I have a bash script which needs to go through files in a directory in an iOS device and remove files one by one.
To list files from command line I use the following command:
ios-deploy --id UUID --bundle_id BUNDLE -l | grep Documents
and to go one by one on each file I use the following for loop in my script
for line in $(ios-deploy --id UUID --bundle_id BUNDLE -l | grep Documents); do
echo "${line}"
done
Now the problem is that there are files which names have spaces in them, and in such cases the for loop treats them as 2 separate lines.
How can I escape that whitespace in for loop definition so that I get one line per each file?
This might solve your issue:
while IFS= read -r -d $'\n'
do
echo "${REPLY}"
done < <(ios-deploy --id UUID --bundle_id BUNDLE -l | grep Documents)
Edit per Charles Duffy recommendation:
while IFS= read -r line
do
echo "${line}"
done < <(ios-deploy --id UUID --bundle_id BUNDLE -l | grep Documents)
Related
How do you use a command line argument as a file path and check for file existence in Bash?
I have the simple Bash script test.sh:
#!/bin/bash
set -e
echo "arg1=$1"
if [ ! -f "$1" ]
then
echo "File $1 does not exist."
exit 1
fi
echo "File exists!"
and in the same directory, I have a data folder containing stuff.txt.
If I run ./test.sh data/stuff.txt I see the expected output:
arg1=data/stuff.txt
"File exists!"
However, if I call this script from a second script test2.sh, in the same directory, like:
#!/bin/bash
fn="data/stuff.txt"
./test.sh $fn
I get the mangled output:
arg1=data/stuff.txt
does not exist
Why does the call work when I run it manually from a terminal, but not when I run it through another Bash script, even though both are receiving the same file path? What am I doing wrong?
Edit: The filename does not have spaces. Both scripts are executable. I'm running this on Ubuntu 18.04.
The filename was getting an extra whitespace character added to it as a result of how I was retrieving it in my second script. I didn't note this in my question, but I was retrieving the filename from folder list over SSH, like:
fn=$(ssh -t "cd /project/; ls -t data | head -n1" | head -n1)
Essentially, I wanted to get the filename of the most recent file in a directory on a remote server. Apparently, head includes the trailing newline character. I fixed it by changing it to:
fn=$(ssh -t "cd /project/; ls -t data | head -n1" | head -n1 | tr -d '\n' | tr -d '\r')
Thanks to #bigdataolddriver for hinting at the problem likely being an extra character.
This question already has answers here:
Run script on multiple files
(3 answers)
Closed 3 years ago.
I'm very new to bash. I have ten text files that I want to edit with the same line of code.
#!/bin/bash
sed -i -e 's/.\{6\}/&\n/g' -e 's/edit/edit2/g' | tr -d "\n" | sed 's/edit2/edit/g'| grep -o "here.*there" | sed -r '/^.{,100}$/d'
< files 1-10
I know I could use sed -f sed.sh <file1 >file1 but that only works with sed commands and it only works one file at a time?
Do I have to run a loop?
There's some great existing answers on the Unix stack exchange that help deal with your problem. Specifically, from this post, they use a loop to recursively loop through all the files in a particular directory, as follows:
( shopt -s globstar dotglob;
for file in **; do
if [[ -f $file ]] && [[ -w $file ]]; then
sed -i -- 's/foo/bar/g' "$file"
fi
done
)
Note the line, shopt -s globstar dotglob;, which allows us to use globbing patterns in the for loop. We also enclose the code in brackets, to prevent the shopt -s globstar dotglob; line option from becoming a global setting.
If you would like to apply this example to your file, you can just place your files in the current directory, and the code would probably look something like this:
( shopt -s globstar dotglob;
for file in **; do
if [[ -f $file ]] && [[ -w $file ]]; then
sed -i -e 's/.\{6\}/&\n/g' -e 's/edit/edit2/g' | tr -d "\n" | sed 's/edit2/edit/g' | grep -o "here.*there" | sed -r '/^.{,100}$/d' "$file"
fi
done
)
Note that we have placed a "$file" variable beside each of the seds that you used in your code, this replaces the name of the file for each command.
There is another example given in the code that allows you to pick which files to run on, rather than all the files in a directory, which you can also re-purpose for your code, as given here:
( shopt -s globstar dotglob
sed -i -- 's/foo/bar/g' **baz*
sed -i -- 's/foo/bar/g' **.baz
)
To answer your question of doing a loop on each line, you will need to put a loop for each line inside your for loop, like so:
while read line ; do
: sed -i -e 's/.\{6\}/&\n/g' -e 's/edit/edit2/g' | tr -d "\n" | sed 's/edit2/edit/g' | grep -o "here.*there" | sed -r '/^.{,100}$/d' "$line”
done
)
Although the for loop can be useful for dealing with files in recursive directories, I would recommend against also using another loop to grab lines, since it muddies your code, and it’s possible there is a better way to do it without parsing line by line.
The linked question is a fairly complete guide to many of the cases you may come across, and is also worth a read if you want to learn more.
Hope that helps!
You could use a for loop.
You could use the tool parallel.
Example
Create a set of test files using a for-loop
mkdir -p /tmp/so58333536
cd /tmp/so58333536
for i in 1.txt 2.txt 3.txt 4.txt 5.txt;do echo "The answer is 41" > $i;done
cat /tmp/so58333536/*
Now correct your mistake using parallel [1].
mkdir /tmp/so58333536.new
ls /tmp/so58333536/* |parallel "sed 's/41/42/' {} > /tmp/so58333536.new/{/}"
cat /tmp/so58333536.new/*
{}:: refers to the current file
{/}:: refers to name of the current file (path is removed)
Reads: List all files in so58333536 and apply the following sed command to each file and write the output to so58333536.new.
