I'm trying to change multiple file names with a for loop.
This works to send the output to the screen:
for i in *.gz; do echo $i | sed 's/\-//g'; done
However, when I try to overwrite the file name using sed -i, I get this error:
for i in *.gz; do echo $i | sed -i 's/\-//g'; done
sed: no input files
Any suggestions?
there is a command for this
$ rename - '' *.gz
NB. this is the standard one, not the advanced perl version.
Use Perl rename instead:
rename 's/-//g' *.gz
Or use simple parameter expansion:
for i in *.gz; do mv -- "$i" "${i//-}"; done
Related
I have a folder with several hundred .txt files with numbers. The files are named in a format word1_word2_word3_word4.txt. The words are different for each of the .txt files.
I would like to add a header to each of those text files based on the filename such that the header is of the format:
'>c_word1_word2'
Is there a way to do this for all the .txt files using the command line or a bash script?
Use this in bash and sed using some parameter expansion builtins :
for i in *.txt; do j="${i%_*}"; sed "1i >c_${j%_*}" "$i"; done
Output :
$ ls -1
word1_word2_word3_word4.txt
word2_word2_word3_word4.txt
word3_word2_word3_word4.txt
$ cat word1_word2_word3_word4.txt
foo
bar
base
$ for i in *.txt; do j="${i%_*}"; sed -i "1i >c_${j%_*}" $i; done
$ cat word1_word2_word3_word4.txt
>c_word1_word2
foo
bar
base
$
-i switch edit the file in place
This is just an alternative to Gilles' answer, using a regular expression
for f in *.txt; do
if [[ $f =~ ^([^_]+_[^_]+) ]]; then
sed -i "1i>c_${BASH_REMATCH[1]}" "$f"
fi
done
I have a list of filenames like this in bash
UTSHoS10_Other_CAAGCC-TTAGGA_R_160418.R1.fq.gz
UTSHoS10_Other_CAAGCC-TTAGGA_R_160418.R2.fq.gz
UTSHoS11_Other_AGGCCT-TTAGGA_R_160418.R2.fq.gz
UTSHoS11_Other_AGGCCT-TTAGGA_R_160418.R2.fq.gz
UTSHoS12_Other_GGCAAG-TTAGGA_R_160418.R1.fq.gz
UTSHoS12_Other_GGCAAG-TTAGGA_R_160418.R2.fq.gz
And I want them to look like this
UTSHoS10_R1.fq.gz
UTSHoS10_R2.fq.gz
UTSHoS11_R1.fq.gz
UTSHoS11_R2.fq.gz
UTSHoS12_R1.fq.gz
UTSHoS12_R2.fq.gz
I do not have the perl rename command and sed 's/_Other*160418./_/' *.gz
is not doing anything. I've tried other rename scripts on here but either nothing occurs or my shell starts printing huge amounts of code to the console and freezes.
This post (Removing Middle of Filename) is similar however the answers given do not explain what specific parts of the command are doing so I could not apply it to my problem.
Parameter expansions in bash can perform string substitutions based on glob-like patterns, which allows for a more efficient solution than calling an extra external utility such as sed in each loop iteration:
for f in *.gz; do echo mv "$f" "${f/_Other_*-TTAGGA_R_160418./_}"; done
Remove the echo before mv to perform actual renaming.
You can do something like this in the directory which contains the files to be renamed:
for file_name in *.gz
do
new_file_name=$(sed 's/_[^.]*\./_/g' <<< "$file_name");
mv "$file_name" "$new_file_name";
done
The pattern (_[^.]*\.) starts matching from the FIRST _ till the FIRST . (both inclusive). [^.]* means 0 or more non-dot (or non-period) characters.
Example:
AMD$ ls
UTSHoS10_Other_CAAGCC-TTAGGA_R_160418.R1.fq.gz UTSHoS12_Other_GGCAAG-TTAGGA_R_160418.R1.fq.gz
UTSHoS10_Other_CAAGCC-TTAGGA_R_160418.R2.fq.gz UTSHoS12_Other_GGCAAG-TTAGGA_R_160418.R2.fq.gz
UTSHoS11_Other_AGGCCT-TTAGGA_R_160418.R2.fq.gz
AMD$ for file_name in *.gz
> do new_file_name=$(sed 's/_[^.]*\./_/g' <<< "$file_name")
> mv "$file_name" "$new_file_name"
> done
AMD$ ls
UTSHoS10_R1.fq.gz UTSHoS10_R2.fq.gz UTSHoS11_R2.fq.gz UTSHoS12_R1.fq.gz UTSHoS12_R2.fq.gz
Pure Bash, using substring operation and assuming that all file names have the same length:
for file in UTS*.gz; do
echo mv -i "$file" "${file:0:9}${file:38:8}"
done
Outputs:
mv -i UTSHoS10_Other_CAAGCC-TTAGGA_R_160418.R1.fq.gz UTSHoS10_R1.fq.gz
mv -i UTSHoS10_Other_CAAGCC-TTAGGA_R_160418.R2.fq.gz UTSHoS10_R2.fq.gz
mv -i UTSHoS11_Other_AGGCCT-TTAGGA_R_160418.R2.fq.gz UTSHoS11_R2.fq.gz
mv -i UTSHoS11_Other_AGGCCT-TTAGGA_R_160418.R2.fq.gz UTSHoS11_R2.fq.gz
mv -i UTSHoS12_Other_GGCAAG-TTAGGA_R_160418.R1.fq.gz UTSHoS12_R1.fq.gz
mv -i UTSHoS12_Other_GGCAAG-TTAGGA_R_160418.R2.fq.gz UTSHoS12_R2.fq.gz
Once verified, remove echo from the line inside the loop and run again.
