I am completely new to bash script. I am trying to do something really basic before using it for my actual requirement. I have written a simple code, which should print test code as many times as the number of files in the folder.
My code:
for variable in `ls test_folder`; do
echo test code
done
"test_folder" is a folder which exist in the same directory where the bash.sh file lies.
PROBLEM: If the number of files are one then, it prints single time but if the number of files are more than 1 then, it prints a different count. For example, if there are 2 files in "test_folder" then, test code gets printed 3 times.
Just use a shell pattern (aka glob):
for variable in test_folder/*; do
# ...
done
You will have to adjust your code to compensate for the fact that variable will contain something like test_folder/foo.txt instead of just foo.txt. Luckily, that's fairly easy; one approach is to start the loop body with
variable=${variable#test_folder/}
to strip the leading directory introduced by the glob.
Never loop over the output of ls! Because of word splitting files having spaces in their names will be a problem. Sure, you could set IFS to $\n, but files in UNIX can also have newlines in their names.
Use find instead:
find test_folder -maxdepth 1 -mindepth 1 -exec echo test \;
This should work:
cd "test_folder"
for variable in *; do
#your code here
done
cd ..
variable will contain only the file names
Related
I have a script that I call with an application, I can't run it from command line. I derive the directory where the script is called and in the next variable go up 1 level where my files are stored. From there I have 3 variables with the full path and file names (with wildcard), which I will refer to as "masks".
I need to find and "do something with" (copy/write their names to a new file, whatever else) to each of these masks. The do something part isn't my obstacle as I've done this fine when I'm working with a single mask, but I would like to do it cleanly in a single loop instead of duplicating loop and just referencing each mask separately if possible.
Assume in my $FILESFOLDER directory below that I have 2 existing files, aaa0.csv & bbb0.csv, but no file matching the ccc*.csv mask.
#!/bin/bash
SCRIPTFOLDER=${0%/*}
FILESFOLDER="$(dirname "$SCRIPTFOLDER")"
ARCHIVEFOLDER="$FILESFOLDER"/archive
LOGFILE="$SCRIPTFOLDER"/log.txt
FILES1="$FILESFOLDER"/"aaa*.csv"
FILES2="$FILESFOLDER"/"bbb*.csv"
FILES3="$FILESFOLDER"/"ccc*.csv"
ALLFILES="$FILES1
$FILES2
$FILES3"
#here as an example I would like to do a loop through $ALLFILES and copy anything that matches to $ARCHIVEFOLDER.
for f in $ALLFILES; do
cp -v "$f" "$ARCHIVEFOLDER" > "$LOGFILE"
done
echo "$ALLFILES" >> "$LOGFILE"
The thing that really spins my head is when I run something like this (I haven't done it with the copy command in place) that log file at the end shows:
filesfolder/aaa0.csv filesfolder/bbb0.csv filesfolder/ccc*.csv
Where I would expect echoing $ALLFILES just to show me the masks
filesfolder/aaa*.csv filesfolder/bbb*.csv filesfolder/ccc*.csv
In my "do something" area, I need to be able to use whatever method to find the files by their full path/name with the wildcard if at all possible. Sometimes my network is down for maintenance and I don't want to risk failing a change directory. I rarely work in linux (primarily SQL background) so feel free to poke holes in everything I've done wrong. Thanks in advance!
Here's a light refactoring with significantly fewer distracting variables.
#!/bin/bash
script=${0%/*}
folder="$(dirname "$script")"
archive="$folder"/archive
log="$folder"/log.txt # you would certainly want this in the folder, not $script/log.txt
shopt -s nullglob
all=()
for prefix in aaa bbb ccc; do
cp -v "$folder/$prefix"*.csv "$archive" >>"$log" # append, don't overwrite
all+=("$folder/$prefix"*.csv)
done
echo "${all[#]}" >> "$log"
The change in the loop to append the output or cp -v instead of overwrite is a bug fix; otherwise the log would only contain the output from the last loop iteration.
