I changed all column in the expected output. If you see in column 4 I have change P201835 to P201829. For column 2 and 10. It is like changing the YYYY , If you look I changed 1832 to 1826. And in Column 10 for last row I changed from RHS_12_1515FF_0706
to RHS_12_1510FF_0706
Ran it and created a same ouput as original
Last edited by arunkumar_mca; 11-02-2018 at 01:17 PM..
Hi i am trying to subtract days from current date. For example todays date is 10/03/2006. If i subtract 2 days it should give 8/03/2006. I am also trying to find the access date of a file in dd/mm/yyyy format. Can any one please help in how to do this.
Ramesh (1 Reply)
I have looked through the forums and found many date / time manipulation tools, but cannot seem to find something that fits my needs for the following.
I have a log file with date time stamps like this:
Jun 21 17:21:52
Jun 21 17:24:56
Jun 21 17:27:59
Jun 21 17:31:03
Jun 21 17:34:07
Jun... (0 Replies)
Dear Expert,
Is there a command to do that in Unix?
In such a way that we don't need to actually "write" or
modified the content.
-- monkfan (4 Replies)
how can we add or subtract days from the output of date command in unix...
like if i want to subtract a day from the result of date command like this..
v_date=`date +%Y%m%d`
this wud give me 20080519
now i want to subtract one day from this.. so tht it wud give me 20080518..
how do i do... (1 Reply)
Hi all,
I'm using Red Hat Linux and want to move some folders and files around but not change the modified date. Is this possible?
I know cp has a -p flag which seems to do what I want, but this is a large volume of data so copying and deleting would not be feasible. (13 Replies)
I got a statement like below to subtract 1 from given date using teradata. I am looking for a one line unix command to perform the same.
select 'parse_this_record', (DATE '${FILE_DATE}' - 1) (FORMAT 'YYYY-MM-DD');
Input: 2012-02-21
Expected Output: 2012-02-20
PS: One liner because I am... (2 Replies)
I have a CSV file with a date format like this;
11/19/2012 17:37:00,1.372,121.6
11/19/2012 17:38:00,0.743,121.6
Want to change the time stamp to seconds after 1970 so I can get the data in rrdtool. For anyone interested, this is data from a TED5000 unit and is Kwatts and volts.
Needs to... (3 Replies)
Hi All,
I am getting a date from environment variable and want to do some processing by subtracting 2 months from the date passed through the environment variable.
I am trying the following syntax :
date_var=2014-08-31
date_2M_ago='$date_var+"%d%m%y" --$date_var="2 months ago" '... (3 Replies)
I am trying to achieve to get only the month and the day. Example Feb 5 (as you can see if it is feb 1-9) the space is 2. If it is feb 10-28, the space is only 1. I am trying to right a script that will list a directory and shoot an email if there is an activity in last 7 days. I dont really trust... (5 Replies)
Discussion started by: invinzin21
5 Replies
LEARN ABOUT MOJAVE
colrm
COLRM(1) BSD General Commands Manual COLRM(1)NAME
colrm -- remove columns from a file
SYNOPSIS
colrm [start [stop]]
DESCRIPTION
The colrm utility removes selected columns from the lines of a file. A column is defined as a single character in a line. Input is read
from the standard input. Output is written to the standard output.
If only the start column is specified, columns numbered less than the start column will be written. If both start and stop columns are spec-
ified, columns numbered less than the start column or greater than the stop column will be written. Column numbering starts with one, not
zero.
Tab characters increment the column count to the next multiple of eight. Backspace characters decrement the column count by one.
ENVIRONMENT
The LANG, LC_ALL and LC_CTYPE environment variables affect the execution of colrm as described in environ(7).
EXIT STATUS
The colrm utility exits 0 on success, and >0 if an error occurs.
SEE ALSO awk(1), column(1), cut(1), paste(1)HISTORY
The colrm command appeared in 3.0BSD.
BSD August 4, 2004 BSD