I'm storing Dates as string in the database with this format DD-MM-YYYY.
When I tried to make a select query with an orderby on the date column. I didn't get the expected result.
example of result :
28/02/2013
27/02/2013
01/03/2013
My sql query :
SELECT * FROM data ORDER BY strftime('%s', date_column)
Thank you.
The problem is that you store dates as DD-MM-YYYY strings, which does not only prevent natural ordering of dates as strings, but also parsing them with SQLite's date and time functions. Click the link and scroll down to 'Time Strings' section.
SQLite expects date/time strings in the natural order, most significant digit to least significant, that is, YYYY-MM-DD. You can use string operations to transform your DD-MM-YYYY strings into that form. For instance:
select
substr(reversed_date, 7,4) || '-' ||
substr(reversed_date, 4, 2)|| '-' ||
substr(reversed_date, 1, 2) as proper_date
from (
select '12-03-2000' as reversed_date
)
;
You can either transform your date column into this format (as #peterm suggests) or just use the value of proper_date for sorting. You don't need to use strftime for that, but date-related functions will work with such values.
IMHO you need to change the format you store dates in from
DD-MM-YYYY
to
YYYY-MM-DD
From docs
Time Strings
A time string can be in any of the following formats:
YYYY-MM-DD
YYYY-MM-DD HH:MM
YYYY-MM-DD HH:MM:SS
YYYY-MM-DD HH:MM:SS.SSS
YYYY-MM-DDTHH:MM
YYYY-MM-DDTHH:MM:SS
YYYY-MM-DDTHH:MM:SS.SSS
...
Then your original query and this one will work as expected
SELECT * FROM Table1 ORDER BY date(date_column);
SELECT * FROM Table1 ORDER BY strftime('%s', date_column);
Output:
| date_column |
---------------
| 2013-02-27 |
| 2013-02-28 |
| 2013-03-01 |
sqlfiddle
According to the documentation the following should work
SELECT *
FROM data
ORDER BY strftime('%Y-%m-%d', date_column)
try:
SELECT * FROM data ORDER BY to_date(date_column)
probably this might solve you problem as it is going for string comparison rather than date comparison so
01/03/2013 appears smaller than 28/02/2013 or 27/02/2013
thus output is :
01/03/2013, 27/02/2013, 28/02/2013
This query worked for me filter dates
SELECT inst.*,
Substr(columnname, 4, 2) AS newdate,
Substr(columnname, 0, 3) AS newday,
Substr(columnname, 12, 5) AS newtime
FROM table_name AS inst
WHERE child_id = id
ORDER BY newdate DESC, newday DESC, newtime DESC
Related
I have a column "date_time" and many other.
date_time is a string date (because in SQLite there is no date format), e.g. "2020.2.1" means 1st of February of 2020th.
I want to sort by this date 2020.1.1-2020.1.2-...-2020.12.31
I am using:
SELECT *
FROM Table
ORDER BY date_time
And I am getting sorted as a string:
here is an example:
2020.1.10
...
2020.1.29
2020.1.3
...
How I can sort it as a date if I am using SQLite and my date is string?
select *
from Table
order by cast(date_time as datetime)
I need concat 2 columns char in 1 column date.
I tried it:
INSERT INTO tb_teste PARTITION (dt_originacao_fcdr)
SELECT
tp_registro_fcdr,
seq_registro_fcdr,
tp_cdr_fcdr,
dt_atendimento_fcdr,
date_dt_atendimento_fcdr,
hr_atendimento_fcdr,
timestamp(from_unixtime(unix_timestamp(CONCAT(dt_atendimento_fcdr, hr_atendimento_fcdr), 'yyyyMMddHHmmss')), "yyyy-MM-dd HH:mm:ss") as date_hr_atendimento_fcdr,
duracao_atend_fcdr,
hr_originacao_fcdr,
duracao_total_fcdr,
duracao_chamada_tarifada_fcdr,
st_chamada_fcdr,
fim_sel_orig_fcdr,
numero_a_fcdr,
tp_numero_a_fcdr,
numero_b_fcdr,
tp_numero_b_fcdr,
numero_b_orig_fcdr,
numero_c_fcdr,
tp_numero_c_fcdr,
tp_trafego_fcdr,
esn_fcdr,
central_fcdr,
erb_fcdr,
tp_erb_fcdr,
face_erb_inici_fcdr,
erb_final_fcdr,
face_erb_final_fcdr,
erb_original_fcdr,
imsi_fcdr,
imei_fcdr,
tecnologia_fcdr,
cd_oper_ass_a_fcdr,
cd_oper_ass_b_fcdr,
cgi_fcdr,
nu_tlfn_fcdr,
tp_tlfn_fcdr,
tp_tarifa_fcdr,
ident_num_a_fcdr,
ident_num_b_fcdr,
cd_prestadora_fcdr,
cna_orig_ar_erb_fcdr
FROM tb_op_nor;
Result: date_hr_atendimento_fcdr 2019-03-03
The column containing the time and date is not null or empty.
