There are files like compile.x86.log compile.x86.log-1 compile.x86.log-2 compile.x86.log-3 and error.log error.log_1 error.log_2 error.log_3 want to use locate command to locate only compile.x86.log and error.log among them.
So far I tried
echo $(/usr/bin/locate -ir '^/\([^.][^/]\+/\)\+compile.x86\.log$')
echo $(/usr/bin/locate -ir '^/\([^.][^/]\+/\)\+error\.log$')
With above individual approach it taking search/execution time as 0m18.068s.
How to combine above two?
Also please provide if some other better solution available only with locate command preferably with locate -b option to search exact names as (compile.x86.log and error.log) in less time.
I have tried echo $(/usr/bin/locate -i -b "compile.x86.log")
It's taking command execution time 0m1.887s only but returning compile.x86.log-1 compile.x86.log-2 compile.x86.log-3 in result instead of returning only compile.x86.log which I don't want.
Is there any way to grep the locate result to return only (compile.x86.log and error.log) in this approach.
Because the locate database outputs entries as absolute path names, and the pattern match test applies to the whole path name (the globbing character * does not treat / specially),
locate -i '*/compile.x86.log' '*/error.log'
does what you want.
By the way, the echo $(…) around a command seems waste.
Following your logic, the most simple answer is:
echo $(/usr/bin/locate -ir '^/\([^.][^/]\+/\)\+(compile\.x86|error)\.log$')
where (compile\.x86|error) is the combination that means "this or this" pattern.
Otherwise, using find command would be better:
find -type f -name "compile.x86.log" -o -name "error.log"
Related
The rest of my bash script works, just having trouble using grep. On each file I am using the following command:
ls -l $filepath | grep "^.r..r..r.*${2}$"
How can I properly use the second argument in the regular expression? What I am trying to do is print the file if it can be read by anyone and the owner is who is passed by the second argument.
Using:
ls -l $filepath | grep "^.r..r..r"
Will print the information successfully based on the read permissions. What I am trying to do is print based on... [read permission][any characters in between][ending with the owner's name]
The immediate problem with your attempt is the final $ which anchors the search to the end of the line, which is the end of the file name, not the owner field. A better solution would replace grep with Awk instead, which has built-in support for examining only specific fields. But actually don't use ls for this, or really in scripts at all.
Unfortuntately, the stat command's options are not entirely portable, but for Linux, try
case $(stat -c %a:%u "$filepath") in
[4-7][4-7][4-7]:"$2") ls -l "$filepath";;
esac
or maybe more portably
find "$filepath" -user "$2" -perm /444 -ls
Sadly, the -perm /444 predicate is not entirely portable, either.
Paradoxically, the de facto most portable replacement for stat to get a file's permissions might actually be
perl -le '#s = stat($ARGV[0]); printf "%03o\n", $s[2]' "$filepath"
The stat call returns a list of fields; if you want the owner, too, the numeric UID is in $s[4] and getpwuid($s[4]) gets the user name.
I have been trying to learn how to adequately perform a single command multiple times using the command line. Although I have learned how to do a single command with no input and output files, it gets more complicated when it needs these.
The cp command requires this so lets use this as an example. I look for all images with .png extension and copy them. The way I have come up with after using google is:
find -regex ".*\.\(png\)" -exec cp {} {}3 \;
The only problem with that is that I have to rename the file with any figure after the name, so it gets renamed to something like file.png3 instead of file.png. I can't figure out how to do if differently as I can't put the new figure before the name as it doesn't seem to work.
Is there a better way to do this or am I going about it completely the wrong way?
I'm not sure how you might do that in a single find command, but you could split it out. First, find the files with find. Then use sed to remove the .png extension. Finally, use xargs to run the copy function on each file. Like this:
find -regex ".*\.\(png\)" | sed -r 's/.png//g' | xargs -I {} cp {}.png {}_copy.png
If you didn't know, the pipe "|" will send the output of one program into the next.
