When i am trying to run below code it is giving error of cp: target "Featurespath" is not a directory
I have tried multiple option but not working.
Featurespath=/permanent/jag/media-*/*/print/cooked/*Features.xml
for file in $(ls $Featurespath);
do
cat $Featurespath | sed "/pB-/s/Direction=\"unidir\"/Direction=\"bidir\"/" $Featurespath > /permanent/jag/temp.xml
cp -rf /permanent/jag/temp.xml $Featurespath
rm /permanent/jag/temp.xml
done
i want modified xml to be pasted in same xml file.
The error you received was because of the cp line: bash expands$Featurespath into a list of files. When cp sees more than 2 parameters, it assumes the last parameter to be a directory, which is not in this case. Here is my suggested fix:
Featurespath=/permanent/jag/media-*/*/print/cooked/*Features.xml
for file in $Featurespath
do
sed "/pB-/s/Direction=\"unidir\"/Direction=\"bidir\"/" "$file" > /permanent/jag/temp.xml
mv -f /permanent/jag/temp.xml "$file"
done
Notes
Do not use ls: bash can expand the wildcards just fine
Within the loop, you are now dealing with individual files $file, not the list of file $Featurespath
Do not need to use the cat command, the sed command can take a file name
sed has an inline editing option, which eliminate the need for temp file. You might want to look into it.
Replace cp/rm combination with mv
Ultimately, like others have said, sed is not the right tool to edit XML contents, but it might work for simple cases
Related
I'd love to have a more elegant solution for a mass rename of files, as shown below. Files were of format DEV_XYZ_TIMESTAMP.dat and we needed them as T-XYZ-TIMESTAMP.dat.
In the end, I copied them all (to be on the same side) into renamed folder:
ls -l *dat|awk '{system("cp " $10 " renamed/T-" substr($10, index($10, "_")+1))}'
So, first I listed all dat files, then picked up 10th column (file name) and executed a command using awk's system function.
The command was essentially copying of original filename into renamed folder with new file name.
New file name was created by removing (awk substring function) prefix before (including) _ and adding "T-" prefix.
Effectively:
cp DEV_file.dat renamed/T-file.dat
Is there a way to use cp or mv together with some regex rules to achieve the same in a bit more elegant way?
Thx
You may use this script:
for file in *.dat; do
f="${file//_/-}"
mv "$file" renamed/T-"${f#*-}"
done
You must avoid parsing output of ls command.
If you have rename utilitity
rename -E "s/[^_]*/T/" -e "s/_/-/g" *dat
Demo
$ls -1
ABC_DEF_TIMESTAMP.dat
DEV_XYZ_TIMESTAMP.dat
$rename -E "s/[^_]*/T/" -e "s/_/-/g" *
$ls -1
T-DEF-TIMESTAMP.dat
T-XYZ-TIMESTAMP.dat
$
This is how I would do it:
cpdir=renamed
for file in *dat; do
newfile=$(echo "$file" | sed -e "s/[^_]*/T/" -e "y/_/-/")
cp "$file" "$cpdir/$newfile"
done
The sed scripts transforms every non-underscore leading characters in a single T and then replaces every _ with -. If cpdir is not sure to exist before execution, you can simply add mkdir "$cpdir" after first line.
I have a directory full of sub-directories that look like this:
Track_0000111
Track_0004444
Track_0022222
Track_0333333
Track_5555555
I would like to remove certain directories if they are contained within a list in the file "RemoveFromTop6000_reformatted.txt"
The contents of the text file look like this:
Track_0000111
Track_0022222
Track_0333333
I tried to write a small script to handle this, but it does not seem to work:
#!/bin/bash
for file in cat RemoveFromTop6000_reformatted.txt; do
rm -rfv $file
done
Unfortunately this simply removes the text files, rather than the directories. Any tips?
Thanks!
You forgot backquotes around your call to cat. Without them, rm will simply delete the files cat (which probably doesn't exist, but you might not notice because you're using rm -f) and RemoveFromTop6000_reformatted.txt
Try this:
#!/bin/bash
for file in `cat RemoveFromTop6000_reformatted.txt`; do
rm -rv "$file"
done
or, more simply,
rm -rv `cat $file`
(but this will only work if the directory names don't contain whitespace).
No need to for, for something like this you can do a while read ...; do ... done < file just like this:
#!/bin/bash
while read file
rm -rfv "$file"
done < RemoveFromTop6000_reformatted.txt
you can try below command,
Command:
sed 's/^/"/g' sample.txt | sed 's/$/"/g' | xargs rm -rfv
Description:
Command will remove files as well as directories mentioned in "sample.txt".
NOTE:
In your case,make sure that "RemoveFromTop6000_reformatted.txt"
contains only directories name.
Command will also work if the directories name contains whitespace.
I want to write a script that do specific thing:
I have a txt file e.g.
from1/from2/from3/apple.file;/to1/to2/to3;some not important stuff
from1/from2/banana.file;/to1/to5;some not important stuff
from1/from10/plum.file;/to1//to5/to100;some not important stuff
Now i want to copy file from each line (e.g. apple.file), from original directory tree to new, non existing directories, after first semicolon (;).
