I have a bunch of .txt files in a directory.
I m looking for a command to copy all .txt files and save it with <filename>_2.txt.
Eg: abc.txt -> abc_2.txt (After copy)
Thanks in tons in advance
EDIT: As per OP's extension request adding following code now.
for file in *.txt
do
if [[ ! -f "${file%.*}_MED_2.txt" ]]
then
cp "$file" "${file%.*}_MED_2.txt"
fi
done
Try following.
for file in *.txt
do
echo "cp $file ${file%.*}_2"
done
Above will print cp commands, if you are ok with them then run following.
for file in *.txt
do
if [[ ! -f "${file%.*}_2" ]]
then
cp "$file" "${file%.*}_2"
fi
done
Related
I wanted to uncompress a tar.gz, with appending a suffix to each of the file name. So, for example abc.tar.gz contains files 'first' and 'second', so, after extracting, if I want to append suffix '.append'the file name of each files should be 'first.append' and 'second.append'. Is there a command or way to this?
Note: Files with name 'first' and 'second' are already there, and I wanted to decompress without affecting already available files.
One thing I can think of is uncompress in a temp dir and then, copy all files one by one. But, I wanted to do it in a one shot, if possible, so that, it will save time.
Thanks in advance.
try this:
#!/bin/bash
COMPRESS_F=abc.tar.gz
if [ $# -ne 0 ]; then
COMPRESS_F=$1
fi
for i in `tar -tf "$COMPRESS_F"`; do
if [ -f $i ]; then
echo "mv ${i} ${i}.orig"
mv ${i} ${i}.orig
fi
done
for i in `tar -xvzf "$COMPRESS_F"`; do
echo "mv ${i} ${i}.append"
mv ${i} ${i}.append
if [ -f ${i}.orig ]; then
echo "mv ${i}.orig ${i};"
mv ${i}.orig ${i};
fi
done
Got the solution, we can do this using '--transform' option of tar command, as shown below:
tar -(x/c)vf archive.tar --transform 's,/abc$,/abc.append,'
this will convert all abc files to abc.append.
Good Morning,
I am trying to write a korn shell script to look inside a directory that contains loads of files and check that each file also exists with .orig on the end.
For example if a file inside the directory is called 'mercury_1' there must also be a file called 'mercury_1.orig'
If there isn't, it needs to move the mercury_1 file to another location. However if the .orig file exists do nothing and move onto the next file.
I am sure it is really simple but I am not that experienced in writing Linux scripts and help would be greatly appreciated!!
Here's a small ksh snippet to check if a file exists in the current directory
fname=mercury_1
if [ -f $fname ]
then
echo "file exists"
else
echo "file doesn't exit"
fi
Edit:
The updated script that does the said functionality
#/usr/bin/ksh
if [ ! $# -eq 1 ]
then
echo "provide dir"
exit
fi
dir=$1
cd $dir
#process file names not ending with orig
for fname in `ls | grep -v ".orig$"`
do
echo processing file $fname
if [ -d $fname ] #skip directory
then
continue
fi
if [ -f "$fname.orig" ] #if equiv. orig file present
then
echo "file exist"
continue
else
echo "moving"
mv $fname /tmp
fi
done
Hope its of help!
You can use the below script
script.sh :
#!/bin/sh
if [ ! $# -eq 2 ]; then
echo "error";
exit;
fi
for File in $1/*
do
Tfile=${File%%.*}
if [ ! -f $Tfile.orig ]; then
echo "$File"
mv $File $2/
fi
done
Usage:
./script.sh <search directory> <destination dir if file not present>
Here, for each file with extension stripped check if "*.orig" is present, if not then move file to different directory, else do nothing.
Extension is stripped because you don't want to repeat the same steps for *.orig files.
I tested this on OSX (basically mv should not differ to much from linux). My test directory is zbar and destination is /tmp directory
#!/bin/bash
FILES=zbar
cd $FILES
array=$(ls -p |grep -v "/") # we search for file without extension so put them in array and ignore directory
echo $array
for f in $array #loop in array and find .orig file
do
#echo $f
if [ -e "$f.orig" ]
then
echo "found $f.orig"
else
mv -f "$f" "/tmp"
fi
done
How to write simple shell script to create zip file.
I want to create zip file by collecting files with same string pattern in their file names from a folder.
For example, there may be many files under a folder.
xxxxx_20140502_xxx.txt
xxxxx_20140502_xxx.txt
xxxxx_20140503_xxx.txt
xxxxx_20140503_xxx.txt
xxxxx_20140504_xxx.txt
xxxxx_20140504_xxx.txt
After running the shell script, the result must be following three zip files.
