Say for example, in the source directory I have the following files:
abc.r
xyz.sh
pqr.fam
lmn.bim
uvw.r
ttt.sh
Now I need to link only the items 1,2 and 5 only (listed above). Most importantly I need to link all the 3 files together (i.e. link all the 3 files at the same time).
I know how to link 1 file at a time (ln -s sourceDirectory/fileName targetDirectory/), but not multiple files at once.
I found ways to do this when the file name prefixes has some pattern (for example, link all the files where the names start with letter "f"), but in my case, I do not have any such pattern. My file names are different.
Try this:
#!/bin/bash
for file in a.txt b.txt c.txt
do
ln -s /sourcedir/"${file}" /targetdir/
done
Since you only have a list, you have to iterate through the list.
Related
I have a folder with many files that look like:
A1_R1.fastq
A2_R1.fastq
A3_R1.fastq
I would like to rename the files based on a text file keeping the _R1.fastq but changing the A# to a specific samples name (example):
A1_R1.fastq KUG_R1.fastq
A2_R1.fastq AUG_R1.fastq
A3_R1.fastq TRY_R1.fastq
I'd also like an output directory which contains all my newly names .fastq files.
I tried this to no avail (only a few were renamed):
ls *.fastq| paste -d' ' - $PATH/txt | xargs -n2 mv
Thank you.
I need to rename hundreds of files in Linux to change the unique identifier of each from the command line. For sake of examples, I have a file containing:
old_name1 new_name1
old_name2 new_name2
and need to change the names from new to old IDs. The file names contain the IDs, but have extra characters as well. My plan is therefore to end up with:
abcd_old_name1_1234.txt ==> abcd_new_name1_1234.txt
abcd_old_name2_1234.txt ==> abcd_new_name2_1234.txt
Use of rename is obviously fairly helpful here, but I am struggling to work out how to iterate through the file of the desired name changes and pass this as input into rename?
Edit: To clarify, I am looking to make hundreds of different rename commands, the different changes that need to be made are listed in a text file.
Apologies if this is already answered, I've has a good hunt, but can't find a similar case.
rename 's/^(abcd_)old_name(\d+_1234\.txt)$/$1new_name$2/' *.txt
Should work, depending on whether you have that package installed. Also have a look at qmv (rename-utils)
If you want more options, use e.g.
shopt -s globstart
rename 's/^(abcd_)old_name(\d+_1234\.txt)$/$1new_name$2/' folder/**/*.txt
(finds all txt files in subdirectories of folder), or
find folder -type f -iname '*.txt' -exec rename 's/^(abcd_)old_name(\d+_1234\.txt)$/$1new_name$2/' {} \+
To do then same using GNU find
while read -r old_name new_name; do
rename "s/$old_name/$new_name/" *$old_name*.txt
done < file_with_names
In this way, you read the IDs from file_with_names and rename the files replacing $old_name with $new_name leaving the rest of the filename untouched.
I was about to write a php function to do this for myself, but I came upon a faster method:
ls and copy & paste the directory contents into excel from the terminal window. Perhaps you may need to use on online line break removal or addition tool. Assume that your file names are in column A In excel, use the following formula in another column:
="mv "&A1&" prefix"&A1&"suffix"
or
="mv "&A1&" "&substitute(A1,"jpeg","jpg")&"suffix"
or
="mv olddirectory/"&A1&" newdirectory/"&A1
back in Linux, create a new file with
nano rename.txt and paste in the values from excel. They should look something like this:
mv oldname1.jpg newname1.jpg
mv oldname1.jpg newname2.jpg
then close out of nano and run the following command:
bash rename.txt. Bash just runs every line in the file as if you had typed it.
and you are done! This method gives verbose output on errors, which is handy.
I need to unzip a bunch of student assignment (jar) files so that I can use a script to submit the contents to the Moss (Stanford) plagiarism detection server. I did the same thing in Java which was trivial but I'm trying to re-implement to as a bash script.
I am trying to do the following:
Get a list of student names (each student has a directory).
In each student directory, sub-directories exist numbered from 1 to the
latest submission. I need to get the directory with the highest
number.
Inside of each of those submission directories contains a
jar file that I need. I copy each jar into a temp directory with the
same name as the student and unzip it.
I need that temp directory listing formatted as a string in the form
/tempDir/studentName1/.languageExt /tempDir/studentName2/.languageExt
The student directory has the basic structure:
Student_Root_Directory:
Student1
Student2
Student1
Sub-Directories: 1 2 3 4 5
1: student1.jar
2: student1.jar
...
Student2
Sub-Directories: 1 2 3
1. student2.jar
...
To do the first 3 steps above I did:
#!/bin/bash
# Extract all jar files into a temp directory called /home/moss/tempJarFiles/studentName
# $1 is the command line argument that contains the path to the institution submission dir.
# $2 is the language extension: .c, .cpp, .java, .py
students=`ls $1`
student_dir=$1
languageExt=$2
mossDir="/home/moss"
tempDir="/home/moss/tempJarStorage"
for student in $students
do
latestSubmissionDir=`ls -t $student_dir/$student | head -1`
for jarDir in $latestSubmissionDir
do
mkdir $tempDir/$student
cp $student_dir/$student/$jarDir/*.jar $tempDir/$student
unzip -d $tempDir/$student/ -o -j $tempDir/$student/$student.jar *.$languageExt
rm $tempDir/$student/$student.jar
done
done
...which results in a number of student directories being created in a temp directory that contains only the unzipped contents for the student submissions.
