I have a large list of files, and I need to check to see whether they are somewhere on my linux server. Some of them may be and some of them may not.
Is there a command line tool to do this?
Or must I resort to looping find in a shell script?
There is another alternative, which relies on using find. The idea is to run find once, save all the filenames and then compare them to the list of files.
First, the list of files must be sorted: let us called sortedFiles.txt
run
find / -type f | xargs -n1 -I# basename '#' | sort -u > /tmp/foundFiles.txt
now compare them, and print only those in the first file but not in the second
comm -23 /tmp/sortedFiles.txt /tmp/foundFiles.txt
This will tell you the ones that are not in the computer.
if you want the ones in the computer then use:
comm -12 /tmp/sortedFiles.txt /tmp/foundFiles.txt
this will tell you the ones that are in the computer. The disadvantage is that you don't know where they are. :)
Alternatively run find:
find / -type f > /tmp/allFiles.txt
then iterate using grep, making sure you match the end of the line from the last /
cat /tmp/filesToFind.txt | xargs -n1 -I# egrep '/#$' /tmp/allFiles.txt
This will print only the locations of the files found, but will not print those that are not found.
--dmg
I assume you have a list of filenames without path (all unique). I would suggest to use locate
assuming you have the file with the filenames: files.txt
cat files.txt | xargs -n1 -I# locate -b '\#' | xargs -n1 -I# basename # | uniq > found.txt
then just diff the files.
diff files.txt found.txt
oh, one clarification. This will tell you if the files EXIST in your computer, not where :)
if you want to know where simple run:
cat files.txt | xargs -n1 -I# locate -b '\#'
--dmg
If you do the loop, it's better to use locate instead of find. It's faster!
If lista contains file names you can use:
cat lista | xargs locate
Related
I try to find a way to list all the files in the directory tree (recursively) that contain several words.
While searching I found example such as egrep -R -l 'toto|tata' . but | induce OR. I would like AND...
Thank you for your help
Using GNU grep with GNU xargs,
grep -ERl 'toto' | xargs -r grep 'tata'
The first grep lists those files containing the pattern toto which is then fed to xargs and with the second grep those files containing tata is retrieved. The -r flag is to ensure second grep doesn't run on an empty output.
The -r flag in xargs from the man page,
-r, --no-run-if-empty
If the standard input does not contain any nonblanks, do not run the command.
Normally, the command is run once even if there is no input. This option is a GNU
extension.
agrep tool is designed for providing AND to grep with usage:
agrep 'pattern1;pattern2' file
In your case you could run
find . -type f -exec agrep 'toto;tata' {} \; #apply -l to display the file names
PS1: For current directory you can just agrep 'pattern1;pattern2' *.*
PS2: Unfortunatelly agrep does not support -R option.
There is a directory which contains 100 text files. I used grep to search a given text in the directory as follow:
cat *.txt | grep Ya_Mahdi
and grep shows Ya_Mahdi.
I need to know which file holds the text. Is it possible?
Just get rid of cat and provide the list of files to grep:
grep Ya_Mahdi *.txt
While this would generally work, depending on the number of .txt files in that folder, the argument list for grep might get too large.
You can use find for a bullet proof solution:
find --maxdepth 1 -name '*.txt' -exec grep -H Ya_Mahdi {} +
Line1: .................
Line2: #hello1 #hello2 #hello3
Line3: .................
Line4: .................
Line5: #hello1 #hello4 #hello3
Line6: #hello1 #hello2 #hello3
Line7: .................
I have files that look similar in terms of lines on one of my project directories. I want to get the counts of all the lines that contain #hello1 and #hello2. In this case I would get 2 as a result only for this file. However, I want to do this recursively.
The canonical way to "do something recursively" is to use the find command. If you want to find lines that have two words on them, a simple regex will do:
grep -lr '#hello1.*#hello2' .
The option -l instructs grep to show us only filenames rather than file content, and the option -r tells grep to traverse the filesystem recursively. The start of the search is the path at the end of the line. Once you have the list of files, you can parse that list using commands run by xargs.
