Im wondering if someone can help me out.
Im currently using the following to find all PHP files in a certain directory
find /home/mywebsite -type f -name "*.php"
How would i extend that to search through those PHP files and get all files with the string base64_decode?
Any help would be great.
Cheers,
find /home/mywebsite -type f -name '*.php' -exec grep -l base64_decode {} +
The -exec option to find executes a command on the files found. {} is replaced by the filename, and the + means that it should keep repeating this for all the filenames. grep looks for a string in the file, and the -l option tells it to print just the filename when there's a match, not all the matching lines.
If you're getting an error from find, you may have an old version that doesn't support the + feature of -exec. Use this command instead:
find /home/mywebsite -type f -name '*.php' | xargs grep -l base64_decode
xargs reads its standard input and turns them into arguments for the command line in its arguments.
Related
I'm trying to write one line of code that finds all .sh files in the current directory and its subdirectories, and print them without the .sh extension (preferably without the path too).
I think I got the find command down. I tried using the output of
find . -type f -iname "*.sh" -print
as input for echo, and formatting it along these lines
echo "${find_output%.sh}"
However, I cannot get it to work in one line, without variable assigment.
I got inspiration from this answer on stackoverflow https://stackoverflow.com/a/18639136/15124805
to use this line:
echo "${$( find . -type f -iname "*.sh" -print)%.sh}"
But I get this error:
ash: ${$( find . -type f -iname "*.sh" -print)%.sh}: bad substitution
I also tried using xargs
find . -type f -iname "*.sh" -print |"${xargs%.sh}" echo
But I get a "command not found error" -probably I didn't use xargs correctly, but I'm not sure how I could improve this or if it's the right way to go.
How can I make this work?
That's the classic useless use of echo. You simply want
find . -type f -iname "*.sh" -exec basename {} .sh \;
If you have GNU find, you can also do this with -printf.
However, basename only matches .sh literally, so if you really expect extensions with different variants of capitalization, you need a different approach.
For the record, the syntax you tried to use for xargs would attempt to use the value of a variable named xargs. The correct syntax would be something like
find . -type f -iname "*.sh" -print |
xargs -n 1 sh -c 'echo "${1%.[Ss][Hh]}"' _
but that's obviously rather convoluted. In some more detail, you need sh because the parameter expansion you are trying to use is a feature of the shell, not of echo (or xargs, or etc).
(You can slightly optimize by using a loop:
find . -type f -iname "*.sh" -print |
xargs sh -c 'for f; do
echo "${f%.[Ss][Hh]}"
done' _
but this is still not robust for all file names; see also https://mywiki.wooledge.org/BashFAQ/020 for probably more than you realized you needed to know about this topic. If you have GNU find and GNU xargs, you can use find ... -print0 | xargs -r0)
I am trying to do a simple search through files.
Find all files that match a name pattern
Grep through results of step 1 and find only files whose contents have a specific string
I tried,
find . -name rio.yml -exec grep "my pattern" \;
Whats best practice for something like this.
If you just want the paths that contain the match, do:
find . -name rio.yml -type f -exec grep -q "my pattern" {} \; -print
(Given that you're already filtering on the name, the -type f may be redundant, but I find it helpful when grepping.) You can use grep -l, but it's often convenient to build a pipeline to xargs with -print0, so this is a good pattern.
To get the filename which contains some string, you need to use grep -l
find . -name rio.yml -exec grep -l "my pattern" {} \;
To get full path of the files; you can use $(pwd) in place of search directory.
I am a noob in bash and have a very basic question about bash.
I have files like:
a_lsst_z1.5_000.txt
a_lsst_z1.5_001.txt
a_lsst90_z1.5_001.txt
a_lsst_mono_z1.5_000.txt
a_lsst_mono_z1.5_001.txt
a_lsst_mono90_z1.5_000.txt
a_lsst_mono90_z1.5_001.txt
and so on
I would like to list ONLY files having lsst not (lsst90 or lsst_mono
or lsst_mono90.
