There will be directory which will have any number folder and may be files, I just need pick one random folder and need to process it ( move the folders , etc ..) I need process folder one by one. Need to ignore if there is any files.
I am tiring with below code able to get folder name but , seems there some hidden character or some thing which not giving proper output.
PROCESSING_FOLDER_NAME= ls -l /ecom/bin/catalogUpload/input/TNF-EU/ | grep '^d' | cut -d ' ' -f23 | head -1
#PROCESSING_FOLDER_NAME= echo $PROCESSING_FOLDER_NAME | tr -d '\n\r'
#PROCESSING_FOLDER_NAME=${PROCESSING_FOLDER_NAME%$'\n'}
#echo "PROCESSING_FOLDER_NAME is/$PROCESSING_FOLDER_NAME "
echo "/ecom/bin/catalogUpload/input/TNF-EU/$PROCESSING_FOLDER_NAME/"
output
Thanks_giving_Dec_08
/ecom/bin/catalogUpload/input/TNF-EU//
I am expecting the output should be /ecom/bin/catalogUpload/input/TNF-EU/Thanks_giving_Dec_08/
Here is my bash version.
GNU bash, version 4.2.50(1)-release (powerpc-ibm-aix6.1.2.0)
I mainly need the folder name (not full path) in variable, As the folder name which is processing need be use for emails to notify other, etc.
To get a random folder from a list of folders,
first put the list of folders in an array:
list=(/ecom/bin/catalogUpload/input/TNF-EU/*/)
Next, get a random index using the $RANDOM variable of the shell,
modulo the size of the list:
((index = RANDOM % ${#list[#]}))
Print the value at the selected index:
echo "${list[index]}"
To get just the name of the directory without the full path, you can use the basename command:
basename "${list[index]}"
As for what's wrong with the original script:
To store the result of a command in a variable, the syntax is name=$(cmd) instead of name= cmd
Do not parse the output of ls, it's not reliable
To get directories in a directory, you can use glob patterns like * ending with /, as in the above example */.
Related
I have two directories, one called clients and another called test, inside the directory called clients I have some folders, I need a shell script that reads the name of the folders inside clients and creates .txt files with the same name inside the folder test, I am very new to shell and I have no idea how to do this, could you guys help me please?
Try using xargs with ls. ls -F displays all files in the directory client, but then displays the folders with an extra / at the end. the grep uses the extra / in the output of ls -F to only pass folders to the next command. Then, sed 's/\///g removes the extra / from grep, and passes the names to xargs. xargs will then pass the folders to the % symbol, and then make text files with the names.
ls -F client | grep / | sed 's/\///g' | xargs -I % touch tests/%.txt
I want to have a command in a variable that runs a program and specifies the output filename for it depending on the number of files exits (to work on a new file each time).
Here is what I have:
export MY_COMMAND="myprogram -o ./dir/outfile-0.txt"
However I would like to make this outfile number increases each time MY_COMMAND is being executed. You may suppose myprogram creates the file soon enough before the next call. So the number can be retrieved from the number of files exists in the directory ./dir/. I do not have access to change myprogram itself or the use of MY_COMMAND.
Thanks in advance.
Given that you can't change myprogram — its -o option will always write to the file given on the command line, and assuming that something also out of your control is running MY_COMMAND so you can't change the way that MY_COMMAND gets called, you still have control of MY_COMMAND
For the rest of this answer I'm going to change the name MY_COMMAND to callprog mostly because it's easier to type.
You can define callprog as a variable as in your example export callprog="myprogram -o ./dir/outfile-0.txt", but you could instead write a shell script and name that callprog, and a shell script can do pretty much anything you want.
So, you have a directory full of outfile-<num>.txt files and you want to output to the next non-colliding outfile-<num+1>.txt.
Your shell script can get the numbers by listing the files, cutting out only the numbers, sorting them, then take the highest number.
If we have these files in dir:
outfile-0.txt
outfile-1.txt
outfile-5.txt
outfile-10.txt
ls -1 ./dir/outfile*.txt produces the list
./dir/outfile-0.txt
./dir/outfile-1.txt
./dir/outfile-10.txt
./dir/outfile-5.txt
(using outfile and .txt means this will work even if there are other files not name outfile)
Scrape out the number by piping it through the stream editor sed … capture the number and keep only that part:
ls -1 ./dir/outfile*.txt | sed -e 's:^.*dir/outfile-\([0-9][0-9]*\)\.txt$:\1:'
(I'm using colon : instead of the standard slash / so I don't have to escape the directory separator in dir/outfile)
Now you just need to pick the highest number. Sort the numbers and take the top
| sort -rn | head -1
Sorting with -n is numeric, not lexigraphic sorting, -r reverses so the highest number will be first, not last.
