Pass string to script - linux

I have a script, download, that takes a string and checks if a file has the filename of the string. If it doesn't, it then downloads it. All the filenames are in a file.
This command is not working:
cat filenames | ./download
Download source:
filename=$1
if [ ! -f $1 ];
then
wget -q http://www.example.com/nature/life/${filename}.rdf
fi
Sample filename file:
file1
file2
file3
file4
How do I pass the command output from the cat to the download script?

In your script $1 is the positional arg on the command line. ./download somefile would work, but cat filename | ./download streams the data into download, which you ignore.
You should read the advanced bash scripting guide, which will give you a good base for how bash scripting works. To fix this, change your command to:
cat filename | xargs -n 1 ./download
This will run ./download for each filename in your list. However, the filenames may have spaces or other special characters in them, which would break your script. You should look into alternatives ways of doing this, to avoid these problems.
Specifically, use a while loop to read your file. This properly escapes your filenames on each line, if they were input into the file correctly. That way, you avoid the problems cat would have with filenames like: fi/\nle.

You can pass a filename to a file that contains file names to your script:
./download filenames
And then loop through file names from the file name in $1:
$!/bin/bash
# Do sanity check
fname=$1
for f in $(<$fname); do
if [ ! -f "$f.rdf" ]; then
wget -q http://www.example.com/nature/life/${f}.rdf
fi
done

Related

How to rename file based on parent and child folder name in bash script

I would like to rename file based on parent/subparent directories name.
For example:
test.xml file located at
/usr/local/data/A/20180101
/usr/local/data/A/20180102
/usr/local/data/B/20180101
how to save test.xml file in /usr/local/data/output as
A_20180101_test.xml
A_20180102_test.xml
b_20180101_test.xml
tried shall script as below but does not help.
#!/usr/bin/env bash
target_dir_path="/usr/local/data/output"
for file in /usr/local/data/*/*/test.xml; do
l1="${file%%/*}"
l2="${file#*/}"
l2="${l2%%/*}"
filename="${file##*/}"
target_file_name="${l1}_${l2}_${filename}"
echo cp "$file" "${target_dir_path}/${target_file_name}"
done
Anything i am doing wrong in this shall script?
You can use the following command to do this operation:
source_folder="usr/local/data/";target_folder="target"; find $source_folder -type f -name test.xml | awk -v targetF=$target_folder 'BEGIN{FS="/"; OFS="_"}{printf $0" "; print targetF"/"$(NF-2),$(NF-1),$NF}' | xargs -n2 cp;
or on several lines for readibility:
source_folder="usr/local/data/";
target_folder="target";
find $source_folder -type f -name test.xml |\
awk -v targetF=$target_folder 'BEGIN{FS="/"; OFS="_"}{printf $0" "; print targetF"/"$(NF-2),$(NF-1),$NF}' |\
xargs -n2 cp;
where
target_folder is your target folder
source_folder is your source folder
the find command will search for all the test.xml named files present under this source folder
then the awk command will receive the target folder as a variable to be able to use it, then in the BEGIN bloc you define the field separator and output field separator, then you just print the initial filename as well as the new one
you use xargs to pass the result output grouped by 2 to the cp command and the trick is done
TESTED:
TODO:
you will just need to set up your source_folder and target_folder variables with what is on your environment and eventually put it in a script and you are good to go!
I've modified your code a little to get it to work. See comments in code
target_dir_path=""/usr/local/data/output"
for file in /usr/local/data/*/*/test.xml; do
tmp=${file%/*/*/*}
curr="${file#"$tmp/"}" # Extract wanted part of the filename
mod=${curr//[\/]/_} # Replace forward slash with underscore
mv "$file" "$target_dir_path$mod" # Move the file
done
if you have perl based rename command
$ for f in tst/*/*/test.xml; do
rename -n 's|.*/([^/]+)/([^/]+)/(test.xml)|./$1_$2_$3|' "$f"
done
rename(tst/A/20180101/test.xml, ./A_20180101_test.xml)
rename(tst/A/20180102/test.xml, ./A_20180102_test.xml)
rename(tst/B/20180101/test.xml, ./B_20180101_test.xml)
-n option is for dry run, remove it after testing
change tst to /usr/local/data and ./ to /usr/local/data/output/ for your usecase
.*/ to ignore file path
([^/]+)/([^/]+)/(test.xml) capture required portions
$1_$2_$3 re-arrange as required

Shell Script With sed and Random number

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.

