I have a tar.gz file.
Because of space issues and the time required extract is longer, I need to extract only the selected file.
I have tried the below
grep -l '<text>' *
file1
file2
only file1,file2 should be extracted.
What should I Do to SAVE all the tail -f data to a FILE swa3?
I have swa1.out which has list of online data inputs.
swa2 is a file which should skip the keywords from swa1.
swa3 is a file where it should write the data.
Can anyone help in this?
I have tried below commnad, but I'm not able to get it
tail -f SWA1.out |grep -vf SWA2 >> swa3
You can use do this with --extract option like this
tar --extract --file=test.tar.gz main.c
Here in --file , specify the .gz filename and at the end specify the
filename you want to extract.
Related
I'm trying to batch process a folder full of text files with pandoc, and I'd like to maintain the current filenames. How do I call the filename as a variable in the output? For example, I want to write a command like this:
pandoc -s notes/*.txt -o rtf/$1.rtf
Where $1 represents the filename grabbed with the * character.
I'm sure this is a simple question, but I don't quite know the right language to search for it properly.
Thanks for any help!
Try
for file in notes/*txt
do
file_base_name=$(basename "${file}" | cut -d'.' -f1)
pandoc -s "$file" -o rtf/${file_base_name}.rtf
done
I am trying to grep the files inside tar.gz.
tar.gz can have folders and files so I want to search through all and print the o/p like grep prints.
I tried following (file_path has many tar.gz archives)
zgrep -B 4 "Token to Search" /file_path/*
I am getting o/p as :
abc.tar.gz:Binary file (standard input) matches
I want lines that matches zgrep and also 4 lines before that matched line,
tar.gz can have /Folder1/file.log etc.
Is it possible to do such zgrep ?
You should use -a option. In man zgrep it says:
All options specified are passed directly to grep.
And in man grep:
-a, --text
Process a binary file as if it were text; this is equivalent to the
--binary-files=text option.
I need to create a script that loops through files in a directory, check if the filename is in "list.txt" then process it. My problem is that the filenames are dynamic since it has timestamp.
Is there a way to grep approximate match in unix?
Sample.
list.txt
SAMPLE_REPORT_1
SAMPLE_REPORT_2
Report Filenames
SAMPLE_REPORT_1_20180416121345.csv
SAMPLE_REPORT_2_20180416121645.csv
I need to check if the filenames are in list.txt
bash + grep solution:
for f in *.csv; do
if grep -qx "${f%_*.csv}" list.txt; then
# processing file $f
fi
done
I use UNIX fairly infrequently so I apologize if this seems like an easy question. I am trying to loop through subdirectories and files, then generate an output from the specific files that the loop grabs, then pipe an output to a file in another directory whos name will be identifiable from the input file. SO far I have:
for file in /home/sub_directory1/samples/SSTC*/
do
samtools depth -r chr9:218026635-21994999 < $file > /home/sub_directory_2/level_2/${file}_out
done
I was hoping to generate an output from file_1_novoalign.bam in sub_directory1/samples/SSTC*/ and to send that output to /home/sub_directory_2/level_2/ as an output file called file_1_novoalign_out.bam however it doesn't work - it says 'bash: /home/sub_directory_2/level_2/file_1_novoalign.bam.out: No such file or directory'.
I would ideally like to be able to strip off the '_novoalign.bam' part of the outfile and replace with '_out.txt'. I'm sure this will be easy for a regular unix user but I have searched and can't find a quick answer and don't really have time to spend ages searching. Thanks in advance for any suggestions building on the code I have so far or any alternate suggestions are welcome.
p.s. I don't have permission to write files to the directory containing the input folders
Beneath an explanation for filenames without spaces, keeping it simple.
When you want files, not directories, you should end your for-loop with * and not */.
When you only want to process files ending with _novoalign.bam, you should tell this to unix.
The easiest way is using sed for replacing a part of the string with sed.
A dollar-sign is for the end of the string. The total script will be
OUTDIR=/home/sub_directory_2/level_2
for file in /home/sub_directory1/samples/SSTC/*_novoalign.bam; do
echo Debug: Inputfile including path: ${file}
OUTPUTFILE=$(basename $file | sed -e 's/_novoalign.bam$/_out.txt/')
echo Debug: Outputfile without path: ${OUTPUTFILE}
samtools depth -r chr9:218026635-21994999 < ${file} > ${OUTDIR}/${OUTPUTFILE}
done
Note 1:
You can use parameter expansion like file=${fullfile##*/} to get the filename without path, but you will forget the syntax in one hour.
Easier to remember are basename and dirname, but you still have to do some processing.
Note 2:
When your script first changes the directory to /home/sub_directory_2/level_2 you can skip the basename call.
When all the files in the dir are to be processed, you can use the asterisk.
When all files have at most one underscore, you can use cut.
You might want to add some error handling. When you want the STDERR from samtools in your outputfile, add 2>&1.
These will turn your script into
OUTDIR=/home/sub_directory_2/level_2
cd /home/sub_directory1/samples/SSTC
for file in *; do
echo Debug: Inputfile: ${file}
OUTPUTFILE="$(basename $file | cut -d_ -f1)_out.txt"
echo Debug: Outputfile: ${OUTPUTFILE}
samtools depth -r chr9:218026635-21994999 < ${file} > ${OUTDIR}/${OUTPUTFILE} 2>&1
done
I have a zipped file containing 10,000 compressed files. Is there a Linux command/bash script to unzip only 1,000 files ? Note that all compressed files have same extension.
unzip -Z1 test.zip | head -1000 | sed 's| |\\ |g' | xargs unzip test.zip
-Z1 provides a raw list of files
sed expression encodes spaces (works everywhere, including MacOS)
You can use wildcards to select a subset of files. E.g.
Extract all contained files beginning with b:
unzip some.zip b*
Extract all contained files whose name ends with y:
unzip some.zip *y.extension
You can either select a wildcard pattern that is close enough, or examine the output of unzip -l some.zip closely to determine a pattern or set of patterns that will get you exactly the right number.
I did this:
unzip -l zipped_files.zip |head -1000 |cut -b 29-100 >list_of_1000_files_to_unzip.txt
I used cut to get only the filenames, first 3 columns are size etc.
Now loop over the filenames :
for files in `cat list_of_1000_files_to_unzip.txt `; do unzip zipped_files.zip $files;done
Some advices:
Execute zip to only list a files, redirect output to some file
Truncate this file to get only top 1000 rows
Pass the file to zip to extract only specified files