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
Related
All of my file names follow this pattern:
abc_001.jpg
def_002.jpg
ghi_003.jpg
I want to replace the characters before the numbers and the underscore (not necessarily letters) with the name of the directory in which those files are located. Let's say this directory is called 'Pictures'. So, it would be:
Pictures_001.jpg
Pictures_002.jpg
Pictures_003.jpg
Normally, the way this website works, is that you show what you have done, what problem you have, and we give you a hint on how to solve it. You didn't show us anything, so I will give you a starting point, but not the complete solution.
You need to know what to replace: you have given the examples abc_001 and def_002, are you sure that the length of the "to-be-replaced" part always is equal to 3? In that case, you might use the cut basic command for deleting this. In other ways, you might use the position of the '_' character or you might use grep -o for this matter, like in this simple example:
ls -ltra | grep -o "_[0-9][0-9][0-9].jpg"
As far as the current directory is concerned, you might find this, using the environment variable $PWD (in case Pictures is the deepest subdirectory, you might use cut, using '/' as a separator and take the last found entry).
You can see the current directory with pwd, but alse with echo "${PWD}".
With ${x#something} you can delete something from the beginning of the variable. something can have wildcards, in which case # deletes the smallest, and ## the largest match.
First try the next command for understanding above explanation:
echo "The last part of the current directory `pwd` is ${PWD##*/}"
The same construction can be used for cutting the filename, so you can do
for f in *_*.jpg; do
mv "$f" "${PWD##*/}_${f#*_}"
done
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
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 $f"
done
& you should not parse ls output.
Take a list of your files)
If you want to take list of your files and see them:
ls ###Takes list###
ls -sh ###Takes list + File size###
...
If you want to send list of files to a file to read and check them later:
ls > FileName.Format ###Takes list and sends them to a file###
ls > FileName.Format ###Takes list with file size and sends them to a file###
I know this question has been asked, but I can't find more than one solution, and it does not work for me. Essentially, I'm looking for a bash script that will take a file list that looks like this:
image1.jpg
image2.jpg
image3.jpg
And then make a copy of each one, but number it sequentially backwards. So, the sequence would have three new files created, being:
image4.jpg
image5.jpg
image6.jpg
And yet, image4.jpg would have been an untouched copy of image3.jpg, and image5.jpg an untouched copy of image2.jpg, and so on. I have already tried the solution outlined in this stackoverflow question with no luck. I am admittedly not very far down the bash scripting path, and if I take the chunk of code in the first listed answer and make a script, I always get "2: Syntax error: "(" unexpected" over and over. I've tried changing the syntax with the ( around a bit, but no success ever. So, either I am doing something wrong or there's a better script around.
Sorry for not posting this earlier, but the code I'm using is:
image=( image*.jpg )
MAX=${#image[*]}
for i in ${image[*]}
do
num=${i:5:3} # grab the digits
compliment=$(printf '%03d' $(echo $MAX-$num | bc))
ln $i copy_of_image$compliment.jpg
done
And I'm taking this code and pasting it into a file with nano, and adding !#/bin/bash as the first line, then chmod +x script and executing in bash via sh script. Of course, in my test runs, I'm using files appropriately titled image1.jpg - but I was also wondering about a way to apply this script to a directory of jpegs, not necessarily titled image(integer).jpg - in my file keeping structure, most of these are a single word, followed by a number, then .jpg, and it would be nice to not have to rewrite the script for each use.
Perhaps something like this. It will work well for something like script image*.jpg where the wildcard matches a set of files which match a regular pattern with monotonously increasing numbers of the same length, and less ideally with a less regular subset of the files in the current directory. It simply assumes that the last file's digit index plus one through the total number of file names is the range of digits to loop over.
#!/bin/sh
# Extract number from final file name
eval lastidx=\$$#
tmp=${lastidx#*[!0-9][0-9]}
lastidx=${lastidx#${lastidx%[0-9]$tmp}}
tmp=${lastidx%[0-9][!0-9]*}
lastidx=${lastidx%${lastidx#$tmp[0-9]}}
num=$(expr $lastidx + $#)
width=${#lastidx}
for f; do
pref=${f%%[0-9]*}
suff=${f##*[0-9]}
# Maybe show a warning if pref, suff, or width changed since the previous file
printf "cp '$f' '$pref%0${width}i$suff'\\n" $num
num=$(expr $num - 1)
done |
sh
This is sh-compatible; the expr stuff and the substring extraction up front is ugly but Bourne-compatible. If you are fine with the built-in arithmetic and string manipulation constructs of Bash, converting to that form should be trivial.
(To be explicit, ${var%foo} returns the value of $var with foo trimmed off the end, and ${var#foo} does similar trimming from the beginning of the value. Regular shell wildcard matching operators are available in the expression for what to trim. ${#var} returns the length of the value of $var.)
Maybe your real test data runs from 001 to 300, but here you have image1 2 3, and therefore you extract one, not three digits from the filename. num=${i:5:1}
Integer arithmetic can be done in the bash without calling bc
${#image[#]} is more robust than ${#image[*]}, but shouldn't be a difference here.
I didn't consult a dictionary, but isn't compliment something for your girl friend? The opposite is complement, isn't it? :)
the other command made links - to make copies, call cp.
Code:
#!/bin/bash
image=( image*.jpg )
MAX=${#image[#]}
for i in ${image[#]}
do
num=${i:5:1}
complement=$((2*$MAX-$num+1))
cp $i image$complement.jpg
done
Most important: If it is bash, call it with bash. Best: do a shebang (as you did), make it executable and call it by ./name . Calling it with sh name will force the wrong interpreter. If you don't make it executable, call it bash name.
I'm trying to make a script that will go into a directory and run my own application with each file matching a regular expression, specifically Test[0-9]*.txt.
My input filenames look like this TestXX.txt. Now, I could just use cut and chop off the Test and .txt, but how would I do this if XX wasn't predefined to be two digits? What would I do if I had Test1.txt, ..., Test10.txt? In other words, How would I get the [0-9]* part?
Just so you know, I want to be able to make a OutputXX.txt :)
EDIT:
I have files with filename Test[0-9]*.txt and I want to manipulate the string into Output[0-9]*.txt
Would something like this help?
#!/bin/bash
for f in Test*.txt ;
do
process < $f > ${f/Test/Output}
done
Bash Shell Parameter Expansion
A good tutorial on regexes in bash is here. Summarizing, you need something like:
if [[$filenamein =~ "^Test([0-9]*).txt$"]]; then
filenameout = "Output${BASH_REMATCH[1]}.txt"
and so on. The key is that, when you perform the =~" regex-match, the "sub-matches" to parentheses-enclosed groups in the RE are set in the entries of arrayBASH_REMATCH(the[0]entry is the whole match,1` the first parentheses-enclosed group, etc).
You need to use rounded brackets around the part you want to keep.
i.e. "Test([0-9]*).txt"
The syntax for replacing these bracketed groups varies between programs, but you'll probably find you can use \1 , something like this:
s/Test(0-9*).txt/Output\1.txt/
If you're using a unix shell, then 'sed' might be your best bet for performing the transformation.
http://www.grymoire.com/Unix/Sed.html#uh-4
Hope that helps
for file in Test[0-9]*.txt;
do
num=${file//[^0-9]/}
process $file > "Output${num}.txt"
done