i have a date column that reads as string value [2018/04/09].i want to read it as date column 2018/04/09.How to do it in postgresql?
Use the function to_date with the date formatter.
Include additional string characters that appear on all dates as part of the format string, and the date can be parsed correctly.
WITH example (dt) AS (
VALUES ('[2018/04/09]')
)
SELECT to_date(dt, '[YYYY/MM/DD]') FROM example
Alternatively, if the object is to clean up the data as well, e.g. some dates have the square brackets while other dates don't, then it is better to replace all invalid characters and then parse as date.
example:
WITH example (dt) AS (
VALUES ('[2018/04/09]')
)
SELECT to_date(trim(both '[]' from dt), 'YYYY/MM/DD') FROM example
Try this:
SELECT TO_DATE(date_column, 'YYYY/MM/DD') as date
FROM tablename
Related
I'm trying to use the below code to convert the 'date_time' values in my table to 10 characters. Currently, the date_time values are in the format '2020:09:08 10:00:00' but I want to get rid of the time portion of the values so that I'm only left with the date. I'm using DB Browser for SQLite and I thought I would have to use the varchar data type to complete this task but its not working. The output of the code is also shown below.
SELECT CAST(date_time AS VARCHAR(10)) FROM stocks;
I would just use string functions:
substr(date_time, 1, 10)
Your datetime values are not valid datetimes for SQLite which recognizes only the format 'YYYY-MM-DD hh:mm:ss'.
If you used the above format then all you need would be the function DATE():
SELECT DATE(date_time)
FROM stocks;
For your current format you must use string functions to extract the first 10 chars and then replace all : with -:
SELECT REPLACE(SUBSTR(date_time, 1, 10), ':', '-')
FROM stocks;
Using Google BIGQUERY, I need to check that the values in a column called birth_day_col are the correct and desired date format: YYYY-MM-DD. The values in this column are defined as STRING. Also the values in this column are currently of the following format: YYYY-MM-DD.
I researched a lot on the internet and found an interesting workaround. The following query:
SELECT
DISTINCT birth_day_col
FROM `project.dataset.datatable`
WHERE birth_day_col LIKE '[1-2][0-9][0-9][0-9]/[0-1][0-9]/[0-3][0-9]'
AND country_code = 'country1'
But the result is: "This query returned no results."
I then checked with NOT, using the following code:
SELECT
DISTINCT birth_day_col
FROM `project.dataset.datatable`
WHERE NOT(birth_day_col LIKE '[1-2][0-9][0-9][0-9]/[0-1][0-9]/[0-3][0-9]')
AND country_code = 'country1'
Surprisingly it gave all the values in birth_dat_col, which I have verified and are of the correct date format, but this result coud very much be a coincidence.
And it is very strange (wrong) that I used a query that should result only the wrong format dates, but it actually gives me the correct ones. Everything about these two queries seems like an inversation of each one's role.
The expected result of any query for this business case is to make a count of all incorrect formatted dates (even if currently this is 0).
Thank you for your help!
Robert
A couple of things here:
Read the documentation for the LIKE operator if you want to understand how to use it. It looks like you're trying to use regular expression syntax, but the LIKE operator does not take a regular expression as input.
The standard format for BigQuery's dates is YYYY-MM-DD, so you can just try casting and see if the result is a valid date, e.g.:
SELECT SAFE_CAST(birth_day_col AS DATE) AS birth_day_col
FROM `project`.dataset.table
This will return null for any values that don't have the correct format. If you want to find all of the ones that don't have the correct format, you can use SAFE_CAST inside a filter:
SELECT DISTINCT birth_day_col AS invalid_date
FROM `project`.dataset.table
WHERE SAFE_CAST(birth_day_col AS DATE) IS NULL
The result of this query will be all of the date strings that don't use YYYY-MM-DD format. If you want to check for slashes instead, you can use REGEXP_CONTAINS, e.g. try this:
SELECT
date,
REGEXP_CONTAINS(date, r'^[0-9]{4}/[0-9]{2}/[0-9]{2}$')
FROM (
SELECT '2019/05/10' AS date UNION ALL
SELECT '2019-05-10' UNION ALL
SELECT '05/10/2019'
)
If you want to find all dates with either YYYY-MM-DD format or YYYY/MM/DD format, you can use a query like this:
SELECT
DISTINCT date
FROM `project`.dataset.table
WHERE REGEXP_CONTAINS(date, r'^[0-9]{4}[/\-][0-9]{2}[/\-][0-9]{2}$')
For example:
SELECT
DISTINCT date
FROM (
SELECT '2019/05/10' AS date UNION ALL
SELECT '2019-05-10' UNION ALL
SELECT '05/10/2019'
)
WHERE REGEXP_CONTAINS(date, r'^[0-9]{4}[/\-][0-9]{2}[/\-][0-9]{2}$')
Yet another example for BigQuery Standrad SQL - with use of SAFE.PARSE_DATE
#standardSQL
WITH `project.dataset.table` AS (
SELECT '1980/08/10' AS birth_day_col UNION ALL
SELECT '1980-08-10' UNION ALL
SELECT '08/10/1980'
)
SELECT birth_day_col
FROM `project.dataset.table`
WHERE SAFE.PARSE_DATE('%Y-%m-%d', birth_day_col) IS NULL
with result of list of all dates which are not formatted as yyyy-mm-dd
Row birth_day_col
1 1980/08/10
2 08/10/1980
Google BigQuery's LIKE operator does not support matching digits nor does it uses the [ character in its syntax (I don't think ISO standard SQL does either - LIKE is nowhere near as powerful as Regex).
