Suppose I have file which contains only a text like below:
Test transition to drned-internal-asr9k-rt24711
load drned-internal-asr9k-rt24711
commit**
Now on the terminal if I do
cat filename | grep load
I would get output something like
load drned-internal-asr9k-rt24711
But how can I modify my grep command to get output as
drned-internal-asr9k-rt24711.txt
i.e. remove "load " and add ".txt" at the end. So how to do that??
May be not the best solution but :
cat | grep load | cut -c4- | sed 's/$/.txt/'
cut -c4- will delete the 4 first characters
sed 's/$/.txt/' will add the ".txt" at the end of output
This can be achieved with the following:
sed -nr 's/.*load\s+(.*)/\1.txt/p' file.txt
This matches anything after load (plus one or more spaces) and returns it, adding .txt on the end.
awk '{for(i=1;i<NF;i++){if(tolower($i)~/^load$/){print $(i+1) ".txt"}}}' file.txt
This matches next column after load and append .txt to it in output.
Related
I'd like to populate a shell variable with the file contents of the file that follow the # (but leave the file as-is) from beginning of one single-line file
I tried
file=$(head -n1 $another_filename | sed 's/^#//')
but output is the same. I don't want to delete the whole line, just hash.
Found it with your help of course.
file=$(head -n1 $filename |sed 's/^#//')
I hv a file like below, i need to get the Value of title(which occurs multiple times in file] and store it in separate file
{"card":{"cardName":"43SCX4","portSignalRates":["43SCX4-L-OTU3","43SCX4-C-TENGIGE","43SCX4-C-OTU2","43SCX4-C-FC8G","43SCX4-C-STM64","43SCX4-C-OC192"],"listOfPort":{"43SCX4-L-OTU3":{"portAid":"43SCX4-L-OTU3","signalType":"OTU3","tabNames":["PortDetails","OTU3e2Details"],"title":"OperationalMode",{"label":"Regen","value":"regen"}],"label":"Regen","value":"regen","checked":"","enabled":"true","selected":""},{"type":"dijit.form.Select","name":"Frequency","title":"Transmit Frequency "}}}}
I tried with "awk -F, '{}' sample", i'm able to split,but not able to iterate and put it to another file only "title":"****"
Through grep with -oE option.
$ grep -oE '\"title\":\"[^"]*"' file
"title":"OperationalMode"
"title":"Transmit Frequency "
If you want to get the value of Title then use a lookbehind,
$ grep -oP '(?<=\"title\":\")[^"]*' file
OperationalMode
Transmit Frequency
If you want to save it another file then redirect the output to another file using output redirection > operator. Example grep -oP '(?<=\"title\":\")[^"]*' file > outfile
i have several files in which i want to replace a certain word with the name of the file itself..
for example i have 2 files named test1.txt and test2.txt
both files are equal and look like
bla1,bla2,temp
bla2,bla3,temp
with the sed i want to replace the word temp with the name of the file itself
so after the sed operation i have 2 different files
test1.txt , which looks like :
bla1,bla2,test1
bla2,bla3,test1
test2.txt, which looks like :
bla1,bla2,test2
bla2,bla3,test2
so my question ... how do i use the actual name of the input file itself as part of the replace command?
sed "s/temp/ ??filename??/ ??? " *.txt
thanks for your suggestions
I'm not sure you can reference the filename using sed although I could be wrong. You would probably use a shell hack. A better aproach to substitute all occurrences of temp with the filename would be the following awk script:
$ awk '{gsub(/temp/,FILENAME)}1' file
use awk, awk has FILENAME variable:
awk '{sub(/temp/,FILENAME)}7' yourfile
awk 'BEGIN{FS=OFS=","} {$NF=FILENAME}1' file
The difference between this and the sub() solutions is that this will work even if the word "temp" exists elsewhere in your file, e.g. if "bla1" contains the word "temperature".
If you need to strip ".txt" from the file name as it appears from your posted desired output, tweak it to:
awk 'BEGIN{FS=OFS=","} {t=FILENAME; sub(/\.txt$/,"",t); $NF=t}1' file
You can probably edit FILENAME itself but I find it best not to mess with the builtin variables if you don't have to.
You could do it with a little bit of bash to help you out, if that's available.
find . -name "test*.txt" -type f | awk -F '/' '{print $2;}' | while read file; do sed -i "s|temp|$file|" ./$file; done
That's a kind of hacky adaptation of a script I have to do something similar. It can undoubtedly be shortened.
no sed internal variable for the file name so you need some previous batch command for a generic process
for FileName in MyFileShellFilter
do
cat <> ${FileName} | sed "s|,temp$|,${FileName}|"
done
just be carrefull with file name used, they normaly don't have \ but could have & that are s// special meaning. I use | as separator to allow / in file name but for this reason, no unescaped | are allowed in file name (normaly not)
with xargs:
printf "%s\n" *.txt | xargs -I FILE -L 1 sed 's/temp/FILE/' FILE
The filename cannot have: newlines, slashes, ampersand, single quote.
