KDB: convert millisecond epoc to date
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I have a table t with a string column tDate.
select tDate from t
tDate
----------
1542776400000
1542776400000
1542776400000
1542776400000
1542776400000
1542776400000
1540958400000
1542776400000
1542776400000
1542776400000
1542776400000
I want to convert this to a string with MM/dd/YYYY format.
but I need to drop the last 3 characters then convert it to date. But somehow, the below statement does not work correctly:
select "P"$-3_tDate from t
How is this wrong and how do I translate it to the "MM/dd/YYYY" string.
kdb
asked Nov 23 '18 at 20:37
user3682563user3682563
936
add a comment |
I have a table t with a string column tDate.
select tDate from t
tDate
----------
1542776400000
1542776400000
1542776400000
1542776400000
1542776400000
1542776400000
1540958400000
1542776400000
1542776400000
1542776400000
1542776400000
I want to convert this to a string with MM/dd/YYYY format.
but I need to drop the last 3 characters then convert it to date. But somehow, the below statement does not work correctly:
select "P"$-3_tDate from t
How is this wrong and how do I translate it to the "MM/dd/YYYY" string.
kdb
asked Nov 23 '18 at 20:37
user3682563user3682563
936
add a comment |
I have a table t with a string column tDate.
select tDate from t
tDate
----------
1542776400000
1542776400000
1542776400000
1542776400000
1542776400000
1542776400000
1540958400000
1542776400000
1542776400000
1542776400000
1542776400000
I want to convert this to a string with MM/dd/YYYY format.
but I need to drop the last 3 characters then convert it to date. But somehow, the below statement does not work correctly:
select "P"$-3_tDate from t
How is this wrong and how do I translate it to the "MM/dd/YYYY" string.
kdb
asked Nov 23 '18 at 20:37
user3682563user3682563
936
I have a table t with a string column tDate.
select tDate from t
tDate
----------
1542776400000
1542776400000
1542776400000
1542776400000
1542776400000
1542776400000
1540958400000
1542776400000
1542776400000
1542776400000
1542776400000
I want to convert this to a string with MM/dd/YYYY format.
but I need to drop the last 3 characters then convert it to date. But somehow, the below statement does not work correctly:
select "P"$-3_tDate from t
How is this wrong and how do I translate it to the "MM/dd/YYYY" string.
kdb
kdb
asked Nov 23 '18 at 20:37
user3682563user3682563
936
asked Nov 23 '18 at 20:37
user3682563user3682563
936
asked Nov 23 '18 at 20:37
user3682563user3682563
936
asked Nov 23 '18 at 20:37
user3682563user3682563
936
asked Nov 23 '18 at 20:37
user3682563user3682563
936
936
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Here's how I would go about this:
- Convert to string type using
string- https://code.kx.com/q/ref/casting/#string
- Pad the strings to count of 10 using
$- https://code.kx.com/q/ref/strings/#pad
- Convert to a timestamp type with cast using type p -
"P"$
- Convert that to the necessary temporal types using `mm`dd`year - https://code.kx.com/q/ref/casting/#cast
- Format with zero-colon - https://code.kx.com/q/ref/filenumbers/#prepare-text
````
q)t:(10#1542776400000)
q)select "/"0:`mm`dd`year$:"P"$10$string x from t
x
------------
"11/21/2018"
"11/21/2018"
"11/21/2018"
````
answered Nov 24 '18 at 11:32
Sean O'HaganSean O'Hagan
1,223411
add a comment |
You need to first convert the long values to string :
q)select `date$"P"$-3_/:string tDate from t
tDate
----------
2018.11.23
2018.11.21
2018.11.21
Converting the date to "MM/dd/YYYY" format can be done by:
q)f:{ d:"." vs x; "/" sv (d 1;d 2;d 0)}
q)select f each string `date$"P"$-3_/:string tDate from t
tDate
------------
"11/23/2018"
"11/21/2018"
"11/21/2018"
"11/21/2018"
If you are interested in any other format then you have to manipulate the string be defining a custom function like f above.
Or you can explore datetimeQ GitHub library which supports the excel style formatting.
e.g.
