recursively "normalize" filenames [duplicate] - linux

This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
recursively “normalize” filenames
Q on pastebin: http://pastebin.com/raw.php?i=19iYZpwY
i mean getting rid of special chars in filenames, etc.
i have made a script, that can recursively rename files [http://pastebin.com/raw.php?i=kXeHbDQw]:
e.g.: before:
THIS i.s my file (1).txt
after running the script:
This-i-s-my-file-1.txt
Ok. here it is:
But: when i wanted to test it "fully", with filenames like this [http://pastebin.com/raw.php?i=LQ07ntcS]:
¤¥¦§¨©ª«¬®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÂÃÄÅÆÇÈÊËÌÎÏÐÑÒÔÕ×ØÙUÛUÝÞßàâãäåæçèêëìîïðñòôõ÷øùûýþÿ.txt
áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&'()+,:;<=>?#[]^_{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£.txt<br>
<br>
it fails [http://pastebin.com/raw.php?i=iu8Pwrnr]:<br>
$ sh renamer.sh directorythathasthefiles<br>
mv: cannot stat./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()+,:;<=>?#[]^{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directorymv: cannot stat./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()*+,:;<=>?#[]^{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directorymv: cannot stat./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()+,:;<=>?#[]^_{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directorymv: cannot stat./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()+,:;<=>?#[]^{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directorymv: cannot stat./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()*+,:;<=>?#[]^{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directorymv: cannot stat./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()+,:;<=>?#[]^_{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directorymv: cannot stat./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()+,:;<=>?#[]^_`{|}~€‚ƒ„…†....and so on
$
so "mv" can't handle special chars.. :\
i worked on it for many hours..
does anyone has a working one? [that can handle chars [filenames] in that 2 lines too?]

Reading that script was almost painful...
For one, you should read this.
Then you should read about bash functions. After that you should read about sed and tr
Then you should consider this: do you really want to move the file each time that you perform a transformation on its name?
Then after all this thinking, you should come up with something a bit saner.

Wtf is going on your system? You should consider re setting up and pay attention on sane applicaitons and security.
However its very likely that you are just running into the max length of command arguments if i am looking at that.
If not, well install UTF8 locales and install them as system default.
On debian based systems this is usually just a matter of dpkg-reconfigure locales
also work on your accept rate.

Related

Bash substring extraction purpose in "ls -lashtg ${1:-.}" [duplicate]

This question already has answers here:
Usage of :- (colon dash) in bash
(2 answers)
Closed 1 year ago.
I ran across the following Bash function which was suggested as a useful alias to add to the .bashrc file. It lists the last 13 files in a directory that were modified.
I don't understand what is being done with the ${1:-.} argument, though. It looks like some kind of substring extraction, but I couldn't find the meaning of -. in the Advanced Bash Scripting Manual.
I tried the command in a few directories and didn't notice any difference between the output when I removed this argument. My guess is that it's there to prevent an error when encountering some specific type of file or file name. What is it doing? And what is the purpose of including it in the command?
function lst()
{
ls -lashtg ${1:-.} | head -13
}
$1 is the first command line argument. Im sure you know.
${1:-.} simply puts a . when no first line argument is given.
Thus
lst
Translates to
ls -lashtg . | head -13
It would workout without the substitution I guess. But I also guess that this is just there as a best practice

Is it possible to display a file's contents and delete that file in the same command?