Alternatively, you could just modify the beginning of the filename (so 3img.png instead of img.png3) or copy to a new folder.
What would be the bash script that:
Goes through a directory, and puts all the sub-directories in an array
For each dir, generate an md5 sum of a file inside that dir
Also, the file who's md5sum has to be generated doesn't always have the same name and path. However, the pattern is always the same:
/var/mobile/Applications/{ the dir name here is taken from the array }/{some name}.app/{ binary, who's name is the same as it's parent dir, but without the .app extension }
I've never worked with bash before (and have never needed to) so this may be something really simple and nooby. Anybody got an idea? As can be seen by the path, this is designed to be run on an iDevice.
for dir in /var/mobile/Applications/*; do
for app in "$dir"/*.app; do
appdirname=${app##*/}
appname=${appdirname%.app}
binary="$app/$appname"
if [ -f "$binary" ]; then
echo "I: dir=$dir appbase=$appbase binary=$binary"
fi
done
done
Try this, I hope the code is straight-forward. The two things worth explaining are:
${app##*/}, which uses the ## operator to strip off the longest prefix matching the expression */.
${appdirname%.app}, which uses the % operator to strip off the shortest suffix matching the expression .app. (You could have also used %% (strip longest suffix) instead of %, since the pattern .app is always four characters long.)
Try something like:
ls -1 /Applications/*/Contents/Info.plist | while read name; do md5 -r "$name"; done
the above will show md5 checksum for all Info.plist files for all applications, like:
d3bde2b76489e1ac081b68bbf18a7c29 /Applications/Address Book.app/Contents/Info.plist
6a093349355d20d4af85460340bc72b2 /Applications/Automator.app/Contents/Info.plist
f1c120d6ccc0426a1d3be16c81639ecb /Applications/Calculator.app/Contents/Info.plist
Bash is very easy but you need to know the cli-tools of your system.
For to print the md5 hash of all files of the a directory recursively:
find /yourdirectory/ -type f | xargs md5sum
If you only want to list the tree of directories:
find /tmp/ -type d
You can generate a list with:
MYLIST=$( find /tmp/ -type d )
Use "for" for iterate the list:
for i in $MYLIST; do
echo $i;
done
If you are a newbie in bash:
http://tldp.org/LDP/Bash-Beginners-Guide/html/
http://tldp.org/HOWTO/Bash-Prog-Intro-HOWTO.html
I have a requirement to search for a pattern which is something like :
timeouts = {default = 3.0; };
and replace it with
timeouts = {default = 3000.0;.... };
i.e multiply the timeout by factor of 1000.
Is there any way to do this for all files in a directory
EDIT :
Please note that some of the files are symlinks in the directory.Is there any way to get this done for symlinks also ?
Please note that timeouts exists as a substring also in the files so i want to make sure that only this line gets replaced. Any solution is acceptable using sed awk perl .
Give this a try:
for f in *
do
sed -i 's/\(timeouts = {default = [0-9]\+\)\(\.[0-9]\+;\)\( };\)/\1000\2....\3/' "$f"
done
It will make the replacements in place for each file in the current directory. Some versions of sed require a backup extension after the -i option. You can supply one like this:
sed -i .bak ...
Some versions don't support in-place editing. You can do this:
sed '...' "$f" > tmpfile && mv tmpfile "$f"
Note that this is obviously not actually multiplying by 1000, so if the number is 3.1 it would become "3000.1" instead of 3100.0.
you can do this
perl -pi -e 's/(timeouts\s*=\s*\{default\s*=\s*)([0-9.-]+)/print $1; $2*1000/e' *
One suggestion for whichever solution above you decide to use - it may be worth it to think through how you could refactor to avoid having to modify all of these files for a change like this again.
Do all of these scripts have similar functionality?
Can you create a module that they would all use for shared subroutines?
In the module, could you have a single line that would allow you to have a multiplier?
