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Don't see LINUX or LVM in the output of fdisk -l. Maybe you are using other than GNU Linux?

What I see here...
NewDisk=`fdisk -l | grep Linux | egrep -v '\*|LVM' |awk '{print $1}' | sed...

If you're doing grep | sed | grep -v | awk | sed | cut | kitchen | sink, just use awk and be done with it.

fdisk prints 'Linux' on my system, not LINUX, but adjust to taste.

$ fdisk -l | awk...

In sed you can replace the slash of the substitute command with almost any character .
$ echo "Turn forward // into dashes" | sed 's/\//-/g'
Turn forward -- into dashes

Using | instead of / :...

Tryawk '{gsub("[\t ]*","");$1=$6"\t"$1; sub (/^[^:]*:/,"",$1); $6=""}1' RS= ORS="\n" FS="\n" OFS="," file

Just put the code in one line!
awk 'NF{if($0~/Six/){H=$0;sub(/Six:[ \t]*/,X,H)}else{gsub("\t",X,$0);S=S?S OFS $0:$0}next}!NF{print H,S;S=""}END{print H,S}' OFS=, file

Try:
sed 's/^\(...\) \(...\) \([0-9][0-9]\).*2011/\2 \3 2011/' INPUTFILE
It works only for 2011 year but it's easy enough to fix this.

PS And you should'n have other "2011" in your strings.

Through Sed..
sed 's/ *, */,/g' inputfile

Hi,

Using 'perl':

$ perl -pe 's/\s*,\s*/,/g' infile
Sat Jul 23 16:10:03 EDT 2011,12345678,PROD,xyz_2345677,testuuyt,149693,abtryu
Jul 22 2011,NULL,PROD,yt12345,testuuytoy,162469,abtryu
Jul...

You can use also builtin calculation, tested using ksh and bash

echo $(( 10+RANDOM%21 ))
# or more generic
min=10
max=30
((diff=max-min+1))
echo $(( min + RANDOM%diff ))

echo "10+$RANDOM%10*(2)+2 " | bc

will generate random numbers in 12~30 range, if that suffice.

Oh God,

You changed the order!!!

try


MM="Jul"
DD=20
nawk -v d=$DD -v m=$MM '$2==m && $3==d' input_file

awk -F# '{gsub(".","x",$NF)}1' OFS=# file

You can use od to show the last characacter of a file.
$ echo -n hello > testfile1
$ echo hello > testfile2
$ file=testfile1
$ od -An -tc -j $(( $(ls -l $file | awk '{print $5}') - 1 )) $file
...

This should do it.


awk -F'--' '{print $1}' $ORACLE_CODE_FILENAME
removing blank lines is left as an exercise for the reader ;)

This command should do the job:
awk '/--/;/\/\*/{p=1};p;/\*\//{p=0}' file
This is the output I get:
$ cat file
Tue Mar 25 23:35:00 GMT 2008
-- COMMENT Tue Mar 25 21:34:23 GMT 2008
Tue Mar 25...

Oh my... ok, let me give it a shot:

#!/bin/ksh
#%W% %I% %D% %T% ---- these are comments (starting with #)
#%W%G --- they don't do anything

The beginning of this line here://g'
Seems like...