I have something like this:
package="com.program.interesting.program.aplication"
I want a script that will create folders like this:
midir com
cd com/
mkdir program
cd program/
...
etc
But I want to do this automatically, no matter how many folder I'll have. I guess this is very simple, but I don't know how to parse a string on the shell and I don't want to read a book, just to solve this.
Thanks
You just need to convert . to / and that is all:
mkdir -p "$(echo $package | tr . /)"
The -p key means that mkdir needs to create all necessary parent directories automatically.
Use "" around the command substitution $() if it is possible that $package can contain spaces inside.
The tr command changes the . symbols in the string to slashes.
e.g.
$ echo com.program.interesting.program.aplication |tr . /
com/program/interesting/program/aplication
mkdir -p com/program/interesting/program/aplication
Related
Is there any Windows app that will search for a string of text within fields in a Word (DOCX) document? Apps like Agent Ransack and its big brother FileLocator Pro can find strings in the Word docs but seem incapable of searching within fields.
For example, I would like to be able to find all occurrences of the string "getProposalTranslations" within a collection of Word documents that have fields with syntax like this:
{ AUTOTEXTLIST \t "<wr:out select='$.shared_quote_info' datasource='getProposalTranslations'/>" }
Note that string doesn't appear within the text of the document itself but rather only within a field. Essentially the DOCX file is just a zip file, I believe, so if there's a tool that can grep within archives, that might work. Note also that I need to be able to search across hundreds or perhaps thousands of files in many directories, so unzipping the files one by one isn't feasible. I haven't found anything on my own and thought I'd ask here. Thanks in advance.
This script should accomplish what you are trying to do. Let me know if that isn't the case. I don't usually write entire scripts because it can hurt the learning process, so I have commented each command so that you might learn from it.
#!/bin/sh
# Create ~/tmp/WORDXML folder if it doesn't exist already
mkdir -p ~/tmp/WORDXML
# Change directory to ~/tmp/WORDXML
cd ~/tmp/WORDXML
# Iterate through each file passed to this script
for FILE in $#; do
{
# unzip it into ~/tmp/WORDXML
# 2>&1 > /dev/null discards all output to the terminal
unzip $FILE 2>&1 > /dev/null
# find all of the xml files
find -type f -name '*.xml' | \
# open them in xmllint to make them pretty. Discard errors.
xargs xmllint --recover --format 2> /dev/null | \
# search for and report if found
grep 'getProposalTranslations' && echo " [^ found in file '$FILE']"
# remove the temporary contents
rm -rf ~/tmp/WORDXML/*
}; done
# remove the temporary folder
rm -rf ~/tmp/WORDXML
Save the script wherever you like. Name it whatever you like. I'll name it docxfind. Make it executable by running chmod +x docxfind. Then you can run the script like this (assuming your terminal is running in the same directory): ./docxfind filenames...
I have a number of files with the following naming:
name1.name2.s01.ep01.RANDOMWORD.mp4
name1.name2.s01.ep02.RANDOMWORD.mp4
name1.name2.s01.ep03.RANDOMWORD.mp4
I need to remove everything between the last . and ep# from the file names and only have name1.name2.s01.ep01.mp4 (sometimes the extension can be different)
name1.name2.s01.ep01.mp4
name1.name2.s01.ep02.mp4
name1.name2.s01.ep03.mp4
This is a simpler version of #Jesse's [answer]
for file in /path/to/base_folder/* #Globbing to get the files
do
epno=${file#*.ep}
mv "$file" "${file%.ep*}."ep${epno%%.*}".${file##*.}"
#For the renaming part,see the note below
done
Note : Didn't get a grab of shell parameter expansion yet ? Check [ this ].
Using Linux string manipulation (refer: http://www.tldp.org/LDP/abs/html/string-manipulation.html) you could achieve like so:
You need to do per file-extension type.
for file in <directory>/*
do
name=${file}
firstchar="${name:0:1}"
extension=${name##${firstchar}*.}
lastchar=$(echo ${name} | tail -c 2)
strip1=${name%.*$lastchar}
lastchar=$(echo ${strip1} | tail -c 2)
strip2=${strip1%.*$lastchar}
mv $name "${strip2}.${extension}"
done
You can use rename (you may need to install it). But it works like sed on filenames.