Going with your sed command, this can work as a bash one-liner:
for name in UTSH*fq.gz; do newname=$(echo $name | sed 's/_Other.*160418\./_/'); echo mv $name $newname; done
Notes:
I've adjusted your sed command: it had an * without a preceeding . (sed takes a regular expression, not a globbing pattern). Similarly, the dot needs escaping.
To see if it works, without actually renaming the files, I've left the echo command in. Easy to remove just that to make it functional.
It doesn't have to be a one-liner, obviously. But sometimes, that makes editing and browsing your command-line history easier.
I have three files in a directory that has the structure like this:
file.exe.trace, file.exe.trace.functions and file.exe.trace.netlog
I want to know how can I get file.exe as file name?
In other world I need to get file name that has the .trace extension? I should note that as you can see all the files has the .trace part.
If $FILENAME has the name, the root part can be gotten from ${FILENAME%%.trace*}
for FILENAME in *.trace; do
echo ${FILENAME%%.trace*}
done
You can also use basename:
for f in *.trace; do
basename "$f" ".trace"
done
Update: The previous won't process files with extra extensions besides .trace like .trace.functions, but the following sed will do:
sed -r 's_(.*)\.trace.*_\1_' <(ls -c1)
You can also use it in a for loop instead:
for f in *.trace*; do
sed -r 's_(.*)\.trace.*_\1_' <<< "$f"
done
Try:
for each in *exe*trace* ; do echo $each | awk -F. '{print $1"."$2}' ; done | sort | uniq
I have the following files in the following format:
$ ls CombinedReports_LLL-*'('*.csv
CombinedReports_LLL-20140211144020(Untitled_1).csv
CombinedReports_LLL-20140211144020(Untitled_11).csv
CombinedReports_LLL-20140211144020(Untitled_110).csv
CombinedReports_LLL-20140211144020(Untitled_111).csv
CombinedReports_LLL-20140211144020(Untitled_12).csv
CombinedReports_LLL-20140211144020(Untitled_13).csv
CombinedReports_LLL-20140211144020(Untitled_14).csv
CombinedReports_LLL-20140211144020(Untitled_15).csv
CombinedReports_LLL-20140211144020(Untitled_16).csv
CombinedReports_LLL-20140211144020(Untitled_17).csv
CombinedReports_LLL-20140211144020(Untitled_18).csv
CombinedReports_LLL-20140211144020(Untitled_19).csv
I would like this part removed:
20140211144020 (this is the timestamp the reports were run so this will vary)
and end up with something like:
CombinedReports_LLL-(Untitled_1).csv
CombinedReports_LLL-(Untitled_11).csv
CombinedReports_LLL-(Untitled_110).csv
CombinedReports_LLL-(Untitled_111).csv
CombinedReports_LLL-(Untitled_12).csv
CombinedReports_LLL-(Untitled_13).csv
CombinedReports_LLL-(Untitled_14).csv
CombinedReports_LLL-(Untitled_15).csv
CombinedReports_LLL-(Untitled_16).csv
CombinedReports_LLL-(Untitled_17).csv
CombinedReports_LLL-(Untitled_18).csv
CombinedReports_LLL-(Untitled_19).csv
I was thinking simply along the lines of the mv command, maybe something like this:
$ ls CombinedReports_LLL-*'('*.csv
but maybe a sed command or other would be better
rename is part of the perl package. It renames files according to perl-style regular expressions. To remove the dates from your file names:
rename 's/[0-9]{14}//' CombinedReports_LLL-*.csv
If rename is not available, sed+shell can be used:
for fname in Combined*.csv ; do mv "$fname" "$(echo "$fname" | sed -r 's/[0-9]{14}//')" ; done
The above loops over each of your files. For each file, it performs a mv command: mv "$fname" "$(echo "$fname" | sed -r 's/[0-9]{14}//')" where, in this case, sed is able to use the same regular expression as the rename command above. s/[0-9]{14}// tells sed to look for 14 digits in a row and replace them with an empty string.
Without using an other tools like rename or sed and sticking strictly to bash alone:
for f in CombinedReports_LLL-*.csv
do
newName=${f/LLL-*\(/LLL-(}
mv -i "$f" "$newName"
done
for f in CombinedReports_LLL-* ; do
b=${f:0:20}${f:34:500}
mv "$f" "$b"
done
You can try line by line on shell:
f="CombinedReports_LLL-20140211144020(Untitled_11).csv"
b=${f:0:20}${f:34:500}
echo $b
You can use the rename utility for this. It uses syntax much like sed to change filenames. The following example (from the rename man-page) shows how to remove the trailing '.bak' extension from a list of backup files in the local directory:
rename 's/\.bak$//' *.bak
I'm using the advice given in the top response and have put the following line into a shell script:
ls *.nii | xargs rename 's/[f_]{2}//' f_0*.nii
In terminal, this line works perfectly, but in my script it will not execute and reads * as a literal part of the file name.
I have a folder full of text files. I need to append the same block of text to each of them (and of course overwrite the original file).
I was wondering what the correct Bash shell syntax would be for this. Would I use cat?
I have done a few batch scripts but I'm not a Bash expert. Any suggestions appreciated.
Use append redirection.
for f in *.txt
do
cat footer >> "$f"
done
If you're needing to do this via a script, you can use echo and append redirection to get the extra text into the files.
FILES=pathto/*
for f in $FILES ; do
echo "#extra text" >> $f
done
sed -i.bak "$ a $(<file_block_of_text)" *.txt
Variant of kurumi's answer:
sed -i.bak "\$aTEXTTOINSERT" *.txt
For more details, see SED: insert something to the last line?
very simply one which worked well for me
#!/bin/sh
FILES="./files/*"
for f in $FILES
do
echo '0000000' | cat - $f > temp && mv temp $f
done