I am having problems using sed to change this:
script_summary("Short random text");
script_id(#12345);
Into this:
script_tag(name:"text", value:"Short random text");
script_oid("1.3.6.1.4.1.25623.1.0.#12345");
Could you please, point me out in the right direction?
Thank you very much in advance.
Best regards,
What about:
Using 2 iterations. Testing it:
echo 'script_summary("Short random text");' | sed -e 's/script_summary("\(.*\)");/script_tag(name:"text", value:"\1");/'
echo 'script_id(#12345);' | sed -e 's/script_id(\(#[0-9]\+\));/script_oid("1.3.6.1.4.1.25623.1.0.\1");/'
So you may want to use it with
sed -i -e <regex1> <file>
sed -i -e <regex2> <file>
or
cat <file> | sed -e <regex1> | sed -e <regex2>
depends on how big the files is, and what you want to do with it.
I want sed to give me a single line output irrespective of whether the matched pattern is found and substituted, or even if there is no pattern match, with same command options.
1. echo "700K" | sed -n 's/[A-Z]//gp' // gives one output
2. echo "700" | sed -n 's/[A-Z]//gp' // no output
Is there any way in sed i can get a single output for second case without removing the "-n" option, forcing it to print the input irrespective of substitution made or not?
It is not clear for me why you need to keep the -n option but if you really do need to keep it you can use the following sed command:
echo "700" | sed -n 's/[A-Z]//g;p'
this will first make the substitution if possible then print the line.
output:
You don't need to mess with all these sed options. Use sed in it's simpliest format which will make a substitution if pattern is found:
$ echo "700K" | sed 's/[A-Z]//g'
700
$ echo "700" | sed 's/[A-Z]//g'
700
$ sed --version
sed (GNU sed) 4.4
$ sed 's/[A-Z]//g' <<<$'700\n700K\n500\n3500A'
700
700
500
3500
I am trying to read/grep a particular word or content that is before a period (.).
e.g. file1 has abinaya.ashok and I want to grep whatever is before the period (.) without hardcoding anything.
if I try
grep \.\ file1
it gives abinaya.ashok.
I've tried: grep\*\.\ file1
it doesn't give anything.Can we find it using grep commands or should we do it only using awk command? Any thoughts?
Using GNU grep for PCRE regex (for non-greedy and positive look-ahead), you can do:
echo 'abinaya.ashok' | grep -oP '.*?(?=\.)'
abinaya
Using awk:
echo 'abinaya.ashok' | awk -F\. '{print $1}'
abinaya
Check the following simple examples.
Including the dot:
$ echo abinaya.ashok | grep -o '.*[.]'
abinaya.
Without the dot:
$ echo abinaya.ashok | grep -o '^[^.]\+'
abinaya
Hope I understand you correctly:
sed -n 's/\..*//p' file1 | grep whatever
sed expression will print only part before dot (lines without dot are not printed).
Now use grep to search what you need.
I have a product which has a command called db2level whose output is given below
I need to extract 8.1.1.64 out of it, so far i came up with,
db2level | grep "DB2 v" | awk '{print$5}'
which gave me an output v8.1.1.64",
Please help me to fetch 8.1.1.64. Thanks
grep is enough to do that:
db2level| grep -oP '(?<="DB2 v)[\d.]+(?=", )'
Just with awk:
db2level | awk -F '"' '$2 ~ /^DB2 v/ {print substr($2,6)}'
db2level | grep "DB2 v" | awk '{print$5}' | sed 's/[^0-9\.]//g'
remove all but numbers and dot
sed is your friend for general extraction tasks:
db2level | sed -n -e 's/.*tokens are "DB2 v\([0-9.]*\)".*/\1/p'
The sed line does print no lines (the -n) but those where a replacement with the given regexp can happen. The .* at the beginning and the end of the line ensure that the whole line is matched.
Try grep with -o option:
db2level | grep -E -o "[0-9]+\.[0-9]+\.[0-9]\+[0-9]+"
Another sed solution
db2level | sed -n -e '/v[0-9]/{s/.*DB2 v//;s/".*//;p}'
This one desn't rely on the number being in a particular format, just in a particular place in the output.
db2level | grep -o "v[0-9.]*" | tr -d v
Try s.th. like db2level | grep "DB2 v" | cut -d'"' -f2 | cut -d'v' -f2
cut splits the input in parts, seperated by delimiter -d and outputs field number -f
I have a source input, input.txt
a.txt
b.txt
c.txt
I want to feed these input into a program as the following:
my-program --file=a.txt --file=b.txt --file=c.txt
So I try to use xargs, but with no luck.
cat input.txt | xargs -i echo "my-program --file"{}
It gives
my-program --file=a.txt
my-program --file=b.txt
my-program --file=c.txt
But I want
my-program --file=a.txt --file=b.txt --file=c.txt
Any idea?
Don't listen to all of them. :) Just look at this example:
echo argument1 argument2 argument3 | xargs -l bash -c 'echo this is first:$0 second:$1 third:$2'
Output will be:
this is first:argument1 second:argument2 third:argument3
None of the solutions given so far deals correctly with file names containing space. Some even fail if the file names contain ' or ". If your input files are generated by users, you should be prepared for surprising file names.
