sed --> sort data by date

Shell Programming and Scripting
1

Hi,

i "tried" to sort data by date. So far, i used sed to take the data from the last and the actual month. Now, after changing the year it is not working properly.

i use:
GNU bash, version 4.2.45(1)-release (x86_64-suse-linux-gnu)

sed -n '/[0-3][0-9]\/['$y']/p' $Home/../scripte/pd_0.txt

y is a variable, build bei the shell-command y=`date +%y`

### data-format in pd_0.txt
2985,12/31/13,23:30,92637,9135,10018,9758,8504,6555,4573,2834,...
2986,12/31/13,23:40,97195,9281,10080,9868,8752,7091,4792,2951,...
2987,12/31/13,23:50,95258,9453,10164,10154,9285,7146,4919,2982...
2988,01/01/14,00:00,100358,9745,10622,10376,9291,7475,5222,3395,...
2989,01/01/14,00:10,128834,11704,13456,14235,13857,12485,10119,...
2990,01/01/14,00:20,165685,13613,16164,17818,18713,18242,15764,....

after all i'am very confused about. espacially why it works properly last year.
what i want to filter, ist the data from the last two month - e.g. from today, the data from Dec 2013 and the Data from Jan 2014.
Can someone help me please.

Thanks in advance,
IMPe

2

A best practice is to use the YYYYMMDD format ( date '+%Y%m%d' ) for the date, so that further sorting are made easy.

Not that YYYY is better than YY because retaining only the 2 last digit suppose a loss of information and may lead to ambiguous situations.

1 Like
3

Hi ctsgnb,

thanks for your fast repley.
what i forgot to say: the meassurement data and data-format is build by a device. it is not possible for me to change it. So i choose it this way, because the data-format is fix and csv: MM/DD/YY, hh:mm, 1234, 5678 .

Thanks in advance,
IMPe

4

To keep the same formatting:

sed 's;[:/];,;g' yourfile | sort -t, -k 4,4 -k 2,3 -k 5,6 | sed 's/,/:/5;s:,:/:3;s:,:/:2'

or move the YY at the beginning of the date so that the format will change into YY/MM/DD:

sed 's:\(../..\)/\(..\):\2/\1:' yourfile | sort -t, -k 2,3

As mentionned in my previous post, retaining only YY loose information and may lead to unexpected result (example: 12 may mean 1912 or 2012 or whatever ???12...)

---------- Post updated at 14:43 ---------- Previous update was at 13:07 ----------

I just got aware i've missunderstood your question:

Before running your

sed -n '/[0-3][0-9]\/['$y']/p' $Home/../scripte/pd_0.txt

You need to check wether you are in january : if so, the %y should be 134 rather than 14

if [ $(date +%m) = "01" ]
then
y=$(echo $(date +%y)$( echo $(($(date +%Y)-1 )) | tail -c 3 ) )
else
y=$(date +%y)
fi

should be sufficient since [1413] in regex will give the same result than [134] (it's just a little less clean)

By the way, you will have to care of century jump because your %y will go from 99 to 00, that's why i calculate last year using %Y (uppercase) rather than %y.

1 Like
5

Hi ctsgnb,

thanks a lot for your extensive support. It take some time for me, to understand all your methods and proposals. But again i have learned more about sed and i solved my problem properly.

Thanks a lot,
IMPe

6

Hi All,

Actually I am new to this Forum.Don't know how to post my question.

Can anyone help me :

we can replace dbSQL( with dbSQLWithBind( value by using :
:.,$s/dbSQL(/dbSQLWithBind(/g command inside a script.

For replacing ^M characters from a file, we normally use this command:

perl -i -p -e 's/^M//g' *.txt

But i am failing in replacing dbSQL( with dbSQLWithBind( value through perl command.
Please help me in replacing the value dbSQL( with dbSQLWithBind( globally in all scripts without opening the scripts.

7

for Vara Prasad:

cat file | sed 's/\^M//g'