only ? not more than that ?
– Madhur Bhaiya
Nov 22 '18 at 10:12
1
Madhur Bhaiya, Done! Perfect answer. It works. TY
– Nelson Bustier
Nov 22 '18 at 10:20
Madhur Bhaiya, But my idea is that it would continue for six days. In this case starting on 2018-11-21 the intended output would be: | 2018-11-21 | 2018-11-20 | 2018-11-19 | 2018-11-16 | 2018-11-15 | 2018-11- 14 |; How is this possible ?
– Nelson Bustier
Nov 22 '18 at 10:02
Madhur Bhaiya, But my idea is that it would continue for six days. In this case starting on 2018-11-21 the intended output would be: | 2018-11-21 | 2018-11-20 | 2018-11-19 | 2018-11-16 | 2018-11-15 | 2018-11- 14 |; How is this possible ?
– Nelson Bustier
Nov 22 '18 at 10:02
@NelsonBustier please specify this in the question. I will update answer accordingly.
– Madhur Bhaiya
Nov 22 '18 at 10:03
@NelsonBustier please specify this in the question. I will update answer accordingly.
– Madhur Bhaiya
Nov 22 '18 at 10:03
Madhur Bhaiya, Already done . TY
– Nelson Bustier
Nov 22 '18 at 10:12
Madhur Bhaiya, Already done . TY
– Nelson Bustier
Nov 22 '18 at 10:12
@NelsonBustier you will need 6 days only ? not more than that ?
– Madhur Bhaiya
Nov 22 '18 at 10:12
@NelsonBustier you will need 6 days only ? not more than that ?
– Madhur Bhaiya
Nov 22 '18 at 10:12
1
1
Madhur Bhaiya, Done! Perfect answer. It works. TY
– Nelson Bustier
Nov 22 '18 at 10:20
Madhur Bhaiya, Done! Perfect answer. It works. TY
– Nelson Bustier
Nov 22 '18 at 10:20
|
show 2 more comments
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