Column string to time [duplicate] - time

This question already has an answer here:
Convert STRING to TIME bigquery
(1 answer)
Closed 9 months ago.
New in SQL here! I use BigQuery.
How do I convert an entire string column to time?
The column name is ride_length and it's format is 00:00 (minutes:seconds).
On schema is string but I need it to be time.
Also how can I calculate its average then?
Thank you in advance.

I think there are multiple ways to do it. But, it totally depends on how you need the end result (or even process the intermediate values). One way is you could possibly calculate the total seconds using split, cast & add, do the average & cast it back to time datatype (or maintain in minutes/seconds)
WITH
s AS (
SELECT
'45:00' AS time
UNION ALL
SELECT
'30:00' AS time
UNION ALL
SELECT
'20:00' AS time
UNION ALL
SELECT
'25:00' AS time
UNION ALL
SELECT
'04:00' AS time
UNION ALL
SELECT
'25:30' AS time )
SELECT
avg_s,
time(TIMESTAMP_SECONDS(avg_s))
FROM (
SELECT
CAST(AVG(secs) AS INT64) AS avg_s
FROM (
SELECT
time,
SAFE_CAST(SPLIT(time, ':') [ORDINAL(1)] AS INT64)*60+ SAFE_CAST(SPLIT(time, ':') [ORDINAL(2)] AS INT64) AS secs
FROM
s ) )

Related

How to SELECT the MAX Time Difference Between Any 2 Consecutive Rows Per Value?

Just had a user answer this correctly for TSQL, but wondering how best to achieve this now in SQL Developer/PLSQL seeing as there is no DATEDIFF function.
Table I want to query on has some 'CODE' values, which can naturally have multiple primary key records ('OccsID') in a table 'Occs'. There is also a datetime column called 'CreateDT' for each OccsID.
Just want to find the maximum possible time variance between any 2 consecutive rows in 'Occs', per 'CODE'.
If you subtract the "next" date and "this" date (using the LEAD analytic function), you'll get the date difference. Then fetch the maximum difference per code. Something like this:
with diff as
(select occsid,
code,
nvl(lead(createdt) over (partition by code order by createdt), createdt) - createdt date_diff
from test
)
select code,
max(date_diff)
from diff
group by code;
Assuming that this T-SQL version works for you (from the prior question)
SELECT x.code, MAX(x.diff_sec) FROM
(
SELECT
code,
DATEDIFF(
SECOND,
CreateDT,
LEAD(CreateDT) OVER(PARTITION BY CODE ORDER BY CreateDT) --next row's createdt
) as diff_sec
FROM Occs
)x
GROUP BY x.code
The simplest option is just to subtract the two dates to get a difference in days. You can then multiply to get the difference in hours, minutes, or seconds
SELECT x.code, MAX(x.diff_day), MAX(x.diff_sec)
FROM
(
SELECT
code,
CreateDT -
LEAD(CreateDT) OVER(PARTITION BY CODE ORDER BY CreateDT) as diff_day,
24*60*60* (CreateDT -
LEAD(CreateDT) OVER(PARTITION BY CODE ORDER BY CreateDT)) as diff_sec
FROM Occs
)x
GROUP BY x.code