Example:
Time zone: Brazil.
I need date and time in the same columns.
You can concat two fields then unix_timestamp and by using from_unixtimestamp function we can format the output timestamp.
with cte as (select stack(1,"20190303","131615") as (dt,hr)) --sample data
select
timestamp( --cast to timestamp
from_unixtime(unix_timestamp(concat(dt,hr),'yyyyMMddHHmmss'),"yyyy-MM-dd HH:mm:ss") --concat and change the format
)
from cte
Output:
2019-03-03 13:16:15.0
If you want to convert to brazil time to UTC or vice versa then use to_utc_timestamp/from_utc_timestamp.
I am selecting a date column which is in the format "YYYY-MM-DD".
I want to cast it to a timestamp such that it will be "YYYY-MM-DD HH:MM:SS:MS"
I attempted:
select CAST(mycolumn as timestamp) from mytable;
but this resulted in the format YYYY-MM-DD HH:MM:SS
I also tried
select TO_TIMESTAMP(mycolumn,YYYY-MM-DD HH:MM:SS:MS) from mytable;
but this did not work either. I cannot seem to figure out the correct way to format this. Note that I only want the first digit of the milliseconds.
//////////////second question
I am also trying to select numeric data such that there will not be any trailing zeros.
For example, if I have values in a table such as 1, 2.00, 3.34, 4.50.
I want to be able to select those values as 1, 2, 3.34, 4.5.
I tried using ::float, but I occasionally strange output. I also tried the rounding function, but how could I use it properly without knowing how many decimal points I need before hand?
thanks for your help!
It seems that the functions to_timestamp() and to_char() are unfortunately not perfect.
If you cannot find anything better, use these workarounds:
with example_data(d) as (
values ('2016-02-02')
)
select d, d::timestamp || '.0' tstamp
from example_data;
d | tstamp
------------+-----------------------
2016-02-02 | 2016-02-02 00:00:00.0
(1 row)
create function my_to_char(numeric)
returns text language sql as $$
select case
when strpos($1::text, '.') = 0 then $1::text
else rtrim($1::text, '.0')
end
$$;
with example_data(n) as (
values (100), (2.00), (3.34), (4.50))
select n::text, my_to_char(n)
from example_data;
n | my_to_char
------+------------
100 | 100
2.00 | 2
3.34 | 3.34
4.50 | 4.5
(4 rows)
See also: How to remove the dot in to_char if the number is an integer
SELECT to_char(current_timestamp, 'YYYY-MM-DD HH:MI:SS:MS');
prints
2016-02-05 03:21:18:346
just add ::timestamp without time zone
select mycolumn::timestamp without time zone from mytable;
My table myTab has the column startDate, which has the datatype "DATE". The data in this column are stored like dd.mm.yyyy.
Now I'm trying to get data with this query:
SELECT * FROM myTab WHERE startDate like '%01.2015"
Somehow it doesn't work and I don't know why.
Hope someone can help.
To make a text search on the date you would have to convert the date to text.
It's more efficient if you calculate the first and last date for what you want to find and get everything between them. That way it's done as numeric comparisons instead of a text pattern match, and it can make use of an index if there is one:
SELECT * FROM myTab WHERE startDate >= DATE '2015-01-01' AND startDate < DATE '2015-02-01'
SELECT * FROM myTab WHERE TO_CHAR(startDate,'dd.mm.yyyy') LIKE '%01.2015'
If the field type is "DATE" then the value isn't stored as a string, it's a number managed by Oracle, so you have to convert it to a string:
SELECT * FROM myTab WHERE to_char(startDate, 'MM.YYYY') = '01.2015';
You can also use date ranges in SQL queries:
SELECT * FROM myTab
WHERE startDate
BETWEEN to_date('01.01.2015', 'DD.MM.YYYY')
AND to_date('31.01.2015', 'DD.MM.YYYY');
Regarding you actual question "Somehow it doesn't work and I don't know why."
Oracle make an implicit conversion from DATE to VARHCAR2, however it uses the default NLS_DATE_FORMAT which is probably different to what you use in your query.