I'm trying to display the output of an AWS lambda that is being captured in a temporary text file, and I want to remove that file as I display its contents. Right now I'm doing:
... && cat output.json && rm output.json
Is there a clever way to combine those last two commands into one command? My goal is to make the full combined command string as short as possible.
For cases where
it is possible to control the name of the temporary text file.
If file is not used by other code
Possible to pass "/dev/stdout" as the.name of the output
Regarding portability: see stack exchange how portable ... /dev/stdout
POSIX 7 says they are extensions.
Base Definitions,
Section 2.1.1 Requirements:
The system may provide non-standard extensions. These are features not required by POSIX.1-2008 and may include, but are not limited to:
[...]
• Additional character special files with special properties (for example,  /dev/stdin, /dev/stdout,  and  /dev/stderr)
Using the mandatory supported /dev/tty will force output into “current” terminal, making it impossible to pipe the output of the whole command into different program (or log file), or to use the program when there is no connected terminals (cron job, or other automation tools)
No, you cannot easily remove the lines of a file while displaying them. It would be highly inefficient as it would require removing characters from the beginning of a file each time you read a line. Current filesystems are pretty good at truncating lines at the end of a file, but not at the beginning.
A simple but extremely slow method would look like this:
while [ -s output.json ]
do
head -1 output.json
sed -i 1d output.json
done
While this algorithm is plain and simple, you should know that each time you remove the first line with sed -i 1d it will copy the whole content of the file but the first line into a temporary file, resulting in approximately 0.5*n² lines written in total (where n is the number of lines in your file).
In theory you could avoid this by do something like that:
while [ -s output.json ]
do
line=$(head -1 output.json)
printf -- '%s\n' "$line"
fallocate -c -o 0 -l $((${#len}+1)) output.json
done
But this does not account for variable newline characters (namely DOS-formatted newlines) and fallocate does not always work on xfs, among other issues.
Since you are trying to consume a file alongside its creation without leaving a trace of its existence on disk, you are essentially asking for a pipe functionality. In my opinion you should look into how your output.json file is produced and hopefully you can pipe it to a script of your own.

bash handling of quotation marks in filename

I am trying to remove and replace quotation marks that are present in a file name. For example, I would like to change:
$ ls
abc"def"ghi"jkl"mno
to this
$ ls
abc:def:ghi:jkl:mno
In trying to solve this, I came across How to rename a bunch of files to eliminate quote marks, which is exactly what I want to do. However, it didn't work for my case. To figure out why, I tried creating a test file like this:
$ touch abba\"abba\"cde\"cde\"efef
With this file, the solutions I came across (such as mentioned above) worked. But why didn't it work for the first file?
One thing I discovered was that bash command completion sees them differently. If I type in
$ ls abb<tab>
bash will complete the filename like so:
$ abba\"abba\"cde\"cde\"efef
just as I created it. But for the original file, bash completion went like this:
$ ls abc<tab>
results in
$ abc"def"ghi"jkl"mno
So in the test case file, there is an escape of the quotation marks, and in the other case (the file I really want to rename), there is no escaping of the the quotation marks. I don't know how the original files were named.
Can anyone explain why bash sees these names differently, and how I would go about renaming my file?
Here is two ways to rename a file with "(quotation) mark,
option 1: With escape character \
mv abc\"cdf\"efg\"hij newFileName
option 2: By using '(single quote)
mv 'abc"cdf"efg"hij' newFileName
Note: using special charaters like :(colon) in file name might not be a good idea,
and regarding the auto completion, it usually fill the name with escape character, example
ls abc<tab> will complete the name to ls abc\"cdf\"efg\"hij
unless you start the name with a quote, example
ls 'abc<tab> will complete the name to ls 'abc"cdf"efg"hij'

Renaming files like 20141207_190822.jpg to "2014-12-07 19.08.22.jpg" in linux or MacOS X

How in Linux or MacOS X to rename a bunch of files with names 20141207_190822.jpg and 20141207_190823.mp4 to this format:
2014-12-07 19.08.22.jpg
and
2014-12-07 19.08.23.mp4
?
I've found many examples how to just add something to the beginning of filename, but here I need to change the mask by inserting symbols in the middle of filenames and replacing _ with space " ".
Thank you
Try doing this :
$ rename 's#^(\d{4})(\d{2})(\d{2})_(\d{2})(\d{2})(\d{2})#$1-$2-$3 $4.$5.$6#' *
There are other tools with the same name which may or may not be able to do this, so be careful.
If you run the following command :
$ file $(readlink -f $(type -p rename))
and you have a result like
.../rename: Perl script, ASCII text executable
then this seems to be the right tool =)
If you don't have this command, search your package manager to install it or do it manually
Last but not least, this tool was originally written by Larry Wall, the Perl's dad.

