I need to automate a process using a script and generate output files similar to the name of the input files but with some additions to it.
my process is a Java code. two input arguments and two output arguments.
java #process_class# abc.txt abd.txt abc.1.out abd.a.out
If i want to iterate this for the set of text files in my folder how can i do this
If you have the files a.txt, b.txt, and c.txt in the directory in which this is run, this program will output a_2.txt, b_2.txt, and c_2.txt with foo appended to each (replace the foo line with your processing commands).
for f in *.txt;
do f2=${f%.*}_2.txt;
cp $f $f2;
echo "Processing $f2 file...";
echo "foo" >> $f2; # Your processing command here
done
VAR="INPUTFILENAME"
# One solution this does not use the VAR:
touch INPUTFILENAME{1,2,3,4,5,6,7,8,9,10}
# Another
for i in `seq 1 20` ; do
touch "${VAR}${i}"
done
And there are several other ways.
Related
I would like to know how to access the contents of a variety of txt files by passing arguments into shell scripts. I'll have different files and I'm expecting to execute with this command:
./script.sh FileA.txt
What should I put into my shell script so that I can access and manipulate the contents of the files?
I tried this but it outputs 0:
echo "$#"
I also tried these, but both output nothing:
for i in $1
do
echo "$i"
done
echo "$1"
To sum up the contents see this link to understand bash arguments more https://tecadmin.net/tutorial/bash-scripting/bash-command-arguments/ . Also as #Barmar said, to iterate through a list of arguments of unknown quantity use for i in "$#" .
edit
and as #Barmar said, $1 is simply the name of the argument. So echoing $1 will just echo the name.
I don't understand your question fully. Lets assume you have list of file names in a text file "FileA.txt".
And you wanted to run some commands for each file in the "FileA.txt" file.
Can you try below:
for i in `cat $1`
do
echo $i
done
How to make a shell script that receives one or more text files and removes from them whitespaces and blanklines. After that new files will have a random 2-digit number in front of them.
For example File1.txt generates File1_56.txt
Tried this:
#!/bin/bash
for file in "$*"; do
sed -e '/^$/d;s/[[:blank:]]//g' $* >> "$*_$$.txt"
done
But when I give 2 files as input script merges them into one single file, when I want for each file a separate one.
Try:
#!/bin/bash
for file in "$#"; do
sed -e '/^$/d;s/[[:blank:]]//g' "$file" >> "${file%.txt}_$$.txt"
done
Notes
To loop over each argument without word splitting or other hazards, use for file in "$#" not for file in "$*"
To run the sed command on one file instead of all, specify "$file" as the file, not $*.
To save the output to the correct file, use "${file%.txt}_$$.txt" where ${file%.txt} is an example of suffix removal: it removes the final .txt from the file name.
$$ is the process ID. The title says mentions a "random" number. If you want a random number, replace $$ with $RANDOM.
I run the following code for concatenating files in a directory given as the argument for the script file in bash
for i in $*
do
cat $* > /home/christy/Documents/filetest/catted.txt
done
This produce the error
cat: /home/christy/Documents/filetest/catted.txt: input file is output file
I think there are at least 4 things wrong with your script....
Firstly, your loop will set the value of i to the name of each file in succession, so you would want to actually use i inside your loop, like this:
for i in $*
cat "$i" ....somewhere
done
Secondly, if you use the > redirection, each file will land exactly on top of the previous one, so you should really use the >> redirection will append the current file to the end of the previous one like this
for i in $*
do
cat "$i" >> ...somewhere
done
Thirdly, I think you should use double-quoted "$#" to get all your command-line arguments, rather than plain $*
for i in "$#"
...
Fourthly, you can achieve the exact effect I think you want with this simpler command:
cat "$#" > /home/christy/Documents/filetest/catted.txt
You can't cat a file back onto itself. That's what "input file is output file" means. Because catted.txt shows up in your list of arguments to cat, it is going to try to cat to itself. So, move catted.txt to somewhere other than the source directory.
I need help copying content from various files to others (same name and format, different path).
For example, $HOME/initial/baby.desktop has text which I need to write into $HOME/scripts/baby.desktop. This is very simple for a single file, but I have 2500 files in $HOME/initial/ and the same number in $HOME/scripts/ with corresponding names (same names and format). I want append (copy) the content of file in path A to path B (which have the same name and format), to the end of file in path B without erase the content of file in path B.
Example content of $HOME/initial/*.desktop to final $HOME/scripts/*.desktop. I tried the following, but it don't work:
cd $HOME/initial/
for i in $( ls *.desktop ); do egrep "Icon" $i >> $HOME/scripts/$i; done
Firstly, I would backup $HOME/initial and $HOME/scripts, because there is lots of scope for people misunderstanding your question. Like this:
cd $HOME
tar -cvf initial.tar initial
tar -cvf scripts.tar scripts
That will put all the files in $HOME/initial into a single tarfile called initial.tar and all the files in $HOME/scripts into a single tarfile called scripts.tar.
Now for your question... in general, if you want to put the contents of FileB onto the end of FileA, the command is
cat FileB >> FileA
Note the DOUBLE ">>" which means "append" rather than single ">" which means overwrite.
So, I think you want to do this:
cd $HOME/initial/baby.desktop
cat SomeFile >> $HOME/scripts/baby.desktop/SomeFile
where SomeFile is the name of any file you choose to test with. I would test that has worked and then, if you are happy with that, go ahead and run the same command inside a loop:
cd $HOME/initial/baby.desktop
for SOURCE in *
do
DESTINATION="$HOME/scripts/baby.desktop/$SOURCE"
echo Appending "$SOURCE" to "$DESTINATION"
#cat "$SOURCE" >> "$DESTINATION"
done
When the output looks correct, remove the "#" at the start of the penultimate line and run it again.
I solved it, if some people want learn how to resolve is very simple:
using Sed
I need only the match (or pattern) line "Icon=/usr/share/some_picture.png into $HOME/initial/example.desktop to other with same name and format $HOME/scripts/example.desktop, but I had a lot of .desktop files (2500 files)
cd $HOME/initial
STRING_LINE=`grep -l -R "Icon=" *.desktop`
for i in $STRING_LINE; do sed -ne '/Icon=/ p' $i >> $HOME/scripts/$i ; done
_________
If you need only copy all to other file with same name and format
using cat
cd $HOME/initial
STRING_LINE=`grep -l -R "Icon=" *.desktop`
for i in $STRING_LINE; do cat $i >> $HOME/scripts/$i ; done
I am currently trying to write a script that will loop through multiple directories. The main raw_data directory contains ~150 subdirectories (subj001, subj002,...,subj00n), each of which has several subdirectories.
How can I make sure that the output from the script given bellow will be sent back to the specific subdirectory (e.g. subj0012) the input was taken from, rather than the current directory (raw_data)?
#!/bin/bash
for dir in ~raw_data/*
do
tractor -d -r -b preproc RunStages:1
done
Thank you.
The name of the dir you want to save the output to is in $dir, right? So, just send the output there via redirection:
#!/bin/bash
for dir in ~raw_data/* ; do
tractor -d -r- b preproc RunStages:1 > $dir/output
done
You should make sure that what you are processing really is a directory, though.
You can use output of find to run a loop and append the output at desired location like this:
while read d
do
echo $d >> ~raw_data/subj0012/output
done < <(find ~raw_data -type d)