GNU Parallel deals nicely with these file names and gives you (at least) 3 different solutions. If your program takes 3 and only 3 arguments then this will work:
(echo a1.txt; echo b1.txt; echo c1.txt;
echo a2.txt; echo b2.txt; echo c2.txt;) |
parallel -N 3 my-program --file={1} --file={2} --file={3}
Or:
(echo a1.txt; echo b1.txt; echo c1.txt;
echo a2.txt; echo b2.txt; echo c2.txt;) |
parallel -X -N 3 my-program --file={}
If, however, your program takes as many arguments as will fit on the command line:
(echo a1.txt; echo b1.txt; echo c1.txt;
echo d1.txt; echo e1.txt; echo f1.txt;) |
parallel -X my-program --file={}
Watch the intro video to learn more: http://www.youtube.com/watch?v=OpaiGYxkSuQ
How about:
echo $'a.txt\nb.txt\nc.txt' | xargs -n 3 sh -c '
echo my-program --file="$1" --file="$2" --file="$3"
' argv0
It's simpler if you use two xargs invocations: 1st to transform each line into --file=..., 2nd to actually do the xargs thing ->
$ cat input.txt | xargs -I# echo --file=# | xargs echo my-program
my-program --file=a.txt --file=b.txt --file=c.txt
You can use sed to prefix --file= to each line and then call xargs:
sed -e 's/^/--file=/' input.txt | xargs my-program
Here is a solution using sed for three arguments, but is limited in that it applies the same transform to each argument:
cat input.txt | sed 's/^/--file=/g' | xargs -n3 my-program
Here's a method that will work for two args, but allows more flexibility:
cat input.txt | xargs -n 2 | xargs -I{} sh -c 'V="{}"; my-program -file=${V% *} -file=${V#* }'
I stumbled on a similar problem and found a solution which I think is nicer and cleaner than those presented so far.
The syntax for xargs that I have ended with would be (for your example):
xargs -I X echo --file=X
with a full command line being:
my-program $(cat input.txt | xargs -I X echo --file=X)
which will work as if
my-program --file=a.txt --file=b.txt --file=c.txt
was done (providing input.txt contains data from your example).
Actually, in my case I needed to first find the files and also needed them sorted so my command line looks like this:
my-program $(find base/path -name "some*pattern" -print0 | sort -z | xargs -0 -I X echo --files=X)
Few details that might not be clear (they were not for me):
some*pattern must be quoted since otherwise shell would expand it before passing to find.
-print0, then -z and finally -0 use null-separation to ensure proper handling of files with spaces or other wired names.
Note however that I didn't test it deeply yet. Though it seems to be working.
xargs doesn't work that way. Try:
myprogram $(sed -e 's/^/--file=/' input.txt)
It's because echo prints a newline. Try something like
echo my-program `xargs --arg-file input.txt -i echo -n " --file "{}`
I was looking for a solution for this exact problem and came to the conclution of coding a script in the midle.
to transform the standard output for the next example use the -n '\n' delimeter
example:
user#mybox:~$ echo "file1.txt file2.txt" | xargs -n1 ScriptInTheMiddle.sh
inside the ScriptInTheMidle.sh:
!#/bin/bash
var1=`echo $1 | cut -d ' ' -f1 `
var2=`echo $1 | cut -d ' ' -f2 `
myprogram "--file1="$var1 "--file2="$var2
For this solution to work you need to have a space between those arguments file1.txt and file2.txt, or whatever delimeter you choose, one more thing, inside the script make sure you check -f1 and -f2 as they mean "take the first word and take the second word" depending on the first delimeter's position found (delimeters could be ' ' ';' '.' whatever you wish between single quotes .
Add as many parameters as you wish.
Problem solved using xargs, cut , and some bash scripting.
Cheers!
if you wanna pass by I have some useful tips http://hongouru.blogspot.com
Actually, it's relatively easy:
... | sed 's/^/--prefix=/g' | xargs echo | xargs -I PARAMS your_cmd PARAMS
The sed 's/^/--prefix=/g' is optional, in case you need to prefix each param with some --prefix=.
The xargs echo turns the list of param lines (one param in each line) into a list of params in a single line and the xargs -I PARAMS your_cmd PARAMS allows you to run a command, placing the params where ever you want.
So cat input.txt | sed 's/^/--file=/g' | xargs echo | xargs -I PARAMS my-program PARAMS does what you need (assuming all lines within input.txt are simple and qualify as a single param value each).
There is another nice way of doing this, if you do not know the number of files upront:
my-program $(find . -name '*.txt' -printf "--file=%p ")
Nobody has mentioned echoing out from a loop yet, so I'll put that in for completeness sake (it would be my second approach, the sed one being the first):
for line in $(< input.txt) ; do echo --file=$line ; done | xargs echo my-program
Old but this is a better answer:
cat input.txt | gsed "s/\(.*\)/\-\-file=\1/g" | tr '\n' ' ' | xargs my_program
# i like clean one liners
gsed is just gnu sed to ensure syntax matches version brew install gsed or just sed if your on gnu linux already...
test it:
cat input.txt | gsed "s/\(.*\)/\-\-file=\1/g" | tr '\n' ' ' | xargs echo my_program