I try few code examples from similar questions, but nothing works fine and I'm too weak in bash scripting, to find errors.
Please help :)
need to add some conditions:
file not only need to be copy, but also rename. Example line in file.txt:
from1/from2/from3/apple.file;to1/to2/to3/juice.file;some1
from1/from2/banana.file;to1/to5/fresh.file;something different from above
so apple.file need to be copy and rename to juice.file and put in to1/to2/to3/juice.file
I think thaht cp will also rename file but
mkdir -p "$to"
from answer below will create full folder path with juice.file as folder
In addidtion after second semicolon in each line will be something different, so how to cut it off?
Thanks for all help
EDIT: There will be no spaces in input txt file.
Try this code..
cat file | while IFS=';' read from to some_not_important_stuff
do
to=${to:1} # strip off leading space
mkdir -p "$to" # create parent for 'to' if not existing yet
cp -i "$from" "$to" # option -i to get a warning when it would overwrite something
done
Using awk
(run the awk command first and confirm the output is fine, then add |sh to do the copy)
awk -F";" '{printf "cp %s %s\n",$1,$2}' file |sh
Using shell (get updated that need manually create folder, base on alfe's
while IFS=';' read from to X
do
mkdir -p $to
cp $from $to
done < file
I had this same problem and used tar to solve it! Posted here:
tmpfile=/tmp/myfile.tar
files="/some/folder/file1.txt /some/other/folder/file2.txt"
targetfolder=/home/you/somefolder
tar --file="$tmpfile" "$files"
tar --extract --file="$tmpfile" --directory="$targetfolder"
In this case, tar will automatically create all (sub)folders for you! Best,
Nabi
I'm encountering many files with the same content and the same name on some of my servers. I need to quarantine these files for analysis so I can't just remove the duplicates. The OS is Linux (centos and ubuntu).
I enumerate the file names and locations and put them into a text file.
Then I do a for statement to move the files to quarantine.
for file in $(cat bad-stuff.txt); do mv $file /quarantine ;done
The problem is that they have the same file name and I just need to add something unique to the filename to get it to save properly. I'm sure it's something simple but I'm not good with regex. Thanks for the help.
Since you're using Linux, you can take advantage of GNU mv's --backup.
while read -r file
do
mv --backup=numbered "$file" "/quarantine"
done < "bad-stuff.txt"
Here's an example that shows how it works:
$ cat bad-stuff.txt
./c/foo
./d/foo
./a/foo
./b/foo
$ while read -r file; do mv --backup=numbered "$file" "./quarantine"; done < "bad-stuff.txt"
$ ls quarantine/
foo foo.~1~ foo.~2~ foo.~3~
$
I'd use this
for file in $(cat bad-stuff.txt); do mv $file /quarantine/$file.`date -u +%s%N`; done
You'll get everyfile with a timestamp appended (in nanoseconds).
You can create a new file name composed by the directory and the filename. Thus you can add one more argument in your original code:
for ...; do mv $file /quarantine/$(echo $file | sed 's:/:_:g') ; done
Please note that you should replace the _ with a proper character which is special enough.
On my computer running Ubuntu, I have a folder full of hundreds files all named "index.html.n" where n starts at one and continues upwards. Some of those files are actual html files, some are image files (png and jpg), and some of them are zip files.
My goal is to permanently remove every single file except the zip archives. I assume it's some combination of rm and file, but I'm not sure of the exact syntax.
If it fits into your argument list and no filenames contain colon a simple pipe with xargs should do:
file * | grep -vi zip | cut -d: -f1 | tr '\n' '\0' | xargs -0 rm
First find to find matching file, then file to get file types. sed eliminates other file types and also removes everything but the filenames from the output of file. lastly, rm for deleting:
find -name 'index.html.[0-9]*' | \
xargs file | \
sed -n 's/\([^:]*\): Zip archive.*/\1/p' |
xargs rm
I would run:
for f in in index.html.*
do
file "$f" | grep -qi zip
[ $? -ne 0 ] && rm -i "$f"
done
and remove -i option if you feel confident enough
Here's the approach I'd use; it's not entirely automated, but it's less error-prone than some other approaches.
file * > cleanup.sh
or
file index.html.* > cleanup.sh
This generates a list of all files (excluding dot files), or of all index.html.* files, in your current directory and writes the list to cleanup.sh.
Using your favorite text editor (mine happens to be vim), edit cleanup.sh:
Add #!/bin/sh as the first line
Delete all lines containing the string "Zip archive"
On each line, delete everything from the : to the end of the line (in vim, :%s/:.*$//)
Replace the beginning of each line with "rm" followed by a space
Exit your editor, updating the file.
chmod +x cleanup.sh
You should now have a shell script that will delete everything except zip files.
Carefully inspect the script before running it. Look out for typos, and for files whose names contain shell metacharacters. You might need to add quotation marks to the file names.
(Note that if you do this as a one-line shell command, you don't have the opportunity to inspect the list of files you're going to delete before you actually delete them.)
Once you're satisfied that your script is correct, run
./cleanup.sh
from your shell prompt.
for i in index.html.*
do
$type = file $i;
if [[ ! $file =~ "Zip" ]]
then
rm $file
fi
done
Change the rm to a ls for testing purposes.