Linux: get specific field from filename

I am currently learning linux bash scripting:
I have files in a folder with the following filename-pattern:
ABC01_-12ab_STRINGONE_logicMatches.txt
DEF02_-12ab_STRINGTWO_logicMatches.txt
JKL03_-12ab_STRINGTHREE_logicMatches.txt
I want to extract STRINGONE, STRINGTWO and STRINGTHREE as a list. To see, if my idea works, I wanted to echo my result to bash first.
Code of my bash-script (executed in the folder, where the files are located):
#!/bin/bash
for element in 'folder' do out='cut -d "_" -f2 $element | echo $out' done
Actual result:
error: unexpected end of file
Desired result:
STRINGONE
STRINGTWO
STRINGTHREE
(echoed in bash)
The idea you are doing is right. But the syntax of file globbing (looking for text files) and command substitution (running the cut command) is wrong. You need to do
for file in folder/*.txt;
# This condition handles the loop exit if no .txt files are found, and
# not throw errors
[ -f "$file" ] || continue
# The command-substitution syntax $(..) runs the command and returns the
# result out to the variable 'out'
out=$(cut -d "_" -f3 <<< "$file")
echo "$out"
done

Linux bash output fdirectory files to a text file with xargs and add new line

I want to generate a text file with the list of files present in the folder
ls | xargs echo > text.txt
I want to prepend the IP address to each file so that I can run parallel wget as per this post : Parallel wget in Bash
So my text.txt file content will have these lines :
123.123.123.123/file1
123.123.123.123/file2
123.123.123.123/file3
How can I append a string as the ls feeds xargs? (and also add line break at the end.)
Thank you
Simply printf and globbing to get the filenames:
printf '123.123.123.123/%s\n' * >file.txt
Or longer approach, leverage a for construct with help from globbing:
for f in *; do echo "123.123.123.123/$f"; done >file.txt
Assuming no filename with newline exists.

How to remove the extension of a file?

I have a folder that is full of .bak files and some other files also. I need to remove the extension of all .bak files in that folder. How do I make a command which will accept a folder name and then remove the extension of all .bak files in that folder ?
Thanks.
To remove a string from the end of a BASH variable, use the ${var%ending} syntax. It's one of a number of string manipulations available to you in BASH.
Use it like this:
# Run in the same directory as the files
for FILENAME in *.bak; do mv "$FILENAME" "${FILENAME%.bak}"; done
That works nicely as a one-liner, but you could also wrap it as a script to work in an arbitrary directory:
# If we're passed a parameter, cd into that directory. Otherwise, do nothing.
if [ -n "$1" ]; then
cd "$1"
fi
for FILENAME in *.bak; do mv "$FILENAME" "${FILENAME%.bak}"; done
Note that while quoting your variables is almost always a good practice, the for FILENAME in *.bak is still dangerous if any of your filenames might contain spaces. Read David W.'s answer for a more-robust solution, and this document for alternative solutions.
There are several ways to remove file suffixes:
In BASH and Kornshell, you can use the environment variable filtering. Search for ${parameter%word} in the BASH manpage for complete information. Basically, # is a left filter and % is a right filter. You can remember this because # is to the left of %.
If you use a double filter (i.e. ## or %%, you are trying to filter on the biggest match. If you have a single filter (i.e. # or %, you are trying to filter on the smallest match.
What matches is filtered out and you get the rest of the string:
file="this/is/my/file/name.txt"
echo ${file#*/} #Matches is "this/` and will print out "is/my/file/name.txt"
echo ${file##*/} #Matches "this/is/my/file/" and will print out "name.txt"
echo ${file%/*} #Matches "/name.txt" and will print out "/this/is/my/file"
echo ${file%%/*} #Matches "/is/my/file/name.txt" and will print out "this"
Notice this is a glob match and not a regular expression match!. If you want to remove a file suffix:
file_sans_ext=${file%.*}
The .* will match on the period and all characters after it. Since it is a single %, it will match on the smallest glob on the right side of the string. If the filter can't match anything, it the same as your original string.
You can verify a file suffix with something like this:
if [ "${file}" != "${file%.bak}" ]
then
echo "$file is a type '.bak' file"
else
echo "$file is not a type '.bak' file"
fi
Or you could do this:
file_suffix=$(file##*.}
echo "My file is a file '.$file_suffix'"
Note that this will remove the period of the file extension.
Next, we will loop:
find . -name "*.bak" -print0 | while read -d $'\0' file
do
echo "mv '$file' '${file%.bak}'"
done | tee find.out
The find command finds the files you specify. The -print0 separates out the names of the files with a NUL symbol -- which is one of the few characters not allowed in a file name. The -d $\0means that your input separators are NUL symbols. See how nicely thefind -print0andread -d $'\0'` together?
You should almost never use the for file in $(*.bak) method. This will fail if the files have any white space in the name.
Notice that this command doesn't actually move any files. Instead, it produces a find.out file with a list of all the file renames. You should always do something like this when you do commands that operate on massive amounts of files just to be sure everything is fine.
Once you've determined that all the commands in find.out are correct, you can run it like a shell script:
$ bash find.out
rename .bak '' *.bak
(rename is in the util-linux package)
Caveat: there is no error checking:
#!/bin/bash
cd "$1"
for i in *.bak ; do mv -f "$i" "${i%%.bak}" ; done
You can always use the find command to get all the subdirectories
for FILENAME in `find . -name "*.bak"`; do mv --force "$FILENAME" "${FILENAME%.bak}"; done

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