zip command not working

I am trying to zip a file using shell script command. I am using following command:
zip ./test/step1.zip $FILES
where $FILES contain all the input files. But I am getting a warning as follows
zip warning: name not matched: myfile.dat
and one more thing I observed that the file which is at last in the list of files in a folder has the above warning and that file is not getting zipped.
Can anyone explain me why this is happening? I am new to shell script world.
zip warning: name not matched: myfile.dat
This means the file myfile.dat does not exist.
You will get the same error if the file is a symlink pointing to a non-existent file.
As you say, whatever is the last file at the of $FILES, it will not be added to the zip along with the warning. So I think something's wrong with the way you create $FILES. Chances are there is a newline, carriage return, space, tab, or other invisible character at the end of the last filename, resulting in something that doesn't exist. Try this for example:
for f in $FILES; do echo :$f:; done
I bet the last line will be incorrect, for example:
:myfile.dat :
...or something like that instead of :myfile.dat: with no characters before the last :
UPDATE
If you say the script started working after running dos2unix on it, that confirms what everybody suspected already, that somehow there was a carriage-return at the end of your $FILES list.
od -c shows the \r carriage-return. Try echo $FILES | od -c
Another possible cause that can generate a zip warning: name not matched: error is having any of zip's environment variables set incorrectly.
From the man page:
ENVIRONMENT
The following environment variables are read and used by zip as described.
ZIPOPT
contains default options that will be used when running zip. The contents of this environment variable will get added to the command line just after the zip command.
ZIP
[Not on RISC OS and VMS] see ZIPOPT
Zip$Options
[RISC OS] see ZIPOPT
Zip$Exts
[RISC OS] contains extensions separated by a : that will cause native filenames with one of the specified extensions to be added to the zip file with basename and extension swapped.
ZIP_OPTS
[VMS] see ZIPOPT
In my case, I was using zip in a script and had the binary location in an environment variable ZIP so that we could change to a different zip binary easily without making tonnes of changes in the script.
Example:
ZIP=/usr/bin/zip
...
${ZIP} -r folder.zip folder
This is then processed as:
/usr/bin/zip /usr/bin/zip -r folder.zip folder
And generates the errors:
zip warning: name not matched: folder.zip
zip I/O error: Operation not permitted
zip error: Could not create output file (/usr/bin/zip.zip)
The first because it's now trying to add folder.zip to the archive instead of using it as the archive. The second and third because it's trying to use the file /usr/bin/zip.zip as the archive which is (fortunately) not writable by a normal user.
Note: This is a really old question, but I didn't find this answer anywhere, so I'm posting it to help future searchers (my future self included).
eebbesen hit the nail in his comment for my case (but i cannot vote for comment).
Another possible reason missed in the other comments is file exceeding the file size limit (4GB).
I converted my script for unix environment using dos2unix command and executed my script as ./myscript.sh instead bash myscript.sh.
I just discovered another potential cause for this. If the permissions of the directory/subdirectory don't allow the zip to find the file, it will report this error. Actually, if you run a chmod -R 444 on the directory, and then try to zip it, you will reproduce this error, and also have a "stored 0%" report, like this:
zip warning: name not matched: borrar/enviar
adding: borrar/ (stored 0%)
Hence, try changing the permissions of the file. If you are trying to send them through email, and those email filters (like Gmail's) invent silly filters of not sending executables, don't forget that making permissions very strict when making zip compression can be the cause of the error you are reporting, of "name not matched".
spaces are not allowed:
it would fail if there are more than one files(s) in $FILES unless you put them in loop
I also encountered this issue. In my case, the line separate is CRLF in my zip shell script which causes the problem. Using LF fixed it.

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