I need the ls output of the new temp directories formatted as a string that contains:
/tempDir/studentName1/\*.languageExt /tempDir/studentName2/\*.languageExt
I have tried variations on
find "$tempDir" -iname "*.$languageExt" -printf "%p/*.$languageExt"
using iname and not - but I either have output that contains extra directory information such as $tempDir/*.languageExt (when I just need the subdirectories $tempDir/$studentName/*.languageExt) or I have output where the path for every source file is also listed such as:
$tempDir/$studentName/studentNameA.java
$tempDir/$studentName/studentNameB.java
when I only need
$tempDir/$studentName/*.java
I think this should be really easy and I'm just over thinking it. Any hints for improving the script also appreciated.
Here's a revised version of the script hat may work:
#/bin/bash
# Extract all jar files into a temp directory called /home/moss/tempJarFiles/studentName
# $1 is the command line argument that contains the path to the institution submission dir.
# $2 is the language extension: c, cpp, java, py
students_dir=$1
languageExt=$2
studentPathsT=( "$students_dir"/*/ )
mossDir='/home/moss'
tempDir='/home/moss/tempJarStorage'
for studentPathT in "${studentPathsT[#]}"; do
student=$(basename "$studentPathT")
mkdir "$tempDir/$student"
submissionDirsT=( "$studentPathT"*/ )
latestSubmissionDirT=${submissionDirsT[${#submissionDirsT[#]-1]}
cp "$latestSubmissionDirT"*.jar "$tempDir/$student/"
unzip -d "$tempDir/$student/" -o -j "$tempDir/$student/*.jar" "*.$languageExt"
rm "$tempDir/$student"/*.jar
done
# Note that at this point `"$tempDir"/*/*.$languageExt` would expand
# to all extracted submission files, across all students.
# Finally, output each student's extracted files as an unexpanded glob à la
# /{tempDir}/{studentName1}/*.{languageExt}
for pT in "$tempDir"/*/; do
echo "$pT*.$languageExt"
# Note: If there is a chance that your filenames contain
# embedded newlines (rare in practice) using `echo` won't work properly
# as #Charles Duffy points out.
# If that is a concern, use
# printf '%s\0' "$pT*.$languageExt"
# and process the output with a utility that can process NUL characters
# as separators, such as `xargs -0`.
done
It avoids using ls and only uses pathname expansion and array variables so as to properly deal with paths that contain embedded spaces and other shell metacharacters.
suffix ...T in variable names indicates that a particular path or array of paths is *T*erminated, i.e, that it ends in a /.
The assumption is that the numbered subdirectories do not go beyond 9, as the implicit lexical sorting of pathname expansion is relied upon; if the numbers go higher, explicit numerical sorting must be applied.
Note that the globs (pathname patterns) passed to unzip are intentionally double-quoted, as they should be interpreted by unzip, not the shell.
Note that, based on your original code, I've assumed that $languageExt does NOT start with . (e.g., cpp rather than .cpp), despite what your comment says.
I working with linux, bash.
I have one directory with 100 folders in it, each one named different.
In each of these 100 folders, there is a file called first.bars (so I have 100 files named first.bars). Although all named first.bars, the files are actually slightly different.
I want to get all these files moved to one new folder and rename/number these files so that I know which file comes from which folder. So the first first.bars file must be renamed to 001.bars, the second to 002.bars.. etc.
I have tried the following:
ls -d * >> /home/directorywiththe100folders/list.txt
cat list.txt | while read line;
do cd $line;
mv first.bars /home/newfolder
This does not work because I can't have 100 files, named the same, in one folder. So I only need to know how to rename them. The renaming must be connected to the cat list.txt, because the first line is the folder containing the first file wich is moved and renamed. That file will be called 001.bars.
Try doing this :
$ rename 's/^.*?\./sprintf("%03d.", $c++)/e' *.bar
If you want more information about this command, see this recent response I gave earlier : How do I rename multiple files beginning with a Unix timestamp - imapsync issue
If the rename command is not available,
for d in /home/directorywiththe100folders/*/; do
newfile=$(printf "/home/newfolder/%d.bars" $(( c++ )) )
mv "$d/first.bars" "$newfile"
done
Is it possible to search for a list of strings (100+), for example in a text file and using a command such as findstr to identify which files contain any of the strings? Or is there a better alternative (on Windows) ?
Probably, from the findstr help I found:
/G:file Gets search strings from the specified file(/ stands for console).
and
/S Searches for matching files in the current directory and all
subdirectories.
so:
C:\Temp>copy con strings.txt
test
test1
test2
^Z
1 file(s) copied.
I created (with copy con brings me back) 3 files test.txt test1.txt and test2.txt and placed the strings we have from strings.txt into the respective files and then ran this command:
C:\Temp>findstr /S /G:strings.txt *.txt
strings.txt:test
strings.txt:test1
strings.txt:test2
test.txt:test
test1.txt:test1
test2.txt:test2
It indeed found them, and it even found all three from the source file strings.txt.