I am currently updating an old site where the date used in the database is formated dd/mm/yyyy.
What I am trying to do is group the rows by month and year. Is this possible with this formatting?
You can always call STR_TO_DATE(value,'%d/%m/%Y').
Then you'll be able to group rows using ordinary date-extracting functions like YEAR() and MONTH().
Whats the column type for the column where date is stored?
if its DateTime then you can do this:
group by YEAR(date_col), MONTH(date_col)
Or you can GROUP BY with substr like
SELECT *
FROM table
GROUP BY substr(date, 4, 2)
where substr(date, 4, 2) = MM and if you want dd let's try substr(date, 1, 2)
Related
I'm trying to sort by a column called date which stores various dates in format MMYYYY (for example, 122020, 102019, etc.).
The SQL query that I have looks like this:
SELECT `date`, `invested` FROM `savings` WHERE `id` = 123 ORDER BY STR_TO_DATE(`date`, "%m%Y") ASC but this query does not sort the output correctly.
Any ideas on how to sort it properly?
Important note: Reclassifying/modifying the column's type is not an option at this point.
EDIT: My current SQL query, which is described above, sorts the dates like this: 102019, 102020, 112019, 112020. But my goal is to have it like this: 102019, 112019, 102020, 112020.
Thank you
Try separating the date string into two parts (year and month), cast each part as integer, and then ordering by that number:
SELECT
date_,
invested
FROM
savings
WHERE
id = 123
ORDER BY
CAST(SUBSTRING(date_, 3, 6) AS UNSIGNED),
CAST(SUBSTRING(date_, 1, 2) AS UNSIGNED)
;
This will do the job :) refer to this DB Fiddle for clarification.
I have a column ifd0_DateTime , in a table named photo, which contains date time in following format: 1966:12:22 17:19:57.
I need to count the number of photos month wise for every year.
So far I have this query. but it doesn't work correctly.
SELECT ifd0_DateTime, count(*) FROM photo
group by YEAR(ifd0_DateTime), MONTH(ifd0_DateTime);
Could anyone please fix this ?
SELECT STR_TO_DATE(LEFT(ifd0_DateTime, 7), '%Y:%m'),
COUNT(*) AS dateCnt
FROM photo
GROUP BY STR_TO_DATE(LEFT(ifd0_DateTime, 7), '%Y:%m')
Instead of date select the Year and Month
Try this
SELECT YEAR(ifd0_DateTime), MONTH(ifd0_DateTime), count(*)
FROM photo
group by YEAR(ifd0_DateTime), MONTH(ifd0_DateTime)
The proper way to store a date time is using the built-in data types. However, it looks like you are storing the value as a string. If so, you just want the first 7 characters:
SELECT LEFT(ifd0_DateTime, 7) as yyyymm, count(*)
FROM photo
GROUP BY LEFT(ifd0_DateTime, 7)
ORDER BY yyyymm;
However, you really should fix the data.
my table and fields are like these:
i must find $sy<year<$ey then it must filter only values by $sm<month<$em at last it must find $sd<day<$ed
i need to find records between dates for example like 2010/10/25 , 2010/10/10
at first i tried :
SELECT SUM(barname) allin,SUM(rooz) allhoghogh,user_id FROM work_result
WHERE (`year`>='$sy' and `month`>='$sm' and `day`>='$sd') and (`year`<='$ey' and `month`<='$em' and `day`<='$ed') group by user_id ;
but it cant find records for dates like e like 2010/10/25 , 2010/10/28
than i tried
SELECT * FROM work_result as t1 join work_result as t2 on t1.year<='$sy' and t2.year>='$ey' and t1.month<='$em' and t2.month>='$sm' and t1.day<='$ed' and t2.day>='$sd' WHERE 1 group by t1.wrid
this isnt usful in my case!
i need some thing like priority select first select all between years than month and than day!!
other way is convert mysql records to timestamp by year and month and day and compare it by input date but UNIX_TIMESTAMP('year-month-day 00:00:00') dont worked correct for me.
i used it like :
SELECT * FROM `work_result` WHERE UNIX_TIMESTAMP('year-month-day 00:00:00')>1238921453
If convert to timestamp didn't work for you what about use date_format to convert:
SELECT *
FROM `work_result`
WHERE date_format(concat(year,'-',month,'-',day), '%Y-%m-%d') >
DATE_FORMAT(FROM_UNIXTIME(`yourDateGoesHere`), '%Y-%m-%d')
my dates in my table are strings in the format:
"10/12/2009"
Now how would one get all the records from a month, lets say June (number "6" being provided)?