[1] Another option is to use sed -i for in-place editing.
Be very carefull with this!! Mistakes can cause serious damages!
# !! Do not use -i option regularly !!
ls /tmp/so58333536/* |parallel "sed -i 's/41/42/'"
I want to copy only 20180721 files from Outgoing to Incoming folder. I also want to remove the first numbers from the file name and want to rename from -1 to -3. I want to keep my commands to minimum so I am using pax command below.
Filename:
216118105741_MOM-09330-20180721_102408-1.jar
Output expected:
MOM-09330-20180721_102408-3.jar
I have tried this command and it's doing most of the work apart from removing the number coming in front of the file name. Can anyone help?
Command used:
pax -rw -pe -s/-1/-3/ ./*20180721*.jar ../Incoming/
Try this simple script using just parameter expansion:
for file in *20180721*.jar; do
new=${file#*_}
cp -- "$file" "/path/to/destination/${new%-*}-3.jar"
done
You can try this
In general
for i in `ls files-to-copy-*`; do
cp $i `echo $i | sed "s/rename-from/rename-to/g"`;
done;
In your case
for i in `ls *_MOM*`; do
cp $i `echo $i | sed "s/_MOM/MOM/g" | sed "s/-1/-3/g"`;
done;
pax only applies the first successful substitution even if the -s option is specified more than once. You can pipe the output to a second pax instance, though.
pax -w -s ':^[^_]*_::p' *20180721*.jar | (builtin cd ../Incoming; pax -r -s ':1[.]jar$:3.jar:p')
I want to make UNIX script, which will automatically move my working directory files to newly created directories.
Example: In you dir you got files:
001-file.html,
001-file.rb,
002-file.html,
002-file.rb
And 2 files will be moved to ./NewDir/001-file and another 2 to ./NewDir/002-file
My problem is that after I get correct string from Unix commands I cannot assign it to variable.
Here is my code:
clear
echo "Starting script"
echo "Dir = "$(pwd)
read -p "Please enter count(max '999') of different file groups:" max_i
read -p "Enter new dir name:" outer_dir_name
for ((i=0; i<=$max_i;i++)) do
a1=$(($i/100))
a2=$((($i-$a1*100)/10))
a3=$(($i-($a2*10)-($a1*100)))
inner_dir_name=$((ls *[$a1][$a2][$a3]* 2>/dev/null | head -n 1 | cut -f1 -d"."))
echo $inner_dir_name
echo "--------------"
done
One pair of round parentheses is enough for command substitution.
inner_dir_name=$(ls *[$a1][$a2][$a3]* 2>/dev/null | head -n 1 | cut -f1 -d".")
It looks like you're going about the operation the hard way. I would probably do something like this, assuming that there are no spaces in the file names:
ls | sed 's/\..*$//' | sort -u |
while read prefix
do
mkdir -p $outer_dir_name/$prefix
mv $prefix.* $outer_dir_name/$prefix
done
The ls could be made more precise with:
ls [0-9][0-9][0-9]-file.*
If I was worried about blanks and other odd-ball characters in the file names, I'd have to use something more careful:
for file in [0-9][0-9][0-9]-file.*
do
prefix=${file%%.*}
[ -d "$outer_dir_name/$prefix" ] || mkdir -p "$outer_dir_name/$prefix"
mv "$file" "$outer_dir_name/$prefix"
done
This executes more mv commands, in general.
I'm trying to parse vsftpd logs to do some extra processing on the successfully uploaded files.
username will be the user so I create the home dir
filename is the file name in the log: it gives a wonky result i.e. "/foo.txt" but that doesn't matter
#!/bin/sh
sudo tail -F /var/log/vsftpd.log | while read line; do
if sudo echo "$line" | grep -q 'OK UPLOAD:'; then
username=$(echo "$line" | cut -d" " -f8 | sed 's/\[\(.*\)\]/\1/')
filename=$(echo "$line" | cut -d, -f2 | sed 's/^[ \t]*//')
home="/home/vsftpd/$username"
if sudo ls "$home$filename" &> /dev/null; then
# do something with $filename
echo "some text"
fi
fi
done
When a file is uploaded I expect the text "some text". I never get that instead I can see it reports:
ls: cannot access /home/vsftpd/user1"/foo.txt": No such file or directory
Although I can run the command in the shell:
$ sudo ls /home/vsftpd/user1"/foo.txt"
/home/vsftpd/user1/foo.txt
I'm guessing permissions related but I've got it running as sudo and I've given the directories full access. Any ideas?
Your problem is that you have an extra set of quotes around the file name component that you need to strip. The file name in the vsftpd logs (just verified this for myself) is surrounded with quotes, and unlike with username you're not removing those quotes.
This means that $filename ends up being set to, literally, "/foo.txt" including the quotes. When you construct the file name for ls with "$home$filename", the variables are interpolated, but the shell isn't then going to strip off another level of quotes. The quotes stay in the final file name, and the directory /home/vsftpd/user1" with the trailing quote doesn't exist.
This works when you enter the command from the shell because you aren't quoting the file name, so the shell does another round of quote interpolation and removes the double quotes.
If sudo works from the shell, it's possible that sudo has the NOEXEC flag set, which prevents it from executing scripts. You can read more about NOEXEC here.