X [NOT] LIKE Y
Checks if the STRING in the first operand X matches a pattern specified by the second operand Y. Expressions can contain these characters:
A percent sign "%" matches any number of characters or bytes
An underscore "_" matches a single character or byte
You can escape "\", "_", or "%" using two backslashes. For example, "\%". If you are using raw strings, only a single backslash is required. For example, r"\%".
You should use REGEX_CONTAINS instead.
I note that string format tests won't tell you if a date is valid or not, however. Consider that 2019-02-31 has a valid date format, but an invalid date value. I suggest using a datatype conversion function (to convert the STRING to a DATE value) instead.
In my DB, there is a date field in the format yyyymmdd.
I have to get all the dates in the format dd-mm-yyyy for that particlar date.
ex:
Date
20170130
20170228
20170325
for the above dates, I need the output in the below format with the dates and day of the particular dates
date day
30-01-2017 tuesday
28-02-2017 tuesday
25-03-2017 saturday
If the column is a string, then it can hold invalid date values such as February 31, one way to avoid this is by a small function such as this:
create or replace
function my_to_date( p_str in varchar2 ) return date
is
begin
return to_date( p_str );
exception
when others then
return null;
end;
\\
select to_char(my_to_date('20170231'),'DD-MM-YYYY Day')
from dual
\\
Demo
Try below:
Select to_char(yrdate, 'dd-mm-yyyy'), to_char(yrdate, 'D') from yrtable
It sounds like your dates aren't actually DATE fields but some kind of CHAR field? The best option would be to convert to DATE and then convert back to CHAR:
SELECT TO_CHAR(TO_DATE(mydate, 'YYYYMMDD'), 'DD-MM-YYYY Day')
FROM mytable;
This uses the YYYYMMDD mask to convert your string into a date, then uses the mask DD-MM-YYYY Day to convert it back into a string. Use day if you want the day name in lowercase (as in your OP).
#user2778168 answer will give you the results you want. But why?
Your database does not have dates stored in yyyymmdd format or any other date format for at mater, unless it's defined with a character type definition. Oracle stores all dates in a single internal structure, and with only slight variations timestamps are the same. The format used only tells Oracle how to display the value or to convert a string to a date. Unless a specific format is specified Oracle uses the NLS_DATE_FORMAT for this determination. See here and scan down to "Datetime Format Models" for format specifications.
To see this run the following:
select value
from nls_session_parameters
where parameter = 'NLS_DATE_FORMAT';
Select yrdate default_format
, to_char(yrdate, 'dd-mm-yyyy') specified_format
, dump(yrdate) unformated
from yrtable;
alter session set nls_date_format = 'Month dd,yyyy';
Rerun the above queries.
It seems you hold date column(date1) in character format. Suppose you have a table named days:
SQL> desc days
date1 varchar2(10)
then,
we should convert it into date, and then into char format, with aliases in quotation marks to provide lowercase aliases as you wanted.
perhaps your database's date language is non-english like mine(turkish), then you need to convert it to english.
lastly, it'a appropriate to order the results according to converted date, seen from your output. So we can use the following SQL :
select to_char(to_date(date1,'yyyymmdd'),'dd-mm-yyyy') "date",
to_char(to_date(date1,'yyyymmdd'),'day','nls_date_language=english') "day"
from days
order by to_date(date1,'yyyymmdd');
D e m o
I want to select a varchar field as a date field
For example a field has this value "30.12.2011 21:15:03"
and when i select this
select DATE from TABLE where DATE = '30.12.2011'
i get no result.
You ask about getting the date part of a timestamp field, but what your question is actually about is filtering on the date of a timestamp field. There is a much simpler method of accomplishing that: you can use the knowledge that all the possible timestamps on a specific date won't have any timestamps for different dates between them.
select DATE
from TABLE
where DATE >= '30.12.2011' and DATE < '31.12.2011'
Your edit explains that you haven't got a timestamp field at all. Nevertheless, a similar approach may still work:
select DATE
from TABLE
where DATE LIKE '30.12.2011 %'
Or the Firebird-specific
select DATE
from TABLE
where DATE starting with '30.12.2011 '
Assuming the field is a date field, use the DATE introducer combined with yyyy-mm-dd (or TIMESTAMP with time as well).