For example, this will count all the lines in files matching the pattern you specified.
grep -lr '#hello1.*#hello2' . | xargs -n 1 wc -l
This uses xargs to run the wc command on each of the files listed by grep. You could probably also run this without the -n 1, unless you're dealing with many many thousands of files that would exceed your maximum command line length.
Or, if I'm interpreting your question correctly, the following will count just the patterns in those files.
grep -lr '#hello1.*#hello2' . | xargs -n 1 grep -Hc '#hello1.*#hello2'
This runs a similar grep to the one used to generate your recursive list of files, and presents the output with filename (-H) and count (-c).
But if you want complex rules like finding two patterns possibly on different lines in the file, then grep probably is not the optimal tool, unless you use multiple greps launched by find:
find /path/to/base -type f \
-exec grep -q '#hello1' {} \; \
-exec grep -q '#hello2' {} \; \
-print
(Lines split for easier reading.)
This is somewhat costly, as find needs to launch up to two children for each file. So another approach would be to use awk instead:
find /path/to/base -type f \
-exec awk '/#hello1/{c++} /#hello2/{c++} c==2{r=1} END{exit 1-r}' {} \; \
-print
Alternately, if your shell is bash version 4 or above, you can avoid using find and use the bash option globstar:
$ shopt -s globstar
$ awk 'FNR=1{c=0} /#hello1/{c++} /#hello2/{c++} c==2{print FILENAME;nextfile}' **/*
Note: none of this is tested.
If you are not nterested in the number of files also,
then just something along:
find $BASEDIRECTORY -type f -print0 | xargs -0 grep -h PATTERN | wc -l
If you want to count lines containing #hello1 and #hello2 separated by space in a specific file you can:
$ grep -c '#hello1 #hello2' file
If you want to count in more than one file:
$ grep -c '#hello1 #hello2' file1 file2 ...
And if you want to get the gran total:
$ grep -c '#hello1 #hello2' file1 file2 ... | paste -s -d+ - | bc
of course you can let your shell expanding file names. So, for example:
$ grep -c '#hello1 #hello2' *.txt | paste -s -d+ - | bc
or so...
find . -type f | xargs -1 awk '/#hello1/ && /#hello2/{c++} END{print FILENAME, c+0}'
I have a list containing about 1000 file names to search under a directory and its subdirectories. There are hundreds of subdirs with more than 1,000,000 files. The following command will run find for 1000 times:
cat filelist.txt | while read f; do find /dir -name $f; done
Is there a much faster way to do it?
If filelist.txt has a single filename per line:
find /dir | grep -f <(sed 's#^#/#; s/$/$/; s/\([\.[\*]\|\]\)/\\\1/g' filelist.txt)
(The -f option means that grep searches for all the patterns in the given file.)
Explanation of <(sed 's#^#/#; s/$/$/; s/\([\.[\*]\|\]\)/\\\1/g' filelist.txt):
The <( ... ) is called a process subsitution, and is a little similar to $( ... ). The situation is equivalent to (but using the process substitution is neater and possibly a little faster):
sed 's#^#/#; s/$/$/; s/\([\.[\*]\|\]\)/\\\1/g' filelist.txt > processed_filelist.txt
find /dir | grep -f processed_filelist.txt
The call to sed runs the commands s#^#/#, s/$/$/ and s/\([\.[\*]\|\]\)/\\\1/g on each line of filelist.txt and prints them out. These commands convert the filenames into a format that will work better with grep.
s#^#/# means put a / at the before each filename. (The ^ means "start of line" in a regex)
s/$/$/ means put a $ at the end of each filename. (The first $ means "end of line", the second is just a literal $ which is then interpreted by grep to mean "end of line").
The combination of these two rules means that grep will only look for matches like .../<filename>, so that a.txt doesn't match ./a.txt.backup or ./abba.txt.
s/\([\.[\*]\|\]\)/\\\1/g puts a \ before each occurrence of . [ ] or *. Grep uses regexes and those characters are considered special, but we want them to be plain so we need to escape them (if we didn't escape them, then a file name like a.txt would match files like abtxt).