I would probably prefer to have the files echoed from inside the loop as well, one per line, instead of collect them all on one humongous line. Then you can remove the array all and instead simply
printf '%s\n' "$folder/$prefix"*.csv >>"$log"
shopt -s nullglob is a Bash extension (so won't work with sh) which says to discard any wildcard which doesn't match any files (the default behavior is to leave globs unexpanded if they don't match anything). If you want a different solution, perhaps see Test whether a glob has any matches in Bash
You should use lower case for your private variables so I changed that, too. Notice also how the script variable doesn't actually contain a folder name (or "directory" as we adults prefer to call it); fixing that uncovered a bug in your attempt.
If your wildcards are more complex, you might want to create an array for each pattern.
tmpspaces=(/tmp/*\ *)
homequest=($HOME/*\?*)
for file in "${tmpspaces[#]}" "${homequest[#]}"; do
: stuff with "$file", with proper quoting
done
The only robust way to handle file names which could contain shell metacharacters is to use an array variable; using string variables for file names is notoriously brittle.
Perhaps see also https://mywiki.wooledge.org/BashFAQ/020
Its my first time to use BASH scripting and been looking to some tutorials but cant figure out some codes. I just want to list all the files in a folder, but i cant do it.
Heres my code so far.
#!/bin/bash
# My first script
echo "Printing files..."
FILES="/Bash/sample/*"
for f in $FILES
do
echo "this is $f"
done
and here is my output..
Printing files...
this is /Bash/sample/*
What is wrong with my code?
You misunderstood what bash means by the word "in". The statement for f in $FILES simply iterates over (space-delimited) words in the string $FILES, whose value is "/Bash/sample" (one word). You seemingly want the files that are "in" the named directory, a spatial metaphor that bash's syntax doesn't assume, so you would have to explicitly tell it to list the files.
for f in `ls $FILES` # illustrates the problem - but don't actually do this (see below)
...
might do it. This converts the output of the ls command into a string, "in" which there will be one word per file.
NB: this example is to help understand what "in" means but is not a good general solution. It will run into trouble as soon as one of the files has a space in its name—such files will contribute two or more words to the list, each of which taken alone may not be a valid filename. This highlights (a) that you should always take extra steps to program around the whitespace problem in bash and similar shells, and (b) that you should avoid spaces in your own file and directory names, because you'll come across plenty of otherwise useful third-party scripts and utilities that have not made the effort to comply with (a). Unfortunately, proper compliance can often lead to quite obfuscated syntax in bash.
I think problem in path "/Bash/sample/*".
U need change this location to absolute, for example:
/home/username/Bash/sample/*
Or use relative path, for example:
~/Bash/sample/*
On most systems this is fully equivalent for:
/home/username/Bash/sample/*
Where username is your current username, use whoami to see your current username.
Best place for learning Bash: http://www.tldp.org/LDP/abs/html/index.html
This should work:
echo "Printing files..."
FILES=(/Bash/sample/*) # create an array.
# Works with filenames containing spaces.
# String variable does not work for that case.
for f in "${FILES[#]}" # iterate over the array.
do
echo "this is $f"
done
& you should not parse ls output.
Take a list of your files)
If you want to take list of your files and see them:
ls ###Takes list###
ls -sh ###Takes list + File size###
...
If you want to send list of files to a file to read and check them later:
ls > FileName.Format ###Takes list and sends them to a file###
ls > FileName.Format ###Takes list with file size and sends them to a file###
I have a bunch of folders containing images that are in order but are not sequential like this:
/root
/f1
img21.jpg
img24.jpg
img26.jpg
img27.jpg
/f2
img06.jpg
img14.jpg
img36.jpg
img57.jpg
and I want to get them looking like this, having the folder title as well as having all the images in sequential order:
/root
/f1
f1_01.jpg
f1_02.jpg
f1_03.jpg
f1_04.jpg
/f2
f2_01.jpg
f2_02.jpg
f2_03.jpg
f2_04.jpg
I'm not sure how to do this using shell script.
Thanks in advance!
Use a for loop to iterate over the directories and another for loop to iterate over the files. Maintain a counter that you increment by 1 for each file.