Difference in two dates not coming as expected in Oracle 11g

I get some data through a OSB Proxy Service and have to transform that using Xquery. Earlier the transformation was done on the database but now it is to be done on the proxy itself. So I have been given the SQL queries which were used and have to generate Xquery expressions corresponding to those.
Here is the SQL query which is supposed to find the difference between 2 dates.
SELECT ROUND((CAST(DATEATTRIBUTE2 AS DATE) -
CAST(DATEATTRIBUTE1 AS DATE) ) * 86400 ) AS result
FROM SONY_TEST_TABLE;
DATEATTRIBUTE1 and DATEATTRIBUTE2 are both of TIMESTAMP type.
As per my understanding this query first casts the TIMESTAMP to DATE so that the time part is stripped then subtracts the dates. That difference in days in multiplied with 86400 to get the duration in seconds.
However, when I take DATEATTRIBUTE2 as 23-02-17 01:17:19.399000000 AM and DATEATTRIBUTE1 as 23-02-17 01:17:18.755000000 AM the result should ideally be 0 as the dates are same and i'm ignoring the time difference but surprisingly the result comes as 1. After checking I found that the ( CAST(DATEATTRIBUTE2 AS DATE) - CAST(DATEATTRIBUTE1 AS DATE) ) part aparently does not give an integer value but a fractional one. How does this work?? o_O
Any help is appreciated. Cheers!
EDIT : So got the problem thanks to all the answers! Even after casting to DATE it still has time so the time difference is also calculated. Now how do I implement this in XQuery? See this other question.
Oracle DATE datatype is actually a datetime. So casting something as a date doesn't remove the time element. To do that we need to truncate the value:
( trunc(DATEATTRIBUTE2) - trunc(DATEATTRIBUTE1) )
you should try this to find difference by day
SELECT (trunc(DATEATTRIBUTE2) -
trunc(DATEATTRIBUTE1) ) AS result
FROM SONY_TEST_TABLE;
alternative 2
you can use extract like below:
SELECT ROUND (
EXTRACT (MINUTE FROM INTERVAL_DIFFERENCE) / (24 * 60)
+ EXTRACT (HOUR FROM INTERVAL_DIFFERENCE) / 24
+ EXTRACT (DAY FROM INTERVAL_DIFFERENCE))
FROM (SELECT ( TO_TIMESTAMP ('23-02-17 01:17:19', 'dd-mm-yy hh24:mi:ss')
- TO_TIMESTAMP ('23-02-17 01:17:17', 'dd-mm-yy hh24:mi:ss'))
INTERVAL_DIFFERENCE
FROM DUAL)

How to generate diff between TIMESTAMP and DATE in SELECT in oracle 10

I need to query 2 tables, one contains a TIMESTAMP(6) column, other contains a DATE column. I want to write a select statement that prints both values and diff between these two in third column.
SB_BATCH.B_CREATE_DT - timestamp
SB_MESSAGE.M_START_TIME - date
SELECT SB_BATCH.B_UID, SB_BATCH.B_CREATE_DT, SB_MESSAGE.M_START_TIME,
to_date(to_char(SB_BATCH.B_CREATE_DT), 'DD-MON-RR HH24:MI:SS') as time_in_minutes
FROM SB_BATCH, SB_MESSAGE
WHERE
SB_BATCH.B_UID = SB_MESSAGE.M_B_UID;
Result:
Error report -
SQL Error: ORA-01830: date format picture ends before converting entire input string
01830. 00000 - "date format picture ends before converting entire input string"
You can subtract two timestamps to get an INTERVAL DAY TO SECOND, from which you calculate how many minutes elapsed between the two timestamps. In order to convert SB_MESSAGE.M_START_TIME to a timestamp you can use CAST.
Note that I have also removed your implicit table join with an explicit INNER JOIN, moving the join condition to the ON clause.
SELECT t.B_UID,
t.B_CREATE_DT,
t.M_START_TIME,
EXTRACT(DAY FROM t.diff)*24*60 +
EXTRACT(HOUR FROM t.diff)*60 +
EXTRACT(MINUTE FROM t.diff) +
ROUND(EXTRACT(SECOND FROM t.diff) / 60.0) AS diff_in_minutes
FROM
(
SELECT SB_BATCH.B_UID,
SB_BATCH.B_CREATE_DT,
SB_MESSAGE.M_START_TIME,
SB_BATCH.B_CREATE_DT - CAST(SB_MESSAGE.M_START_TIME AS TIMESTAMP) AS diff
FROM SB_BATCH
INNER JOIN SB_MESSAGE
ON SB_BATCH.B_UID = SB_MESSAGE.M_B_UID
) t
Convert the timestamp to a date using cast(... as date). Then take the difference between the dates, which is a number - expressed in days, so if you want it in minutes, multiply by 24*60. Then round the result as needed. I made up a small example below to isolate just the steps needed to answer your question. (Note that your query has many other problems, for example you didn't actually take a difference of anything anywhere. If you need help with your query in general, please post it as a separate question.)
select ts, dt, round( (sysdate - cast(ts as date))*24*60, 2) as time_diff_in_minutes
from (select to_timestamp('2016-08-23 03:22:44.734000', 'yyyy-mm-dd hh24:mi:ss.ff') as ts,
sysdate as dt from dual )
;
TS DT TIME_DIFF_IN_MINUTES
-------------------------------- ------------------- --------------------
2016-08-23 03:22:44.734000000 2016-08-23 08:09:15 286.52

oracle insert only time [duplicate]

