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
Related
I have the following CODE
$thedate = "2018-06-04"; // YYYY-MM-DD
SELECT *
FROM (`leadactivity`)
WHERE statusdate = $thedate
ORDER BY id DESC
LIMIT 25
I am trying to only show the results where statusdate is = to the date provided. However because in sql the column autoupdates with Timestamp, it's including the time, and for some reason is just not giving any results.
Any ideas on what I am doing wrong?
Your code looks like MySQL. One simple way is:
SELECT la.*
FROM leadactivity la
WHERE DATE(la.statusdate) = $thedate
ORDER BY id DESC
LIMIT 25;
However, the use of the DATE() function prevents the use of an index. So, a better approach is:
SELECT la.*
FROM leadactivity la
WHERE la.statusdate >= $thedate AND la.statusdate < $thedate + interval 1 day
ORDER BY la.id DESC
LIMIT 25;
try this:
SELECT * FROM (`leadactivity`)
WHERE statusdate like '%"$thedate"%'
ORDER BY id DESC LIMIT 25
When comparing timestamps with date values, a time-of-day of midnight (ie 00:00:00) is assumed. Specify a range of timestamps based on the date from 00:00:00 to 23:59:59, like this:
SELECT * FROM (`leadactivity`)
WHERE statusdate between '$thedate 00:00:00' and '$thedate 23:59:59'
ORDER BY id DESC LIMIT 25
Depending on your language and database, the syntax to achieve this will vary.
Although converting the timestamp to a date before comparing is simpler, your database won’t use the index on the timestamp column (if one exists), resulting in the worst performance.
i created query!
SELECT install_time, Count(id)
FROM (SELECT DISTINCT install_time, id
FROM mytab
where event='run'
and install_time>='09.01.2017' and install_time<='09.05.2017')
GROUP BY install_time
but install_time has that format like 09-05-2017 5:34:23
but i need this format without hours,minutes and seconds
09-05-2017
How change this date format?
First, you don't need a subquery.
Second, date/time functions vary by database. Here is one method that assumes a date() function for removing the time component:
SELECT DATE(install_time), COUNT(DISTINCT id)
FROM mytab
WHERE event = 'run' AND
install_time >= '2017-09-01' AND install_time < '2017-09-06'
GROUP BY DATE(install_time)
ORDER BY DATE(install_time);
Note that I also changed the date format to a standard format -- this works in most databases. And, I changed the second comparison for install_time to take the time component into account.
Here are some equivalents to the date() function (which works in MySQL, SQLite, BigQuery, and DB2 at least):
SQL Server: cast(install_time as date)
Postgres: date_trunc('day', install_time)
Oracle: trunc(install_time)
Teradata: cast(install_time as format 'YYYYMMDD')
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';
I have a set of dates that are in the format DD-MMM-YYYY. I need to be able to compare dates by using only the DD-MMM part of the date, since the year isn't important.
How would I achieve this?
I have tried reading up on the DATEPART function (edit: which evidently wouldn't work) but I can only theoretically get that to return either the DD or the MMM parts, not both of them at once.
Edit: added oracle tag. Sorry.
Example of date field: 01-MAR-1994
If your column is of type DATE then it doesn't have a format.
If I understand you right, then you want to view the mon-dd part only, so you need to convert it with TO_CHAR function,
i.e.:
select to_char(your_date_column, 'mon-dd') from your_table
Convert your dates using the following format, it will only month and the date part. You have to replace getdate() with you date fields.:
select convert(varchar(5),getdate(),110)
Assuming that you are using SQL Server or Oracle since you attempted using DATEPART, you can just get the day and month using the DAY() and MONTH() functions. Assuming, again, that the dates you are comparing are in two different tables, it would look similar to this:
SELECT MONTH(t1.date), DAY(t2.date)
FROM table AS t1
INNER JOIN table2 AS t2
ON t1.key = t2.key
WHERE MONTH(t1.date) = MONTH(t2.date)
AND DAY(t1.date) = DAY(t2.date)
EDIT: If you are just comparing rows in the same table, you only need a very simple query.
SQLFiddle
select id, TO_CHAR(most_recent, 'mon-dd')
from (
select id, MAX(date1) AS most_recent
from table1
group by id
)
You can also combine month and day into one integer:
EXTRACT(MONTH FROM datecol) * 100 + EXTRACT(DAY FROM datecol) AS MonthDay
Then it's easier to sort and compare.
select FORMAT(yourcoulmn_name, 'dd/MM') from table
This should do the trick
`select CONVERT(varchar(7),datetime_column,100) from your_table`
date_default_timezone_set("Asia/Kolkata");
$m = date("m");//Month
$d = date("d");//Day
$sql = "SELECT * FROM contactdata WHERE MONTH(date) = '$m' AND DAY(date) = '$d' ";
only checks day and month and returns today, day and month from database
SELECT LEFT(REPLACE(CONVERT(varchar(10),GETDATE()-1,3),'/',''),4)
WOuld this work for you?
FROMAT(DATETIME, 'dd-MMM') = FROMAT(DATETIME, 'dd-MMM') use any format you want
Suppose I have a date 2010-07-29. Now I would like to check the result of one day ahead. how to do that
For example,
SELECT *
from table
where date = date("2010-07-29")
How to do one day before without changing the string "2010-07-29"?
I searched and get some suggestion from web and I tried
SELECT *
from table
where date = (date("2010-07-29") - 1 Day)
but failed.
MySQL
SELECT *
FROM TABLE t
WHERE t.date BETWEEN DATE_SUB('2010-07-29', INTERVAL 1 DAY)
AND '2010-07-29'
Change DATE_SUB to DATE_ADD if you want to add a day (and reverse the BETWEEN parameters).
SQL Server
SELECT *
FROM TABLE t
WHERE t.date BETWEEN DATEADD(dd, -1, '2010-07-29')
AND '2010-07-29'
Oracle
SELECT *
FROM TABLE t
WHERE t.date BETWEEN TO_DATE('2010-07-29', 'YYYY-MM-DD') - 1
AND TO_DATE('2010-07-29', 'YYYY-MM-DD')
I used BETWEEN because the date column is likely DATETIME (on MySQL & SQL Server, vs DATE on Oracle), which includes the time portion so equals means the value has to equal exactly. These queries give you the span of a day.
If you're using Oracle, you can use the + and - operators to add a number of days to a date.
http://psoug.org/reference/date_func.html
Example:
SELECT SYSDATE + 1 FROM dual;
Will yield tomorrow's date.
If you're not using Oracle, please tell use what you ARE using so we can give better answers. This sort of thing depends on the database you are using. It will NOT be the same across different databases.
Depends of the DateTime Functions available on the RDBMS
For Mysql you can try:
mysql> SELECT DATE_ADD('1997-12-31',
-> INTERVAL 1 DAY);
mysql> SELECT DATE_SUB('1998-01-02', INTERVAL 31 DAY);
-> '1997-12-02'
If youre using MSSQL, you're looking for DateAdd() I'm a little fuzzy on the syntax, but its something like:
Select * //not really, call out your columns
From [table]
Where date = DateAdd(dd, -1, "2010-07-29",)
Edit: This syntax should be correct: it has been updated in response to a comment.
I may have the specific parameters in the wrong order, but that should get you there.
In PL SQL : select sysdate+1 from dual;