As an example:
$ cat filelist.txt
file1.txt
file2.txt
blah[2012].txt
blah[2011].txt
lastfile
$ sed 's#^#/#; s/$/$/; s/\([\.[\*]\|\]\)/\\\1/g' filelist.txt
/file1\.txt$
/file2\.txt$
/blah\[2012\]\.txt$
/blah\[2011\]\.txt$
/lastfile$
Grep then uses each line of that output as a pattern when it is searching the output of find.
If filelist.txt is a plain list:
$ find /dir | grep -F -f filelist.txt
If filelist.txt is a pattern list:
$ find /dir | grep -f filelist.txt
Use xargs(1) for the while loop can be a bit faster than in bash.
Like this
xargs -a filelist.txt -I filename find /dir -name filename
Be careful if the file names in filelist.txt contains whitespaces, read the second paragraph in the DESCRIPTION section of xargs(1) manpage about this problem.
An improvement based on some assumptions. For example, a.txt is in filelist.txt, and you can make sure there is only one a.txt in /dir. Then you can tell find(1) to exit early when it finds the instance.
xargs -a filelist.txt -I filename find /dir -name filename -print -quit
Another solution. You can pre-process the filelist.txt, make it into a find(1) arguments list like this. This will reduce find(1) invocations:
find /dir -name 'a.txt' -or -name 'b.txt' -or -name 'c.txt'
I'm not entirely sure of the question here, but I came to this page after trying to find a way to discover which 4 of 13000 files had failed to copy.
Neither of the answers did it for me so I did this:
cp file-list file-list2
find dir/ >> file-list2
sort file-list2 | uniq -u
Which resulted with a list of the 4 files I needed.
The idea is to combine the two file lists to determine the unique entries.
sort is used to make duplicate entries adjacent to each other which is the only way uniq will filter them out.
I'm using bash.
Suppose I have a log file directory /var/myprogram/logs/.
Under this directory I have many sub-directories and sub-sub-directories that include different types of log files from my program.
I'd like to find the three newest files (modified most recently), whose name starts with 2010, under /var/myprogram/logs/, regardless of sub-directory and copy them to my home directory.
Here's what I would do manually
1. Go through each directory and do ls -lt 2010*
to see which files starting with 2010 are modified most recently.
2. Once I go through all directories, I'd know which three files are the newest. So I copy them manually to my home directory.
This is pretty tedious, so I wondered if maybe I could somehow pipe some commands together to do this in one step, preferably without using shell scripts?
I've been looking into find, ls, head, and awk that I might be able to use but haven't figured the right way to glue them together.
Let me know if I need to clarify. Thanks.
Here's how you can do it:
find -type f -name '2010*' -printf "%C#\t%P\n" |sort -r -k1,1 |head -3 |cut -f 2-
This outputs a list of files prefixed by their last change time, sorts them based on that value, takes the top 3 and removes the timestamp.
Your answers feel very complicated, how about
for FILE in find . -type d; do ls -t -1 -F $FILE | grep -v "/" | head -n3 | xargs -I{} mv {} ..; done;
or laid out nicely
for FILE in `find . -type d`;
do
ls -t -1 -F $FILE | grep -v "/" | grep "^2010" | head -n3 | xargs -I{} mv {} ~;
done;
My "shortest" answer after quickly hacking it up.
for file in $(find . -iname *.php -mtime 1 | xargs ls -l | awk '{ print $6" "$7" "$8" "$9 }' | sort | sed -n '1,3p' | awk '{ print $4 }'); do cp $file ../; done
The main command stored in $() does the following:
Find all files recursively in current directory matching (case insensitive) the name *.php and having been modified in the last 24 hours.
Pipe to ls -l, required to be able to sort by modification date, so we can have the first three
Extract the modification date and file name/path with awk
Sort these files based on datetime
With sed print only the first 3 files
With awk print only their name/path
Used in a for loop and as action copy them to the desired location.
Or use #Hasturkun's variant, which popped as a response while I was editing this post :)