So use:
select datefield from sometable where datefield = DATE '2011-12-30'
Technically you can leave off the introducer, but it is 'correcter' in the light of the SQL standard.
Assuming a TIMESTAMP field, you won't get results unless the timestamp is (always) at 00:00:00.0000 (in which case it should have been a DATE instead).
For the comparison to work, you need to use either BETWEEN, eg:
select timestampfield from sometable
where timestampfield BETWEEN '2011-12-30 00:00:00.0000' AND '2011-12-30 23:59:59.9999'
or truncate the timestamp to a date (this may adversely effect performance if the timestamp is indexed, because then the index can no longer be used), eg:
select timestampfield from sometable
where CAST(timestampfield AS DATE) = '2011-12-30'
If the date is stored in a VARCHAR field (which in itself is a bad idea), there are several solutions, first is to handle it as date manipulation:
select varcharfield from sometable
where CAST(CAST(varcharfield AS TIMESTAMP) AS DATE) = '2011-12-30'
The double cast is required if you have a time-component in VARCHARFIELD as well. This assumes dates in the supported format listed below. If you use BETWEEN as above, you can use a single cast to timestamp)
The other solution (as suggested by hvd) is to treat it purely as string manipulation, for example:
select varcharfield from sometable
where varcharfield STARTING WITH '30.12.2011'
This has its own set of problems if you want to select ranges. Bottomline: use a real TIMESTAMP field!
Note that Firebird supports multiple formats:
yyyy-mm-dd, eg 2014-05-25 (ISO-8601 format, probably best to use as it reduces confusion)
dd.mm.yyyy, eg 25.05.2014
mm/dd/yyyy, eg 05/25/2014
mm-dd-yyyy, eg 05-25-2014
dd mmm yyyy, eg 25 MAY 2014 (+ variations with a -, . or / as separator)
mmm dd yyyy, eg MAY 25 2014 (+ variations with a -, . or / as separator)
select DATE from TABLE where cast(DATE as date) = '30.12.2011'
Date field is a timestamp
Here is the answere to my question:
CAST
(
SUBSTRING
(field FROM 1 FOR 2)
||'.'||
SUBSTRING
(field FROM 4 FOR 2)
||'.'||
SUBSTRING
(field FFROM 7 FOR 4)
AS DATE)
This took me 5 hours to find this out, maybe there should be a "-" instead of "." but it works.
I have a table student with the following columns:
no - integer
name - string
startdate - date
enddate - date.
Date format is MM/DD/YYYY.
I will give a date as input. Now I need a query the inputdate which found in between the start and end date.
For an example I will give 04/14/2012, then the query should return the 1st record as in the figure.
(because input date (04/14/2012) is found in between the 04/10/2012 to 04/20/2012)
Please help me.
The issue you are having is caused by your assumption that sqlite has a date/datetime type when in fact it doesn't.
I suggest you read the following http://www.sqlite.org/datatype3.html to have a better understanding of sqlite types.
The dates in the MM/DD/YYYY format are handled as TEXT by sqlite, and so those dates are compared as strings. For example, 02/01/2012 is considered bigger than 01/02/2012by sqlite if compared directly.
You will need to transform those dates to a format that can be string-compared. Here is an example:
sqlite> create table foo (d TEXT);
sqlite> insert into foo values ('02/01/2012');
sqlite> select substr(d, 7, 4) || substr(d, 1, 2) || substr(d, 4, 2) from foo;
20120201
You should post what you have tried so far.
There should be a between clause that you can use:
select * from table
where inputdate between startdate and enddate
Dates as a date type in SQLite don't exist. There are a number of approaches to dealing with dates - store them as integer seconds since 1 Jan 1970 (unixepoch) or store them as strings, but if you do, then you really need to store them in 'YYYY-MM-DD' format because that is what the date functions require as input.
Assuming you use the string format in the format I suggested then your query would look something like
SELECT * FROM Table WHERE Date(Inputdate) BETWEEEN Date(startDate) AND Date(EndDate);
(although you may want to format the output of the date columns to US date format with
SELECT Strftime("%m/%d/%Y",startDate) As StartDate ...
If you use seconds since 1970 its somewhat easier because the seconds just compare without needing the convert them to dates, although you still might want to output in US date format, so ...
SELECT Strftime("%m/%d/%Y",startDate) As StartDate ... FROM Table WHERE inputDate BETWEEN startDate and EndDate;
sqlite> select *from tbl_node where mydate between '2014-02-02' and '2014-02-06';
it show the output :-
1|1|123|456|12eb-ab|1|1|254|123|19|2014-02-03 16:00:44
2|1|123|456|12eb-ab|1|1|254|123|19|2014-02-03 16:01:03
3|1|123|456|12eb-ab|1|1|254|123|19|2014-02-03 16:00:57
4|1|123|456|12eb-ab|1|1|254|123|19|2014-02-03 16:00:34
Here mydate is column name in tbl_node;
we can also use from current time , using now.
sqlite> select *from tbl_node where mydate between '2014-02-02' and 'now';