q).dtf.format["d mmmm, dddd ,yyyy"; 2018.06.18];
"18 June, Tuesday ,2018"
answered Nov 23 '18 at 20:58
nyinyi
2,37931537
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here's how I would go about this:
- Convert to string type using
string- https://code.kx.com/q/ref/casting/#string
- Pad the strings to count of 10 using
$- https://code.kx.com/q/ref/strings/#pad
- Convert to a timestamp type with cast using type p -
"P"$
- Convert that to the necessary temporal types using `mm`dd`year - https://code.kx.com/q/ref/casting/#cast
- Format with zero-colon - https://code.kx.com/q/ref/filenumbers/#prepare-text
````
q)t:(10#1542776400000)
q)select "/"0:`mm`dd`year$:"P"$10$string x from t
x
------------
"11/21/2018"
"11/21/2018"
"11/21/2018"
````
answered Nov 24 '18 at 11:32
Sean O'HaganSean O'Hagan
1,223411
add a comment |
Here's how I would go about this:
- Convert to string type using
string- https://code.kx.com/q/ref/casting/#string
- Pad the strings to count of 10 using
$- https://code.kx.com/q/ref/strings/#pad
- Convert to a timestamp type with cast using type p -
"P"$
- Convert that to the necessary temporal types using `mm`dd`year - https://code.kx.com/q/ref/casting/#cast
- Format with zero-colon - https://code.kx.com/q/ref/filenumbers/#prepare-text
````
q)t:(10#1542776400000)
q)select "/"0:`mm`dd`year$:"P"$10$string x from t
x
------------
"11/21/2018"
"11/21/2018"
"11/21/2018"
````
answered Nov 24 '18 at 11:32
Sean O'HaganSean O'Hagan
1,223411
add a comment |
Here's how I would go about this:
- Convert to string type using
string- https://code.kx.com/q/ref/casting/#string
- Pad the strings to count of 10 using
$- https://code.kx.com/q/ref/strings/#pad
- Convert to a timestamp type with cast using type p -
"P"$
- Convert that to the necessary temporal types using `mm`dd`year - https://code.kx.com/q/ref/casting/#cast
- Format with zero-colon - https://code.kx.com/q/ref/filenumbers/#prepare-text
````
q)t:(10#1542776400000)
q)select "/"0:`mm`dd`year$:"P"$10$string x from t
x
------------
"11/21/2018"
"11/21/2018"
"11/21/2018"
````
answered Nov 24 '18 at 11:32
Sean O'HaganSean O'Hagan
1,223411
Here's how I would go about this:
- Convert to string type using
string- https://code.kx.com/q/ref/casting/#string
- Pad the strings to count of 10 using
$- https://code.kx.com/q/ref/strings/#pad
- Convert to a timestamp type with cast using type p -
"P"$
- Convert that to the necessary temporal types using `mm`dd`year - https://code.kx.com/q/ref/casting/#cast
- Format with zero-colon - https://code.kx.com/q/ref/filenumbers/#prepare-text
````
q)t:(10#1542776400000)
q)select "/"0:`mm`dd`year$:"P"$10$string x from t
x
------------
"11/21/2018"
"11/21/2018"
"11/21/2018"
````
answered Nov 24 '18 at 11:32
Sean O'HaganSean O'Hagan
1,223411
answered Nov 24 '18 at 11:32
Sean O'HaganSean O'Hagan
1,223411
answered Nov 24 '18 at 11:32
Sean O'HaganSean O'Hagan
1,223411
answered Nov 24 '18 at 11:32
Sean O'HaganSean O'Hagan
1,223411
1,223411
add a comment |
add a comment |
You need to first convert the long values to string :
q)select `date$"P"$-3_/:string tDate from t
tDate
----------
2018.11.23
2018.11.21
2018.11.21
Converting the date to "MM/dd/YYYY" format can be done by:
q)f:{ d:"." vs x; "/" sv (d 1;d 2;d 0)}
q)select f each string `date$"P"$-3_/:string tDate from t
tDate
------------
"11/23/2018"
"11/21/2018"
"11/21/2018"
"11/21/2018"
If you are interested in any other format then you have to manipulate the string be defining a custom function like f above.
Or you can explore datetimeQ GitHub library which supports the excel style formatting.
e.g.
q).dtf.format["d mmmm, dddd ,yyyy"; 2018.06.18];
"18 June, Tuesday ,2018"
answered Nov 23 '18 at 20:58
nyinyi
2,37931537
add a comment |
You need to first convert the long values to string :
q)select `date$"P"$-3_/:string tDate from t
tDate
----------
2018.11.23
2018.11.21
2018.11.21
Converting the date to "MM/dd/YYYY" format can be done by:
q)f:{ d:"." vs x; "/" sv (d 1;d 2;d 0)}
q)select f each string `date$"P"$-3_/:string tDate from t
tDate
------------
"11/23/2018"
"11/21/2018"
"11/21/2018"
"11/21/2018"
If you are interested in any other format then you have to manipulate the string be defining a custom function like f above.
Or you can explore datetimeQ GitHub library which supports the excel style formatting.
e.g.
q).dtf.format["d mmmm, dddd ,yyyy"; 2018.06.18];
"18 June, Tuesday ,2018"
answered Nov 23 '18 at 20:58
nyinyi
2,37931537
add a comment |
You need to first convert the long values to string :
q)select `date$"P"$-3_/:string tDate from t
tDate
----------
2018.11.23
2018.11.21
2018.11.21
Converting the date to "MM/dd/YYYY" format can be done by:
q)f:{ d:"." vs x; "/" sv (d 1;d 2;d 0)}
q)select f each string `date$"P"$-3_/:string tDate from t
tDate
------------
"11/23/2018"
"11/21/2018"
"11/21/2018"
"11/21/2018"
If you are interested in any other format then you have to manipulate the string be defining a custom function like f above.
Or you can explore datetimeQ GitHub library which supports the excel style formatting.
e.g.
q).dtf.format["d mmmm, dddd ,yyyy"; 2018.06.18];
"18 June, Tuesday ,2018"
answered Nov 23 '18 at 20:58
nyinyi
2,37931537
You need to first convert the long values to string :
q)select `date$"P"$-3_/:string tDate from t
tDate
----------
2018.11.23
2018.11.21
2018.11.21
Converting the date to "MM/dd/YYYY" format can be done by:
q)f:{ d:"." vs x; "/" sv (d 1;d 2;d 0)}
q)select f each string `date$"P"$-3_/:string tDate from t
tDate
------------
"11/23/2018"
"11/21/2018"
"11/21/2018"
"11/21/2018"
If you are interested in any other format then you have to manipulate the string be defining a custom function like f above.
Or you can explore datetimeQ GitHub library which supports the excel style formatting.
e.g.
q).dtf.format["d mmmm, dddd ,yyyy"; 2018.06.18];
"18 June, Tuesday ,2018"
answered Nov 23 '18 at 20:58
nyinyi
2,37931537
edited Nov 23 '18 at 21:20
answered Nov 23 '18 at 20:58
nyinyi
2,37931537
answered Nov 23 '18 at 20:58
nyinyi
2,37931537
answered Nov 23 '18 at 20:58
nyinyi
2,37931537
2,37931537
add a comment |
add a comment |
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