As an example
$ for i in `seq 3`; do touch "name1.name2.s01.ep0$i.RANDOMWORD.txt"; done
$ ls -l
name1.name2.s01.ep01.RANDOMWORD.txt
name1.name2.s01.ep02.RANDOMWORD.txt
name1.name2.s01.ep03.RANDOMWORD.txt
$ rename 's/(name1.name2.s01.ep\d{2})\..*(.txt)$/$1$2/' name1.name2.s01.ep0*
$ ls -l
name1.name2.s01.ep01.txt
name1.name2.s01.ep02.txt
name1.name2.s01.ep03.txt
Where this expression matches your filenames, and using two capture groups so that the $1$2 in the replacement operation are the parts outside the "RANDOMWORD"
(name1.name2.s01.ep\d{2})\..*(.txt)$
I would like to ask some help regarding my script below:
#!/bin/bash
year=$(date +%Y)
lastyear=$((year-1))
month=$(date +%m)
log="$lastyear$month"
mkdir -p "/root/temp/$lastyear"
mkdir -p "/root/temp/$lastyear/$month"
cd /root
mv -f "*$log*" "/root/temp/$lastyear/$month"
Error Prompt is
mv: cannot stat `*201602*': No such file or directory
My target was to move all the files that has the specific "201602" string in file name to a specific location.
Sample logs files is OUTXXX-201602XXX, INXXX-201602XXX, 201602XXXX.
This will be implemented through crontab, because there are about 500k+ log files to be transferred and using find will receive a too much argument error T_T.
Any suggestions will help!
You have to put globbing characters outside of quotes for them to be interpreted.
Try instead:
mv -f *"$log"* …
The shell will still keep the resulting matches as one word, so this is OK even if the filenames have spaces.
I have this list of files on a Linux server:
abc.log.2012-03-14
abc.log.2012-03-27
abc.log.2012-03-28
abc.log.2012-03-29
abc.log.2012-03-30
abc.log.2012-04-02
abc.log.2012-04-04
abc.log.2012-04-05
abc.log.2012-04-09
abc.log.2012-04-10
I've been deleting selected log files one by one, using the command rm -rf see below:
rm -rf abc.log.2012-03-14
rm -rf abc.log.2012-03-27
rm -rf abc.log.2012-03-28
Is there another way, so that I can delete the selected files at once?
Bash supports all sorts of wildcards and expansions.
Your exact case would be handled by brace expansion, like so:
$ rm -rf abc.log.2012-03-{14,27,28}
The above would expand to a single command with all three arguments, and be equivalent to typing:
$ rm -rf abc.log.2012-03-14 abc.log.2012-03-27 abc.log.2012-03-28
It's important to note that this expansion is done by the shell, before rm is even loaded.
Use a wildcard (*) to match multiple files.
For example, the command below will delete all files with names beginning with abc.log.2012-03-.
rm -f abc.log.2012-03-*
I'd recommend running ls abc.log.2012-03-* to list the files so that you can see what you are going to delete before running the rm command.
For more details see the Bash man page on filename expansion.
If you want to delete all files whose names match a particular form, a wildcard (glob pattern) is the most straightforward solution. Some examples:
$ rm -f abc.log.* # Remove them all
$ rm -f abc.log.2012* # Remove all logs from 2012
$ rm -f abc.log.2012-0[123]* # Remove all files from the first quarter of 2012
Regular expressions are more powerful than wildcards; you can feed the output of grep to rm -f. For example, if some of the file names start with "abc.log" and some with "ABC.log", grep lets you do a case-insensitive match:
$ rm -f $(ls | grep -i '^abc\.log\.')
This will cause problems if any of the file names contain funny characters, including spaces. Be careful.
When I do this, I run the ls | grep ... command first and check that it produces the output I want -- especially if I'm using rm -f:
$ ls | grep -i '^abc\.log\.'
(check that the list is correct)
$ rm -f $(!!)
where !! expands to the previous command. Or I can type up-arrow or Ctrl-P and edit the previous line to add the rm -f command.
