Input with columns
994475;1832 + S PP1835 10/17/18;S;P201835;115;N;4,4;M;0;xx994475;*;BA7005;10/17/2018 16:48
994477;1836 + S PP1839 10/17/18;S;P201839;115;N;4,4;M;0;xxh994477;*;BA7005;10/17/2018 16:48
994479;CONTROL 1452-1527,1552-1627;S;P201527;115;N;4,4;M;0;RHS_12_1515FF_0706;*;B7005;10/17/2018 16:49
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
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 DEBIAN
numsum
NUMSUM(1) User Contributed Perl Documentation NUMSUM(1)NAME
numsum - numsum program file
SYNOPSIS
numsum [-iIcdhrsvxy] <FILE>
| numsum [-iIcdhrsvxy] (Input on STDIN from pipeline.)
numsum [-iIcdhrsvxy] (Input on STDIN. Use Ctrl-D to stop.)
DESCRIPTION
numsum will take all the numbers on stdin and return the sum of those numbers. Currently it only processes the first number on each line.
Besides positive numbers, it also handles negative numbers and numbers with decimals.
OPTIONS -i Only return the integer portion of the final sum.
-I Only return the decimal portion of the final sum.
-c Print out the sum of each column.
-r Print out the sum of each row.
-x <n> Specify a comma seperated list of columns to print.
-y <n> Specify a comma seperated list of rows to print.
-s <string> Specify a string to use as a seperator for columns.
This defaults to be consecutive whitespace (s+).
-h Help: You're looking at it.
-V Increase verbosity.
-d Debug mode. For developers
-q Quiet mode, don't print any warnings.
EXAMPLES
Simply add up the numbers in a file.
$ numsum numbers.txt
4315
Enter your own numbers on STDIN. The last number is the answer.
$ numsum
4
21
98
100
223
Use it in a command pipeline.
$ ls -1s | grep .mp3 | numsum -c -x 5
72288
Add up the total byte count in a http log file.
$ cat access_log | awk {'print $10'} numsum
or
numsum -c -x 10 access_log
Add up the columns of numbers of a file.
$ cat columns
1 6 11 16 21
2 7 12 17 22
3 8 13 18 23
4 9 14 19 24
5 10 15 20 25
$ numsum -c columns
15 40 65 90 115
Add up the 1st, 2nd and 5th columns only.
$ numsum -c -x 1,2,5 columns
15 40 115
Add up the rows of numbers of a file.
$ numsum -r columns
55
60
65
70
75
Add up the 2nd and 4th rows.
$ numsum -r -y 2,4 columns
60
70
SEE ALSO numaverage(1), numbound(1), numinterval(1), numnormalize(1), numgrep(1), numprocess(1), numrandom(1), numrange(1), numround(1)COPYRIGHT
numsum is part of the num-utils package, which is copyrighted by Suso Banderas and released under the GPL license. Please read the COPYING
and LICENSE files that came with the num-utils package
Developers can read the GOALS file and contact me about providing
submitions or help for the project.
MORE INFO
More info on numsum can be found at:
http://suso.suso.org/programs/num-utils/
perl v5.10.1 2009-10-31 NUMSUM(1)