This question already has answers here:
How to store only time; not date and time?
(5 answers)
Closed 8 years ago.
I would like to insert data to db as time, not date. If I use to_date('2012-08-31 07:39:33', 'YYYY-MM-DD HH24:MI:SS') it adds date too.
If I use to_date('09:34:00', 'HH24:MI:SS') it adds year, month, day as well, from nowhere :|
Later I need to get rows where time is between x and y, not taking in account the year, month or day. How do I do that?
thanks
As an alternative to the date solution Dave shows, you could use an interval data type for the column:
create table t42(id number, t interval day to second);
insert into t42 (id, t) values(123, to_dsinterval('0 07:39:33'));
insert into t42 (id, t) values(456, to_dsinterval('0 09:34:00'));
select id
from t42
where t between to_dsinterval('0 07:00:00') and to_dsinterval('0 07:59:59');
ID
----------
123
Displaying intervals is a little awkward as they don't have format models, but see this question for some ideas if needed. If you only use them for filtering then that may not be an issue at all.
A DATE type always includes the date component.
One option is to continue using DATE and write your code to ignore the date component. In order to make queries on the time efficient, you might want to create a function-based index on something like TO_CHAR( date_field, 'HH24:MI:SS' ) and use that expression in your queries.
Alternatively, you could use a NUMBER field to store the number of seconds since midnight, and write your queries in terms of that.
you can use number column type and insert value as
INSERT INTO table_name (nTime)
VALUES (date - trunc(date));
and then select values
select *
from table_name t
where t.nTime between (10 / 24 + 15 / 24 / 60) and (12 / 24 + 30 / 24 / 60) --between 10:15 and 12:30

Adding one month to saved date(oracle)

I have a table A which contains a Date type attribute. I want to write a query to select the date in another table B with value one month after the value in A.Any one know how to do it in oracle?
uhm... This was the first hit on google:
http://psoug.org/reference/date_func.html
It seems you're looking for the "add_months" function.
You need to use the ADD_MONTHS function in Oracle.
http://www.techonthenet.com/oracle/functions/add_months.php
Additional info: If you want to use this function with today's date you can use ADD_MONTHS(SYSDATE, 1) to get one month from now.
The question is to select a date_field from table b where date_field of table b is one month ahead of a date_field in table a.
An additional requirement must be taken into consideration which is currently unspecified in the question. Are we interested in whole months (days of month not taken into consideration) or do we want to include the days which might disqualify dates that are one month ahead but only by a couple of days (example: a=2011-04-30 and b=2011-05-01, b is 1 month ahead but only by 1 day).
In the first case, we must truncate both dates to their year and month values:
SELECT TRUNC( TO_DATE('2011-04-22','yyyy-mm-dd'), 'mm') as trunc_date
FROM dual;
gives:
trunc_date
----------
2011-04-01
In the second case we don't have to modify the dates.
At least two approaches can be used to solve the initial problem:
First one revolves around adding one month to the date_field in table a and finding a row in table b with a matching date.
SELECT b.date_field
FROM tab_a as a
,tab_b as b
WHERE ADD_MONTHS( TRUNC( a.date_field, 'mm' ), 1) = TRUNC( b.date_field, 'mm' )
;
Note the truncated dates. Leaving this out will require a perfect day to day match between dates.
The second approaches is based on calculating the difference in months between two dates and picking a calculation that gives a 1 month difference.
SELECT b.date_field
FROM tab_a as a
,tab_b as b
WHERE months_between( TRUNC( b.date_field, 'mm') , TRUNC(a.date_field, 'mm') ) = 1
The order of the fields in months_between is important here. In the provided example:
for b.date_field one month ahead of a.date_field the value is 1
for b.date_field one month before a.date_field the value is -1 (negative one)
Reversing the order will also reverse the results.
Hope this answers your question.

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