20140502.zip
20140503.zip
20140504.zip
Please give me right direction to create simple shell script to output the result as above.
#!/bin/bash
for file in *_????????_*.csv *_????????_*.txt; do
[ -f "${file}" ] || continue
date=${file#*_} # adjust this and next line depending
date=${date%_*} # on your actual prefix/suffix
echo "${date}"
done | sort -u | while read date; do
zip "${date}.zip" *${date}*
done
Since zip will update the archive, this will do:
shopt -s nullglob
for file in *.{txt,csv}; do [[ $file =~ _([[:digit:]]{8})_ ]] && zip "${BASH_REMATCH[1]}.zip" "$file"; done
The shopt -s nullglob is because you don't want to have unexpanded globs if there are no matching files.
Everything below this line is my old answer...
First, get all the possible dates. Heuristically, this could be the files ending in .txt and .csv that match the regex _[[:digit:]]{8}_:
#!/bin/bash
shopt -s nullglob
declare -A dates=()
for file in *.{csv,txt}; do
[[ $file =~ _([[:digit:]]{8})_ ]] && dates[${BASH_REMATCH[1]}]=
done
printf "Date found: %s\n" "${!dates[#]}"
This will output to stdout all the dates found in the files. E.g. (I called the previous snipped gorilla and I chmod +x gorilla and touched a few files for demo):
$ ls
banana_20010101_gorilla.csv gorilla_20140502_bonobo.csv
gorilla notthisone_123_lol.txt
gorilla_20140502_banana.txt
$ ./gorilla
Date found: 20140502
Date found: 20010101
Next step, for each date found, get all the files ending in .txt and .csv and zip them in the archive corresponding to the date: appending this to gorilla will do the job:
for date in "${!dates[#]}"; do
zip "$date.zip" *"_${date}_"*.{csv,txt}
done
Full script after removing the flooding part:
#!/bin/bash
shopt -s nullglob
declare -A dates=()
for file in *.{csv,txt}; do
[[ $file =~ _([[:digit:]]{8})_ ]] && dates[${BASH_REMATCH[1]}]=
done
for date in "${!dates[#]}"; do
zip "$date.zip" *"_${date}_"*.{csv,txt}
done
Edit. I overlooked your requirement with one line command. Then here's the one-liner:
shopt -s nullglob; declare -A dates=(); for file in *.{csv,txt}; do [[ $file =~ _([[:digit:]]{8})_ ]] && dates[${BASH_REMATCH[1]}]=; done; for date in "${!dates[#]}"; do zip "$date.zip" *"_${date}_"*.{csv,txt}; done
:)
#! /bin/bash
dates=$(ls ?????_[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]_???.{csv,txt} \
| cut -f2 -d_ | sort -u)
for date in $dates ; do
zip $date.zip ?????_"$date"_???.{csv,txt}
done
This piece of bash code, shows no folder name while there exists many folders.
#!/bin/bash
for file in .; do
if [ -d $file ]; then
echo $file
fi
done
the output is only .
Can you explain why?
it reads . as an array of size one and prints it for you. use something like this instead:
for file in `ls`; do
if [ -d $file ]; then
echo $file
fi
done
I have a directory which has many folder and each folder contains a list of XML files. I am writing a bash script that traverses through the files and renames the extension of the file to "manual" if the size of the file is greater than 65Mb. This is my first writing a shell script and I was able to write the code for traversing the files but I am having difficulty in the renaming part.
for file in $dir
do
size=$(stat -c%s "$file")
if test "$size" -gt "68157440"; then
echo "Before Renaming...."
echo $file
echo "After renaming"
mv *.manual `basename $file`.xml
echo $file
else
echo $file >> outlog.log
fi
done
an example of $file is,
/apps/jAS/dev/products-app/BConverter/data/supplier-data/TF/output/Fiber Optics and Fiber Management Solutions/Fiber Optic Cable Assemblies.xml
mv *.manual `basename $file`.xml
If you want to change the extension of $file from xml to manual, do instead
mv "$file" "${file%.xml}".manual
What exactly is the difficulty you're having?
If it's white space in file names, try
mv *.manual `basename "$file"`.xml
Note that your script will not work if *.manual expands to more than one file name.
No need for a script on this, combination of find and xargs should do the trick:
find . -size +65M | xargs -IQ mv Q Q.manual
The little-used -I option to Xargs:
runs each input as a separate command, and
lets you replace the filename, so you can use it multiple time, ideal for a mv