if i want an alias to do "rgrep pattern *" to search all files from my current location down through any sub directories, what is an alias for rgrep I can add to my bashrc file?
i would also like it to ignore errors and only report positive hits
In order for it to ignore errors (such as "Permission denied"), you'll probably need to use a function instead of an alias:
rgrep () { grep -r "${#}" 2>/dev/null; }
How about:
alias rgrep="grep -r"
This will only show 'positive hits', i.e. lines that contain the pattern you specify.
Small piece of advice, however: you might want to get used to just using grep -r directly. You'll then find it much easier if you ever need to use someone else's workstation, for instance!
Edit: you want to match patterns in file names, not in their contents (and also in directory names too). So how about this instead:
alias rgrep="find | grep"
By default, find will find all files and directories, so then it's just a case of passing that list to grep to find the pattern you're looking for.
Related
All of my file names follow this pattern:
abc_001.jpg
def_002.jpg
ghi_003.jpg
I want to replace the characters before the numbers and the underscore (not necessarily letters) with the name of the directory in which those files are located. Let's say this directory is called 'Pictures'. So, it would be:
Pictures_001.jpg
Pictures_002.jpg
Pictures_003.jpg
Normally, the way this website works, is that you show what you have done, what problem you have, and we give you a hint on how to solve it. You didn't show us anything, so I will give you a starting point, but not the complete solution.
You need to know what to replace: you have given the examples abc_001 and def_002, are you sure that the length of the "to-be-replaced" part always is equal to 3? In that case, you might use the cut basic command for deleting this. In other ways, you might use the position of the '_' character or you might use grep -o for this matter, like in this simple example:
ls -ltra | grep -o "_[0-9][0-9][0-9].jpg"
As far as the current directory is concerned, you might find this, using the environment variable $PWD (in case Pictures is the deepest subdirectory, you might use cut, using '/' as a separator and take the last found entry).
You can see the current directory with pwd, but alse with echo "${PWD}".
With ${x#something} you can delete something from the beginning of the variable. something can have wildcards, in which case # deletes the smallest, and ## the largest match.
First try the next command for understanding above explanation:
echo "The last part of the current directory `pwd` is ${PWD##*/}"
The same construction can be used for cutting the filename, so you can do
for f in *_*.jpg; do
mv "$f" "${PWD##*/}_${f#*_}"
done
Its my first time to use BASH scripting and been looking to some tutorials but cant figure out some codes. I just want to list all the files in a folder, but i cant do it.
Heres my code so far.
#!/bin/bash
# My first script
echo "Printing files..."
FILES="/Bash/sample/*"
for f in $FILES
do
echo "this is $f"
done
and here is my output..
Printing files...
this is /Bash/sample/*
What is wrong with my code?
You misunderstood what bash means by the word "in". The statement for f in $FILES simply iterates over (space-delimited) words in the string $FILES, whose value is "/Bash/sample" (one word). You seemingly want the files that are "in" the named directory, a spatial metaphor that bash's syntax doesn't assume, so you would have to explicitly tell it to list the files.
for f in `ls $FILES` # illustrates the problem - but don't actually do this (see below)
...
might do it. This converts the output of the ls command into a string, "in" which there will be one word per file.
NB: this example is to help understand what "in" means but is not a good general solution. It will run into trouble as soon as one of the files has a space in its name—such files will contribute two or more words to the list, each of which taken alone may not be a valid filename. This highlights (a) that you should always take extra steps to program around the whitespace problem in bash and similar shells, and (b) that you should avoid spaces in your own file and directory names, because you'll come across plenty of otherwise useful third-party scripts and utilities that have not made the effort to comply with (a). Unfortunately, proper compliance can often lead to quite obfuscated syntax in bash.
I think problem in path "/Bash/sample/*".
U need change this location to absolute, for example:
/home/username/Bash/sample/*
Or use relative path, for example:
~/Bash/sample/*
On most systems this is fully equivalent for:
/home/username/Bash/sample/*
Where username is your current username, use whoami to see your current username.
Best place for learning Bash: http://www.tldp.org/LDP/abs/html/index.html
This should work:
echo "Printing files..."
FILES=(/Bash/sample/*) # create an array.
# Works with filenames containing spaces.
# String variable does not work for that case.
for f in "${FILES[#]}" # iterate over the array.
do
echo "this is $f"
done
& you should not parse ls output.
Take a list of your files)
If you want to take list of your files and see them:
ls ###Takes list###
ls -sh ###Takes list + File size###
...
If you want to send list of files to a file to read and check them later:
ls > FileName.Format ###Takes list and sends them to a file###
ls > FileName.Format ###Takes list with file size and sends them to a file###
So I have a directory with ~50 files, and each contain different things. I often find myself not remembering which files contain what. (This is not a problem with the naming -- it is sort of like having a list of programs and not remembering which files contain conditionals).
Anyways, so far, I've been using
cat * | grep "desiredString"
for a string that I know is in there. However, this just gives me the lines which contain the desired string. This is usually enough, but I'd like it to give me the file names instead, if at all possible.
How could I go about doing this?
