1. Not sure what your original code was before you edited it as your updated code is identical.
2. There is a space in the first position of each line on your input file so the condition ^> will not find any lines.
3. What is the requirement as to why you need to use substr as your solution appears to work with a slight modification?
4. Based on your input and output files, length>=7 should be length>7 in your solution after removing the spaces in first position of each line.
What is the more efficient way to do this (awk only and default FS) ?
$ echo "jefe@alm"|awk '{pos = index($0, "@");printf ("USER: %s\n",substr ($0,1,pos-1))}'
USER: jefe
Thx in advance (2 Replies)
Hi,
My input file is
41;2;xxxx;yyyyy....
41;2;xxxx;yyyyy....
41;2;xxxx;yyyyy....
..
..
I need to change the second field value from 2 to 1. i.e.,
41;1;xxxx;yyyyy....
41;1;xxxx;yyyyy....
41;1;xxxx;yyyyy....
..
..
Thanks in advance. (9 Replies)
Hi,
I have a long string like,
aabab|bcbcbcbbc|defgh|paswd123 dedededede|efef|ghijklmn|paswd234 ghghghghgh|ijijii|klllkkk|paswd345 lmlmlmmm|nononononn|opopopopp|paswd456
This string is devided into one space between substrings. This substrings are,
aabab|bcbcbcbbc|defgh|paswd123... (6 Replies)
Hi I am trying to run this command in ksh ...its not working
$line="123356572867116w1671716"
actual_length = 16
cut_line=`awk 'BEGIN{print substr(ARGV,1,actual_length)}' "$line"`
the substr is not giving me an output
how can i make it done
can anyone hwlp me on this
cut_line=`awk... (2 Replies)
Hi, i'm a newbie and i don't know unix...
I'm a dba oracle.
I need to cat the content of a file like this:
> ps -eaf|grep pmon
oracle 221422 1 0 Sep 17 - 7:20 ora_pmon_ORCL
oracle 405626 1 0 Sep 17 - 8:39 ora_pmon_ORCL1
oracle 491534 1 0 ... (3 Replies)
Hello life savers!!
Is there any way to use substr in awk command for returning one part of a string from declared start and stop point?
I mean I know we have this:
substr(string, start, length)
Do we have anything like possible to use in awk ? :
substr(string, start, stop)
... (9 Replies)
I have a command like this:
listdb ID923 -l |gawk '{if (substr($0,37,1)==1 && NR == 3)print "YES" else if (substr ($0,37,1)==0 && NR == 3) print "NO"}'
This syntax doesn't work. But I was able to get this to work:
listdb ID923 -l |gawk '{if (substr($0,37,1)==1 && NR == 3)print "YES"}'
... (4 Replies)
Hi,
- In a file test.wmi
Col1 | firstName | lastName
4003 | toto_titi_CT- | otot_itit
- I want to have only ( colones $7,$13 and $15) with code 4003 and 4002. for colone $13 I want to have the whole name untill _CT- or _GC-
1- I used the command egrep with awk
#egrep -i... (2 Replies)
Discussion started by: georg2014
2 Replies
LEARN ABOUT PLAN9
pr
PR(1) General Commands Manual PR(1)NAME
pr - print file
SYNOPSIS
pr [ option ... ] [ file ... ]
DESCRIPTION
Pr produces a printed listing of one or more files on its standard output. The output is separated into pages headed by a date, the name
of the file or a specified header, and the page number. With no file arguments, pr prints its standard input.
Options apply to all following files but may be reset between files:
-n Produce n-column output.
+n Begin printing with page n.
-b Balance columns on last page, in case of multi-column output.
-d Double space.
-en Set the tab stops for input text every n spaces.
-h Take the next argument as a page header (file by default).
-in Replace sequences of blanks in the output by tabs, using tab stops set every n spaces.
-f Use formfeeds to separate pages.
-ln Take the length of the page to be n lines instead of the default 66.
-m Print all files simultaneously, each in one column.
-n Number the lines of each file.
-on Offset the left margin n character positions.
-sc Separate columns by the single character c instead of aligning them with white space. A missing c is taken to be a tab.
-t Do not print the 5-line header or the 5-line trailer normally supplied for each page.
-wn For multi-column output, take the width of the page to be n characters instead of the default 72.
SOURCE
/sys/src/cmd/pr.c
SEE ALSO cat(1), lp(1)PR(1)