Putting it all together, this will list the files, edit the names keeping only the numeric part, sort, and get just the first entry. You want to assign that to a variable to work with it, so it is:
high=$(ls -1 ./dir/outfile*.txt | sed -e 's:^.*dir/outfile-\([0-9][0-9]*\)\.txt$:\1:' | sort -rn | head -1)
In the shell (I'm using bash) you can do math on that, $[high + 1] so if high is 10, the expression produces 11
You would use that as the numeric part of your filename.
The whole shell script then just needs to use that number in the filename. Here it is, with lines broken for better readability:
#!/bin/sh
high=$(ls -1 ./dir/outfile*.txt \
| sed -e 's:^.*dir/outfile-\([0-9][0-9]*\)\.txt$:\1:' \
| sort -rn | head -1)
echo "myprogram -o ./dir/outfile-$[high + 1].txt"
Of course you wouldn't echo myprogram, you'd just run it.
you could do this in a bash function under your .bashrc by using wc to get the number of files in the dir and then adding 1 to the result
yourfunction () {
dir=/path/to/dir
filenum=$(expr $(ls $dir | wc -w) + 1)
myprogram -o $dir/outfile-${filenum}.txt
}
this should get the number of files in $dir and append 1 to that number to get the number you need for the filename. if you place it in your .bashrc or under .bash_aliases and source .bashrc then it should work like any other shell command
You can try exporting a function for MY_COMMAND to run.
next_outfile () {
my_program -o ./dir/outfile-${_next_number}.txt
((_next_number ++ ))
}
export -f next_outfile
export MY_COMMAND="next_outfile" _next_number=0
This relies on a "private" global variable _next_number being initialized to 0 and not otherwise modified.
I want to add sequential number for each file and its contents in a directory. The sequential number should be prefixed with the filename and for each line of its contents should have the same number prefixed. In this manner, the sequential numbers should be generated for all the files(for names and its contents) in the sub-folders of the directory.
I have tried using maxdepth, rename, print function as a part. but it throws error saying that "-maxdepth" - not a valid option.
I have already a part of code(to print the names and contents of text files in a directory) and this logic should be appended with it.
#!bin/bash
cd home/TESTING
for file in home/TESTING;
do
find home/TESTING/ -type f -name *.txt -exec basename {} ';' -exec cat {} \;
done
P.s - print, rename, maxdepth are not working
If the name of the first file is File1.txt and its contents is mentioned as "Louis" then the output for the filename should be 1File1.txt and the content should be as "1Louis".The same should be replaced with 2 for second file. In this manner, it has to traverse through all the subfolders in the directory and print accordingly. I have already a part of code and this logic should be appended with it.
There should be fail safe if you execute cd in a script. You can execute command in wrong directory if you don't.
In your attempt, the output would be the same even without the for cycle, as for file in home/TESTING only pass home/TESTING as argument to for so it only run once. In case of
for file in home/TESTING/* this would happen else how.
I used find without --maxdepth, so it will look into all subdirectory as well for *.txt files. If you want only the current directory $(find /home/TESTING/* -type f -name "*.txt") could be replaced to $(ls *.txt) as long you do not have directory that end to .txt there will be no problem.
#!/bin/bash
# try cd to directory, do things upon success.
if cd /home/TESTING ;then
# set sequence number
let "x = 1"
# pass every file to for that find matching, sub directories will be also as there is no maxdeapth.
for file in $(find /home/TESTING/* -type f -name "*.txt") ; do
# print sequence number, and base file name, processed by variable substitution.
# basename can be used as well but this is bash built in.
echo "${x}${file##*/}"
# print file content, and put sequence number before each line with stream editor.
sed 's#^#'"${x}"'#g' ${file}
# increase sequence number with one.
let "x++"
done
# unset sequence number
unset 'x'
else
# print error on stderr
echo 'cd to /home/TESTING directory is failed' >&2
fi
Variable Substitution:
There is more i only picked this 4 for now as they similar.
${var#pattern} - Use value of var after removing text that match pattern from the left
${var##pattern} - Same as above but remove the longest matching piece instead the shortest
${var%pattern} - Use value of var after removing text that match pattern from the right
${var%%pattern} - Same as above but remove the longest matching piece instead the shortest
So ${file##*/} will take the variable of $file and drop every caracter * before the last ## slash /. The $file variable value not get modified by this, so it still contain the path and filename.
sed 's#^#'"${x}"'#g' ${file} sed is a stream editor, there is whole books about its usage, for this particular one. It usually placed into single quote, so 's#^#1#g' will add 1 the beginning of every line in a file.s is substitution, ^ is the beginning of the file, 1 is a text, g is global if you not put there the g only first mach will be affected.
# is separator it can be else as well, like / for example. I brake single quote to let variable be used and reopened the single quote.
If you like to replace a text, .txt to .php, you can use sed 's#\.txt#\.php#g' file , . have special meaning, it can replace any singe character, so it need to be escaped \, to use it as a text. else not only file.txt will be matched but file1txt as well.
