Change a decimal string into a timestamp in Impala
How do you convert a string type like
t1.updte_timestamp
2018-06-02-08.18.45.562742
2018-05-26-09.18.16.594824
into a timestamp? SHOULD RESULT IN:
2018-06-02-08.18.45
2018-05-26-09.18.16
ETC
The values had been imported from excel and are in STRING-TYPE
I tried:
SELECT
to_timestamp(cast (t1.updte_timestamp as string), 'yyyy-mm-dd hh:mm:ss') as updted_timestamp FROM OLD;
but results in NULL for all values
thank you
hadoop impala
add a comment |
How do you convert a string type like
t1.updte_timestamp
2018-06-02-08.18.45.562742
2018-05-26-09.18.16.594824
into a timestamp? SHOULD RESULT IN:
2018-06-02-08.18.45
2018-05-26-09.18.16
ETC
The values had been imported from excel and are in STRING-TYPE
I tried:
SELECT
to_timestamp(cast (t1.updte_timestamp as string), 'yyyy-mm-dd hh:mm:ss') as updted_timestamp FROM OLD;
but results in NULL for all values
thank you
hadoop impala
why don't you just take a length=19 prefix out of the timestamp?substr(value, 1, 19)
– mangusta
Nov 16 '18 at 1:23
add a comment |
How do you convert a string type like
t1.updte_timestamp
2018-06-02-08.18.45.562742
2018-05-26-09.18.16.594824
into a timestamp? SHOULD RESULT IN:
2018-06-02-08.18.45
2018-05-26-09.18.16
ETC
The values had been imported from excel and are in STRING-TYPE
I tried:
SELECT
to_timestamp(cast (t1.updte_timestamp as string), 'yyyy-mm-dd hh:mm:ss') as updted_timestamp FROM OLD;
but results in NULL for all values
thank you
hadoop impala
How do you convert a string type like
t1.updte_timestamp
2018-06-02-08.18.45.562742
2018-05-26-09.18.16.594824
into a timestamp? SHOULD RESULT IN:
2018-06-02-08.18.45
2018-05-26-09.18.16
ETC
The values had been imported from excel and are in STRING-TYPE
I tried:
SELECT
to_timestamp(cast (t1.updte_timestamp as string), 'yyyy-mm-dd hh:mm:ss') as updted_timestamp FROM OLD;
but results in NULL for all values
thank you
hadoop impala
hadoop impala
asked Nov 16 '18 at 0:29
Anna Anna
799
799
why don't you just take a length=19 prefix out of the timestamp?substr(value, 1, 19)
– mangusta
Nov 16 '18 at 1:23
add a comment |
why don't you just take a length=19 prefix out of the timestamp?substr(value, 1, 19)
– mangusta
Nov 16 '18 at 1:23
why don't you just take a length=19 prefix out of the timestamp?
substr(value, 1, 19)
– mangusta
Nov 16 '18 at 1:23
why don't you just take a length=19 prefix out of the timestamp?
substr(value, 1, 19)
– mangusta
Nov 16 '18 at 1:23
add a comment |
1 Answer
1
active
oldest
votes
you can substr
your string and apply to_timestamp as follow
select to_timestamp(substr('2018-06-02-08.18.45.562742', 1, 19) , 'yyyy-MM-dd-HH.mm.ss');
Make sure you use MM for month and HH for hour in upper case
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
you can substr
your string and apply to_timestamp as follow
select to_timestamp(substr('2018-06-02-08.18.45.562742', 1, 19) , 'yyyy-MM-dd-HH.mm.ss');
Make sure you use MM for month and HH for hour in upper case
add a comment |
you can substr
your string and apply to_timestamp as follow
select to_timestamp(substr('2018-06-02-08.18.45.562742', 1, 19) , 'yyyy-MM-dd-HH.mm.ss');
Make sure you use MM for month and HH for hour in upper case
add a comment |
you can substr
your string and apply to_timestamp as follow
select to_timestamp(substr('2018-06-02-08.18.45.562742', 1, 19) , 'yyyy-MM-dd-HH.mm.ss');
Make sure you use MM for month and HH for hour in upper case
you can substr
your string and apply to_timestamp as follow
select to_timestamp(substr('2018-06-02-08.18.45.562742', 1, 19) , 'yyyy-MM-dd-HH.mm.ss');
Make sure you use MM for month and HH for hour in upper case
answered Nov 16 '18 at 1:45
hlagoshlagos
4,5281818
4,5281818
add a comment |
add a comment |
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why don't you just take a length=19 prefix out of the timestamp?
substr(value, 1, 19)
– mangusta
Nov 16 '18 at 1:23