It sounds like you want grep -l, which will list the files that contain a particular string. You can also just pass the filename arguments directly to grep and skip cat.
grep -l "desiredString" *
In the directory containing the files among which you want to search:
grep -rn "desiredString" .
This can list all the files matching "desiredString", with file names, matching lines and line numbers.
Problem:
I have a list of filenames, filenames.txt:
Eg.
/usr/share/important-library.c
/usr/share/youneedthis-header.h
/lib/delete/this-at-your-peril.c
I need to rename or delete these files and I need to find references to these files in a project directory tree: /home/noob/my-project/ so I can remove or correct them.
My thought is to use bash to extract the filename: basename filename, then grep for it in the project directory using a for loop.
FILELISTING=listing.txt
PROJECTDIR=/home/noob/my-project/
for f in $(cat "$FILELISTING"); do
extension=$(basename ${f##*.})
filename=$(basename ${f%.*})
pattern="$filename"\\."$extension"
grep -r "$pattern" "$PROJECTDIR"
done
I could royally screw up this project -- does anyone see a flaw in my logic; better: do you see a more reliable scalable way to do this over a huge directory tree? Let's assume that revision control is off the table ( it is, in fact ).
A few comments:
Instead of
for f in $(cat "$FILELISTING") ; do
...
done
it's somewhat safer to write
while IFS= read -r f ; do
...
done < "$FILELISTING"
That way, your code will have no problem with spaces, tabs, asterisks, and so on in the filenames (though it still won't support newlines).
Your goal in separating f into extension and filename, and then reassembling them with \., seems to be that you want the filename to be treated as a literal string; right? Like, you're worried that grep will treat the . as meaning "any character" rather than as "one dot". A more general solution is to use grep's -F option, which tells it to treat the pattern as a fixed string rather than a regex:
grep -r -F "$f" "$PROJECTDIR"
Your introduction mentions using basename, but then you don't actually use it. Is that intentional?
If your non-use of basename is intentional, then filenames.txt really just contains a list of patterns to search for; you don't even need to write a loop, in this case, since grep's -f option tells it to take a newline-separated list of patterns from a file:
grep -r -F -f "$FILELISTING" "$PROJECTDIR"
You should back up your project, using something like tar -czf backup.tar.gz "$PROJECTDIR". "Revision control is off the table" doesn't mean you can't have a rollback strategy!
Edited to add:
To pass all your base-names to grep at once, in the hopes that it can do something smarter with them than just looping over them just as though the calls were separate, you can write something like:
grep -r -F "$(sed 's#.*/##g' "$FILELISTING")" "$PROJECTDIR"
(I used sed rather than while+basename for brevity's sake, but you can an entire loop inside the "$(...)" if you prefer.)
This is a job for an IDE.
You're right that this is a perilous task, and unless you know the build process and the search directories and the order of the directories, you really can't say what header is with which file.
Let's take something as simple as this:
# include "sql.h"
You have a file in the project headers/sql.h. Is that file needed? Maybe it is. Maybe not. There's also a /usr/include/sql.h. Maybe that's the one that's actually used. You can't tell without looking at the Makefile and seeing the order of the include directories which is which.
Then, there are the libraries that get included and may need their own header files in order to be able to compile. And, once you get to the C preprocessor, you really will have a hard time.
This is a task for an IDE (Integrated Development Environment). An IDE builds the project and tracks file and other resource dependencies. In the Java world, most people use Eclipse, and there is a C/C++ plugin for those developers. However, there are over 2 dozen listed in Wikipedia and almost all of them are open source. The best one will depend upon your environment.
I have lots subdirectories containing data, and I want a short list of which jobs (subdirectories) I have. I'm not happy with the following command.
$ ls H2*
H2a:
energy.dat overlap.dat
norm.dat zdip.dat ...
(much more)
H2b:
energy.dat overlap.dat
norm.dat zdip.dat ...
(much more)
This needless clutter defeats the purpose of the wildcard (limiting the output). How can I limit the output to one level deep? I'd like to see the following output
H2a/ H2b/ H2z/
Thanks for your help,
Nick
Try this
ls -d H2*/
The -d option is supposed to list "directories only", but by itself just lists
.
which I personally find kind of strange. The wildcard is needed to get an actual list of directories.
UPDATE: As #Philipp points out, you can do this even more concisely and without leaving bash by saying
echo H2*/
The difference is that ls will print the items on separate lines, which is often useful for piping to other functions.
You should consider using find, like this:
find . -maxdepth 1 -type d -name "H2*"
NOTE: Putting "-type d" before "-maxdepth 1" results in a warning on Debian Linux ("find: warning: you have specified the global option -maxdepth after the argument -type, but global options are not positional, i.e., -maxdepth affects tests specified before it as well as those specified after it. Please specify global options before other arguments.") No such warning is issued on Mac.
echo H2*
It's Bash who does the expansion, so you don't even need ls.
Should you have both files and directories starting with H2, you can append a slash to restrict the glob to directories:
echo H2*/
Perhaps this is what you are looking for?
ls | grep H2*
Use tree by Steve Baker at http://mama.indstate.edu/users/ice/tree/
It fills in for a lot of things that are missing from ls.
To list directories one layer deep:
tree -adi -L 1 H2*