There's no direct convenient way of padding numbers with leading zeroes. You can call printf, but that's a little slow. A useful, fast trick is to start counting at 101 (if you want two-digit numbers — 1000 if you want 3-digit numbers, and so on) and strip the leading 1.
cd /root
for d in */; do
i=100
for f in "$d/"*; do
mv -- "$f" "$d/${d%/}_${i#1}.${f##*.}"
i=$(($i+1))
done
done
${d%/} strips / at the end of $d, ${i#1} strips 1 at the start of $i and ${f##*.} strip everything from $f except what follows the last .. These constructs are documented in the section on parameter expansion in your shell's manual.
Note that this script assumes that the target file names will not clash with the names of existing files. If you have a directory called img, some files will be overwritten. If this may be a problem, the simplest method is to first move all the files to a different directory, then move them back to the original directory as you rename them.
Within a directory, ls will give you files in lexical order, which gets you the correct sort. So you can do something like this:
let i=0
ls *.jpg | while read file; do
mv $file prefix_$(printf "%02d" $i).jpg
let i++
done
This will take all the *.jpg files and rename them starting with prefix_00.jpg, prefix_01.jpg and so forth.
This obviously only works for a single directory, but hopefully with a little work you can use this to build something that will do what you want.
I am trying to make heads or tails of a shell script. Could someone please explain this line?
$FILEDIR is a directory containing files. F is a marker in an array of files that is returned from this command:
files=$( find $FILEDIR -type f | grep -v .rpmsave\$ | grep -v .swp\$ )
The confusing line is within a for loop.
for f in $files; do
target=${f:${#FILEDIR}}
<<do some more stuff>>
done
I've never seen the colon, and the hash before in a shell script for loop. I haven't been able to find any documentation on them... could someone try and enlighten me? I'd appreciate it.
There are no arrays involved here. POSIX sh doesn't have arrays (assuming you're not using another shell based upon the tags).
The colon indicates a Bash/Ksh substring expansion. These are also not POSIX. The # prefix expands to the number of characters in the parameter. I imagine they intended to chop off the directory part and assign it to target.
To explain the rest of that: first find is run and hilariously piped into two greps which do what could have been done with find alone (except breaking on possible filenames containing newlines), and the output saved into files. This is also something that can't really be done correctly if restricted only to POSIX tools, but there are better ways.
Next, files is expanded unquoted and mutalated by the shell in more ridiculous ways for the for loop to iterate over the meaningless results. If the rest of the script is this bad, probably throw it out and start over. There's no way that will do what's expected.
The colon can be as a substring. So:
A=abcdefg
echo ${A:4}
will print the output:
efg
I'm not sure why they would use a file directory as the 2nd parameter though...
If you are having problems understanding the for loop section, try http://www.dreamsyssoft.com/unix-shell-scripting/loop-tutorial.php
I am trying to iterate through files in the same directory with only one 4 in them.
Here is what I have so far. The problem with my current script is that files with any number of 4's get selected, not files with only one 4.
for i in *4*.cpp;
do
...
Sort of like [!4] but for any number of non 4 characters.
*http://www.tuxfiles.org/linuxhelp/wildcards.html
I want to iterate through file names such as me4.cpp, 4.cpp, and hi4hi.cpp
I want to ignore file names such as lala.cpp, 44.cpp, 4hi4.cpp
Thank you!
Figured it out. I tried [!4]* on a whim.
Oops turned out I didn't. That is interpreted as ([!4]) then (*)
The grep style regex you need is:
^[^4]*4[^4]*$
A bunch of not-4's after the start of the line, a 4, and another bunch of not-4's to the end of the line.
In pure shell, consider using a case statement:
for file in *4*.cpp
do
case "$file" in
(*4*4*) : Ignore;;
(*) : Process;;
esac
done
That looks for names containing 4's, and then ignores those containing 2 or more 4's.
How about using find
find ./ -regex "<regular expression>"
Assuming bash:
shopt -s extglob
for file in *([^4])4*([^4]).cpp; ...
where *([^4]) means zero or more characters that are not "4"