It can be piped , you not need to specify file name in that case, else you have to provide at least one filename in our case it was the ${file} variable that contain the filename. As i mentioned variable substitution is not modify variable value so its still contain the filename with path.
I have a folder where inside that contains two folder such as below:
/vobs/code
/vobs/code/item
/vobs/code/image
I would like to obtain its absolute path of the two folder.In my perl script my code looks like this .
#folder = `cd /vobs/code/item;ls -d $PWD/**`; or
#folder = `cd /vobs/code/item;ls -d $PWD/*`;
From online it says that with the two wild card it will display files and directories while one wild card display folder but I tried in the console it gives files and folder also with or extra one wild card.
The result that I get after executing the perl script it does not give the folder path inside /vobs/code.Instead it display the root folder path.I put a pwd in between the cd and ls the pwd shows that it actually cd to the directory and perform the next command.
#folder = `cd /vobs/code/item;ls -d $pwd/*`
The command above produce the same output as the first and 2nd command as the pwd is lower case and it display root folder path.
I tried readlink -f but it requires the folder name else it will prompt error.I can try to get the file name and pwd the path and join it to become a variable but it is kind of hassle and I would take it as final resort if I could not find any solutions.Please help Thanks
You can use glob to expand a wildcard. Use grep to filter a list, -f and -d (see -x) check for files and directories, respectivelly.
my #files_and_directories = glob '/vobs/code/*';
my #files = grep -f, #files_and_directories;
my #directories = grep -d, #files_and_directories;
Have you considered abs_path?
use Cwd 'abs_path';
my $base_path = abs_path('vobs/code');
for my $dir ( qw/item image/ ){
my $path = abs_path(join("/",$base_path,$dir));
# Ofc, this will only work if vobs/code is in working directory.
my #files_in_dir = grep{ -f $_ }<$path/*>; # these will be absolute
}
I would like to create a bash script to list all the directories in a directory provided by the user via input, or all the directories in the current directory (given no input).
Here's what I have thus far, but when I execute it I encounter two problems.
1) The script completely ignores my input. The file is located on my desktop but when I type in "home" as the input, the script simply prints the directories of the Desktop (current directory).
2) The directories are printed on their own lines (intended) but it treats each word in a folder name as its own folder. i.e. is printed as:
this
folder
Here's the code I have so far:
#!/bin/bash
echo -n "Enter a directory to load files: "
read d
if [ $d="" ]; #if input is blank, assume d = current directory
then d=${PWD##*/}
for i in $(ls -d */);
do echo ${i%%/};
done
else #otherwise, print sub-directories of given directory
for i in $(ls -d */);
do echo ${i%%/};
done
fi
Also in your response please explain your answer as I'm very new to bash.
Thanks for looking, I appreciate your time.
EDIT: Thanks to John1024's answer, I came up with the following:
#!/bin/bash
echo -n "Enter a directory to load files: "
IFS= read d
ls -1 -d "${d:-.}"/*/
And it does everything I need. Much appreciated!
I believe that this script accomplishes what you want:
#!/bin/sh
ls -1 -d "${1:-.}"/*/
Usage example:
$ bash ./script.sh /usr/X11R6
/usr/X11R6/bin
/usr/X11R6/man
Explanation:
-1 tells ls to print each file/directory on a separate line
-d tells ls to list directories by name instead of their contents
The shell will ${1:-.} to be the first argument to the script if there is one or . (which means the current directory) if there isn't.
Enhancement
The above script displays a / at the end of each directory name. If you don't want that, we can use sed to remove trailing slashes from the output:
#!/bin/sh
ls -1d ${1:-.}/*/ | sed 's|/$||'
Revised Version of Your Script
Starting with your script, some simplifications can be made:
#!/bin/bash
echo -n "Enter a directory to load files: "
IFS= read d
d=${d:-$PWD}
for i in "$d"/*/
do
echo ${i%%/}
done
Notes:
IFS= read d
Normally leading and trailing white space are stripped before the input is assigned to d. By setting IFS to an empty value, however, leading and trailing white space will be preserved. Thus this will work even if the pathologically strange case where the user specifies a directory whose name begins or ends with white space.
If the user enters a backslash, the shell will try to process it as an escape. If you don't like that, use IFS= read -r d and backslashes will be treated as normal characters, not escapes.
d=${d:-$PWD}
If the user supplied a value for d, this leaves it unchanged. If he didn't, this assigns it to $PWD.
for i in "$d"/*/
This will loop over every subdirectory of $d and will correctly handle subdirectory names with spaces, tabs, or any other odd character.
By contrast, consider:
for i in $(ls -d */)
After ls executes here, the shell will split up the output into individual words. This is called "word splitting" and is why this form of the for loop should be avoided.
Notice the double-quotes in for i in "$d"/*/. They are there to prevent word splitting on $d.