Check the MySQL function STR_TO_DATE.
You should not store dates as string, however. Use the type DATE.
The short answer to your question is that you can use the STR_TO_DATE and MONTH functions to 1) convert the string representation into a DATE, and 2) extract the month component from the date:
SELECT t.*
FROM mytable t
WHERE MONTH(STR_TO_DATE(t.dateasstringcol,'%M/%d/%Y')) = 6
(This is assuming here that by '10/12/2009', you are specifying Oct 12, and not Dec 10. You'd need to adjust the format string if that's not the case.)
Alternatively, if month is indeed the leading part of the date, you could do a simple string comparison, if the month is the leading component:
SELECT t.*
FROM mytable t
WHERE t.dateasstringcol LIKE '6/%'
OR t.dateasstringcol LIKE '06/%'
(You could eliminate one of those predicates, if you have an exact format specified for the striing value representing the date: either if month is always stored as two digits -OR- the month is never stored with a leading zero.)
If you are passing in an argument for the month, e.g. '6', then you could construct your statement something like this:
WHERE t.dateasstringcol LIKE '6' + '/%'
If month is the second component of the date, then you could use:
SELECT t.*
FROM mytable t
WHERE t.dateasstringcol LIKE '%/' + '6' + '/%'
OR t.dateasstringcol LIKE '%/' + '06' + /%'
NOTE:
All of the previous example queries will return rows for June of any year (2009, 2010, 2011)
You can extend those examples, and do something similar with the year, using the YEAR function in place of the MONTH function, or for string comparison
AND t.dateasstringcol LIKE '%/%/2011'
Normally, we'd extract rows for a particular month for a particular year, using a date range, for example:
SELECT t.*
FROM mytable t
WHERE MONTH(STR_TO_DATE(t.dateasstring,'%M/%d/%Y')) >= '2011-06-01'
AND MONTH(STR_TO_DATE(t.dateasstring,'%M/%d/%Y')) < '2011-07-01'
Of course, when the date value is stored as a DATE datatype rather than as a VARCHAR, this means we don't need the STR_TO_DATE and MONTH functions, we can specify a comparison to the native DATE column. This approach allows us to make use of an index on the date column, which can improve query performance on large tables.
SELECT t.*
FROM mytable t
WHERE t.realdatecol >= '2011-06-01'
AND t.realdatecol < '2011-07-01'
The STR_TO_DATE function is your friend here:
SELECT * FROM my_table
WHERE STR_TO_DATE('10/12/2009','%M/%d/%Y') >= '2012-06-01';
MONTH should help here if we want current month or particular month data. e.g:
$month = date('m'); OR particular month.
SELECT * FROM users WHERE MONTH(str_to_date("10/12/2009",'%e/%m/%Y')) = $month;
I have a mysql database that looks like this:
id | userid | timestamp | activity
Timestamp is a datetime data type, I need to get the data grouped by month, day and hour. I am using mysql and php for my scripts.
I am able to do it by month and day with the following query:
$query = "SELECT COUNT(id) as totals FROM security_transactions WHERE YEAR(timestamp) = 2012 GROUP BY MONTH(timestamp), DAY(timestamp)";
I need to do it by month day and hours.
Please help.
Thanks.
You can add , HOUR(TIME(timestamp)) to your group by query providing your column is of DATETIME format
http://dev.mysql.com/doc/refman/5.1/en/date-and-time-functions.html#function_hour
Also, from the error messages put in the comments below, it looks like #Aprentice is not using mysql, but I've improved this answer for others looking for mysql.
I have never used mssql, so I can't test this but the following might work to group by nearest hour:
GROUP BY dateadd(hour, datediff(hour, 0, timestamp, 0)
Just take it one step further and use HOUR() as well. You will first need to extract the time portion of the timestamp. But guess what, there is a function for that as well ;)