For me, anytime I need to make similar changes in more than one file, it's the perfect time to be lazy to save myself time and maintenance issues later.
$ perl -pi.bak -e 's/\w+\s*=\s*{\s*\w+\s*=\s*\K(-?[0-9.]+)/sprintf "%0.1f", 1000 * $1/eg' *
Notes:
The regex matches just the number (see \K in perlre)
The /e means the replacement is evaluated
I include a sprintf in the replacement just in case you need finer control over the formatting
Perl's -i can operate on a bunch of files
EDIT
It has been pointed out that some of the files are shambolic links. Given that this process is not idempotent (running it twice on the same file is bad), you had better generate a unique list of files in case one of the links points to a file that appears elsewhere in the list. Here is an example with find, though the code for a pre-existing list should be obvious.
$ find -L . -type f -exec realpath {} \; | sort -u | xargs -d '\n' perl ...
(Assumes none of your filenames contain a newline!)
I want to apply a certain regular expression substitution globally to about 40 Javascript files in and under a directory. I'm a vim user, but doing this by hand can be tedious and error-prone, so I'd like to automate it with a script.
I tried sed, but handling more than one line at a time is awkward, especially if there is no limit to how many lines the pattern might match.
I also tried this script (on a single file, for testing):
ex $1 <<EOF
gs/,\(\_\s*[\]})]\)/\1/
EOF
The pattern will eliminate a trailing comma in any Perl/Ruby-style list, so that "[a, b, c,]" will come out as "[a, b, c]" in order to satisfy Internet Explorer, which alone among browsers, chokes on such lists.
The pattern works beautifully in vim but does nothing if I run it in ex, as per the above script.
Can anyone see what I might be missing?
You asked for a script, but you mentioned that you are vim user. I tend to do project-wide find and replace inside of vim, like so:
:args **/*.js | argdo %s/,\(\_\s*[\]})]\)/\1/ge | update
This is very similar to the :bufdo solution mentioned by another commenter, but it will use your args list rather than your buflist (and thus doesn't require a brand new vim session nor for you to be careful about closing buffers you don't want touched).
:args **/*.js - sets your arglist to contain all .js files in this directory and subdirectories
| - pipe is vim's command separator, letting us have multiple commands on one line
:argdo - run the following command(s) on all arguments. it will "swallow" subsequent pipes
% - a range representing the whole file
:s - substitute command, which you already know about
:s_flags, ge - global (substitute as many times per line as possible) and suppress errors (i.e. "No match")
| - this pipe is "swallowed" by the :argdo, so the following command also operates once per argument
:update - like :write but only when the buffer has been modified
This pattern will obviously work for any vim command which you want to run on multiple files, so it's a handy one to keep in mind. For example, I like to use it to remove trailing whitespace (%s/\s\+$//), set uniform line-endings (set ff=unix) or file encoding (set filencoding=utf8), and retab my files.
1) Open all the files with vim:
bash$ vim $(find . -name '*.js')
2) Apply substitute command to all files:
:bufdo %s/,\(\_\s*[\]})]\)/\1/ge
3) Save all the files and quit:
:wall
:q
I think you'll need to recheck your search pattern, it doesn't look right. I think where you have \_\s* you should have \_s* instead.
Edit: You should also use the /ge options for the :s... command (I've added these above).
You can automate the actions of both vi and ex by passing the argument +'command' from the command line, which enables them to be used as text filters.
In your situation, the following command should work fine:
find /path/to/dir -name '*.js' | xargs ex +'%s/,\(\_\s*[\]})]\)/\1/g' +'wq!'
you can use a combination of the find command and sed
find /path -type f -iname "*.js" -exec sed -i.bak 's/,[ \t]*]/]/' "{}" +;
If you are on windows, Notepad++ allows you to run simple regexes on all opened files.
Search for ,\s*\] and replace with ]
should work for the type of lists you describe.