How do I split a file to N files using as a filename the first 2 chars on the line.
Ex input file:
AA23409234TEXT
BA23201202Other Text
AA23509234YADA
BA23202202More Text.
C1000000000000000000
Should generate 3 files:
AA.txt
AA23409234TEXT
AA23509234YADA
BA.txt
BA23201202Other Text
BA23202202More Text.
C1.txt
C1000000000000000000
I'm thinking of using a sed script similar to this
/^(..)/w \1
But what that really does is create a file named '\1' instead of the capture group.
Any ideas?
$ awk '{fname=substr($0, 0, 2); print >>fname}' input.txt
Or
$ while read line; do echo "$line" >>"${line:0:2}"; done <input.txt
The first thing you need to do is determine all of your file names:
filenames=$(sed 's/\(..\).*/\1/' listOfStrings.txt | sort | uniq)
Then, loop through those filenames
for filename in $filenames
do
sed -n '/^$filename/ p' listOfStrings.txt > $filename.txt
done
I have not tested this, but I think it should work.
This might work for you:
sed 's/\(..\).*/echo "&" >>\1.txt/' file | sh
or if you have GNU sed:
sed 's/\(..\).*/echo "&" >>\1.txt/e' file
I would like the option of extracting the following string/data:
/work/foo/processed/25
/work/foo/processed/myproxy
/work/foo/processed/sample
=or=
25
myproxy
sample
But it would help if I see both.
From this output using cut or perl or anything else that would work:
Found 3 items
drwxr-xr-x - foo_hd foo_users 0 2011-03-16 18:46 /work/foo/processed/25
drwxr-xr-x - foo_hd foo_users 0 2011-04-05 07:10 /work/foo/processed/myproxy
drwxr-x--- - foo_hd testcont 0 2011-04-08 07:19 /work/foo/processed/sample
Doing a cut -d" " -f6 will get me foo_users, testcont. I tried increasing the field to higher values and I'm just not able to get what I want.
I'm not sure if cut is good for this or something like perl?
The base directories will remain static /work/foo/processed.
Also, I need the first line Found Xn items removed. Thanks.
You can do a substitution from beginning to the first occurrence of / , (non greedily)
$ your_command | ruby -ne 'print $_.sub(/.*?\/(.*)/,"/\\1") if /\//'
/work/foo/processed/25
/work/foo/processed/myproxy
/work/foo/processed/sample
Or you can find a unique separator (field delimiter) to split on. for example, the time portion is unique , so you can split on that and get the last element. (2nd element)
$ ruby -ne 'print $_.split(/\s+\d+:\d+\s+/)[-1] if /\//' file
/work/foo/processed/25
/work/foo/processed/myproxy
/work/foo/processed/sample
With awk,
$ awk -F"[0-9][0-9]:[0-9][0-9]" '/\//{print $NF}' file
/work/foo/processed/25
/work/foo/processed/myproxy
/work/foo/processed/sample
perl -lanF"\s+" -e 'print #F[-1] unless /^Found/' file
Here is an explanation of the command-line switches used:
-l: remove line break from each line of input, then add one back on print
-a: auto-split each line of input into an #F array
-n: loop through each line of input
-F: the regexp pattern to use for the auto-split (with -a)
-e: the perl code to execute (for each line of input if using -n or -p)
If you want to just output the last portion of your directory path, and the basedir is always '/work/foo/processed', I would do this:
perl -nle 'print $1 if m|/work/foo/processed/(\S+)|' file
Try this out :
<Your Command> | grep -P -o '[\/\.\w]+$'
OR if the directory '/work/foo/processed' is always static then:
<Your Command>| grep -P -o '\/work\/foo\/processed\/.+$'
-o : Show only the part of a matching line that matches PATTERN.
-P : Interpret PATTERN as a Perl regular expression.
In this example, the last word in the input will be matched .
(The word can also contain dot(s)),so file names like 'text_file1.txt', can be matched).
Ofcourse, you can change the pattern, as per your requirement.
If you know the columns will be the same, and you always list the full path name, you could try something like:
ls -l | cut -c79-
which would cut out the 79th character until the end. That might work in this exact case, but I think it would be better to find the basename of the last field. You could easily do this in awk or perl. Respond if this is not what you want and I'll add the awk and perl versions.
take the output of your ls command and pipe it to awk
your command|awk -F'/' '{print $NF}'
your_command | perl -pe 's#.*/##'