The data in this column are stored like dd.mm.yyyy.
Oracle does not store date in the format you see. It stores it internally in proprietary format in 7 bytes with each byte storing different components of the datetime value.
WHERE startDate like '%01.2015"
You are comparing a DATE with a STRING, which is pointless.
From performance point of view, you should use a date range condition so that if there is any regular INDEX on the date column, it would be used.
SELECT * FROM table_name WHERE date_column BETWEEN DATE '2015-01-01' AND DATE '2015-02-01'
To understand why a Date range condition is better in terms of performance, have a look at my answer here.
I solved my problem that way. Thank you for suggestions for improvements. Example in C#.
string dd, mm, aa, trc, data;
dd = nData.Text.Substring(0, 2);
mm = nData.Text.Substring(3, 2);
aa = nData.Text.Substring(6, 4);
trc = "-";
data = aa + trc + mm + trc + dd;
"Select * From bdPedidos Where Data Like '%" + data + "%'";
To provide a more detailed answer and address this https://stackoverflow.com/a/42429550/1267661 answer's issue.
In Oracle a column of type "date" is not a number nor a string, it's a "datetime" value with year, month, day, hour, minute and seconds.
The default time is always midnight "00:00:00"
The query:
Select * From bdPedidos Where Data Like '%" + data + "%'"
won't work in all circumstances because a date column is not a string, using "like" forces Oracle to do a conversion from date value to string value.
The string value may be year-month-day-time or month-day-year-time or day-month-year-time, that all depends how a particular Oracle instance has set the parameter NLS_DATE_FORMAT to show dates as strings.
The right way to cover all the possible times in a day is:
Select *
From bdPedidos
Where Data between to_date('" + data + " 00:00:00','yyyy-mm-dd hh24:mi:ss')
and to_date('" + data + " 23:59:59','yyyy-mm-dd hh24:mi:ss')
SELECT * FROM myTab WHERE startDate like '%-%-2015';
This will search for all dates in 2015. If this doesn't work, try:
SELECT * FROM myTab WHERE startDate like '%-%-15';
Every record in my SQLite database contains a field which contains a Date stored as a string in the format 'yyyy-MM-dd HH:mm:ss'.
Is it possible to query the database to get the record which contains the most recent date please?
you can do it like this
SELECT * FROM Table ORDER BY date(dateColumn) DESC Limit 1
For me I had my query this way to solve my problem
select * from Table order by datetime(datetimeColumn) DESC LIMIT 1
Since I was storing it as datetime not date column
When you sure the format of text field is yyyy-MM-dd HH:mm:ss (ex.: 2017-01-02 16:02:55), So It works for me simply:
SELECT * FROM Table ORDER BY dateColumn DESC Limit 1
Without any extra date function!
You need to convert it to unix timestamp, and then compare them:
SELECT * FROM data ORDER BY strftime('%s', date_column) DESC
But this can be pretty slow, if there are lots of rows.
Better approach would be to store unix timestamp by default, and create an index for that column.
You can convert your column sent_date_time to yyyy-MM-dd format and then order by date:
1) substr(sent_date_time,7,4)||"-"||substr(sent_date_time,1,2)||"-"||substr(sent_date_time,4,2) as date
2) order by date desc
In my case everything works fine without casting column to type 'date'. Just by specifying column name with double quotes like that:
SELECT * FROM 'Repair' ORDER BY "Date" DESC;
I think SQLite makes casting by itself or something like that, but when I tried to 'cast' Date column by myself it's not worked. And there was no error messages.
You can also use the following query
"SELECT * FROM Table ORDER BY strftime('%Y-%m-%d %H:%M:%S'," + dateColumn + ") DESC Limit 1"
I found this ugly hack worked.
select *, substr(date_col_name,7,4)as yy,
substr(date_col_name,4,2) as mm,
substr(date_col_name,1,2) as dd
from my_table
order by yy desc,mm desc,dd desc
it would be better to convert the text column to date field type, but I found that did not work reliably for me.
If you do a lot of date sorting/comparison, you may get better results by storing time as ticks rather than strings, here is showing how to get 'now' in ticks with:
((strftime('%s', 'now') - strftime('%S', 'now') + strftime('%f', 'now')) * 1000)
(see https://stackoverflow.com/a/20478329/460084)
Then it's easy to sort, compare, etc ...
This will work for both date and time
SELECT *
FROM Table
ORDER BY
julianday(dateColumn)
DESC Limit 1