This assumes you're using the bash shell. Some other shells, particularly csh and tcsh and some older sh-derived shells, may not support the $(...) syntax. You can use the equivalent backtick syntax:
$ rm -f `ls | grep -i '^abc\.log\.'`
The $(...) syntax is easier to read, and if you're really ambitious it can be nested.
Finally, if the subset of files you want to delete can't be easily expressed with a regular expression, a trick I often use is to list the files to a temporary text file, then edit it:
$ ls > list
$ vi list # Use your favorite text editor
I can then edit the list file manually, leaving only the files I want to remove, and then:
$ rm -f $(<list)
or
$ rm -f `cat list`
(Again, this assumes none of the file names contain funny characters, particularly spaces.)
Or, when editing the list file, I can add rm -f to the beginning of each line and then:
$ . ./list
or
$ source ./list
Editing the file is also an opportunity to add quotes where necessary, for example changing rm -f foo bar to rm -f 'foo bar' .
Just use multiline selection in sublime to combine all of the files into a single line and add a space between each file name and then add rm at the beginning of the list. This is mostly useful when there isn't a pattern in the filenames you want to delete.
[$]> rm abc.log.2012-03-14 abc.log.2012-03-27 abc.log.2012-03-28 abc.log.2012-03-29 abc.log.2012-03-30 abc.log.2012-04-02 abc.log.2012-04-04 abc.log.2012-04-05 abc.log.2012-04-09 abc.log.2012-04-10
A wild card would work nicely for this, although to be safe it would be best to make the use of the wild card as minimal as possible, so something along the lines of this:
rm -rf abc.log.2012-*
Although from the looks of it, are those just single files? The recursive option should not be necessary if none of those items are directories, so best to not use that, just for safety.
I am not a linux guru, but I believe you want to pipe your list of output files to xargs rm -rf. I have used something like this in the past with good results. Test on a sample directory first!
EDIT - I might have misunderstood, based on the other answers that are appearing. If you can use wildcards, great. I assumed that your original list that you displayed was generated by a program to give you your "selection", so I thought piping to xargs would be the way to go.
if you want to delete all files that belong to a directory at once.
For example:
your Directory name is "log" and "log" directory include abc.log.2012-03-14, abc.log.2012-03-15,... etc files. You have to be above the log directory and:
rm -rf /log/*
I know my topic is a little confusing, but here is what I want to do.
I have a file which I would like to create a link to in my home directory ~/bin, but when I execute the file that is symbolically linked, the file requires another file in its directory. Therefore, it fails to run because it cannot find the other file. What can I do?
Well, you have two simple solutions.
edit the shell script to point to the absolute path of the file, not just the the basename.
./path/to/file.sh
VS
file.sh
so something like this should do what your after. sed -i 's|file.sh|./path/to/file.sh|g' ~/bin/script.sh it searches your symlinked file, script.sh in this case, and replaces the call to file.sh to ./path/to/file.sh. note you often see sed use /'s. but it can use just about anything as a delimiter, if you wish to use /'s here you will need to escape them. /. you may want to consider escaping the . (period) as well, but in this case its not necessary. If you are new to sed realize that the -i flag means it will edit the file in place. Lastly, realize its a simple search and replace operation and you may chose to do it by hand.
The second way is to create a ln -s to the file as you did with the other file so there exists a symbolic link between both files.
ln -s /far/off/script.sh ~/bin/script.sh
and
ln -s /far/off/file.sh ~/bin/file.sh
more on symlinking
I would rather create a script file in ~/bin/` that calls your executable from the appropriate directory.
Here is an example using /sbin/ifconfig:
$ cat > ~/bin/file
#!/bin/bash
file=/sbin/ifconfig
cd `dirname $file`
`basename $file`
(ctr+d)
$ chmod +x ~/bin/file
$ file
Here you should see the output of ifconfig but the point is: its get executed from the /sbin directory. So if ifconfig had dependencies it would work properly. Just replace /sbin/ifconfig with your absolute path.
Alternatively, you can modify your script as
pushd ~/bin
##### your script here
popd
Combination of readlink and dirname will get the actual directory of the script:
my_dir=$(dirname "$(readlink -f "$0")")
source "$my_dir/other_file"