I have tried:
ls a_lsst_*.txt # but it gives all files
Required output:
a_lsst_z1.5_000.txt
a_lsst_z1.5_001.txt
How to get only lsst files?
Maybe just match the first character after _ as a number?
echo a_lsst_[0-9]*.txt
After your edit, you could just match the z1.5 part:
echo a_lsst_z1.5_*.txt
try this
ls -ltr a_lsst_z1.5_*.txt
If you want to use ls and exclude certain character, you can try:
ls a_lsst[^9m]*.txt
This will exclude lsst90 and lsst_mono etc files.
find . -iname "a_lsst_*.txt" -type f -printf %P\\n 2>/dev/null
gives:
a_lsst_mono90_z1.5_001.txt
a_lsst_z1.5_000.txt
a_lsst_z1.5_001.txt
a_lsst_mono_z1.5_000.txt
a_lsst_mono_z1.5_001.txt
a_lsst_mono90_z1.5_000.txt
and
find . -iname "a_lsst_z1*.txt" -type f -printf %P\\n 2>/dev/null
gives:
a_lsst_z1.5_000.txt
a_lsst_z1.5_001.txt
with find command in the current dir . and with -iname you'll get expected results when using the pattern a_lsst_*.txt or a_lsst_z1*.txt.
Use -type f to get only match on files (not dirs).
Use -printf with %P to get paths without ./ at the beginning and \\n to have them ended with a new line char.
2>/dev/null prevents from displaying any error including very common Permission denied when using find command.
OK, so simple enough.. I want to recursively search a directory for files with a specific extension - and then perform an action on those files.
# pwdENTER
/dir
# ls -R | grep .txt | xargs -I {} open {} ENTER
The file /dir/reallyinsubfolder.txt does not exist. ⬅ fails (bad)
Not output, but succeeds.. /dir/fileinthisfolder.txt ⬅ opens silently (good)
This does find ALL the files I am interested in… but only OPEN's those which happen to be "1-level" deep. In this case, the attempt to open /dir/reallyinsubfolder.txt fails, as reallyinsubfolder.txt is actually /dir/sub/reallyinsubfolder.txt.
I understand that grep is simply returning the matched filename… which then chokes (in this case), the open command, as it fails to reach down to the correct sub-directory to execute the file..
How do I get grep to return the full path of a match?
How about using the find command -
find /path/to/dir -type f -iname "*.txt" -exec action to perform {} \;
find . -name *.txt -exec open {};
(Decorate with backslashes of your needing)
I believe you're asking the wrong question; parsing ls(1) output in this fashion is far more trouble than it is worth.
What would work far more reliably:
find /dir -name '*.txt' -print0 | xargs -0 open
or
find /dir -name '*.txt' -exec open {} \;
find(1) does not mangle names nearly as much as ls(1) and makes executing programs on matched files far more reliable.
I have a directory with roughly 100000 files in it, and I want to perform some function on all files beginning with a specified string, which may match tens of thousands of files.
I have tried
ls mystring*
but this returns with the bash error 'Too many arguments'. My next plan was to use
find ./mystring* -type f
but this has the same issue.
The code needs to look something like
for FILE in `find ./mystring* -type f`
do
#Some function on the file
done
Use find with a wildcard:
find . -name 'mystring*'
ls | grep "^abc"
will give you all files beginning (which is what the OP specifically required) with the substringabc.
It operates only on the current directory whereas find operates recursively into sub folders.
To use find for only files starting with your string try
find . -name 'abc'*
If you want to restrict your search only to files you should consider to use -type f in your search
try to use also -iname for case-insensitive search
Example:
find /path -iname 'yourstring*' -type f
You could also perform some operations on results without pipe sign or xargs
Example:
Search for files and show their size in MB
find /path -iname 'yourstring*' -type f -exec du -sm {} \;