Hello
I'm looking some script or program that use keywords or pattern search in files ex. php, html, etc and show where is this file
I use command cat /home/* | grep "keyword"
but i have too many folders and files and this command causes big uptime :/
I need this script to find fake websites (paypal, ebay, etc)
find /home -exec grep -s "keyword" {} \; -print
You don't really say what OS (and shell) you are using. You might want to retag your question to help us out.
Because you mention cat | ... , I am assuming you are using a Unix/Linux variant, so here are some pointers for looking at files. (bmargulies solution is good too).
I'm looking some script or program that use keywords or pattern search in files
grep is the basic program for searching files for text strings. Its usage is
grep [-options] 'search target' file1 file2 .... filen
(Note that 'search target' contains a space, if you don't surround spaces in your searchTarget with double or single quotes, you will have a minor error to debug.)
(Also note that 'search target' can use a wide range of wild-card characters, like .,?,+,,., and many more, that is beyond the scope of your question). ... anyway ...
As I guess you have discovered, you can only cram so many files at a time into the comand-line, even when using wild-card filename expansion. Unix/linux almost always have a utiltiyt that can help with that,
startDir=/home
find ${startDir} -print | xargs grep -l 'Search Target'
This, as one person will be happy to remind you, will require further enhancements if your filenames contain whitespace characters or newlines.
The options available for grep can vary wildly based on which OS you are using. If you're lucky, you type the following to get the man page for your local grep.
man grep
If you don't have your page buffer setup for a large size, you might need to do
man grep | page
so you can see the top of the 'document'. Press any key to advance to the next page and when you are at the end of the document, the last key press returns you to the command prompt.
Some options that most greps have that might be useful to you are
-i (ignore case)
-l (list filenames only (where txt is found)
There is also fgrep, which is usually interpretted to mean 'file' grep
becuase you can give it a file of search targets to scan for, and is used like
fgrep [-other_options] -f srchTargetsFile file1 file2 ... filen
I need this script to find fake websites (paypal, ebay, etc)
Final solution
you can make a srchFile like
paypal.fake.com
ebay.fake.com
etc.fake.com
and then combined with above, run the following
startDir=/home
find ${startDir} -print | xargs fgrep -il -f srchFile
Some greps require that the -fsrchFile be run together.
Now you are finding all files starting /home, searching with fgrep for paypay, ebay, etc in all files. The -l says it will ONLY print the filename where a match is found. You can remove the -l and then you will see the output of what is found, prepended with the filename.
IHTH.
Related
I'm attempting to find all .PHP files that are in certain depth of a directory (at least 4 levels down, but not more than 5 levels in).
I'm logged into my Centos server with root authority via shell.
The string I want to search for is:
$slides='';
What I have in front of me.. I would expect it to work. I tried to escape the $ with a \ (I thought perhaps it works like regex, needing special chars excluded). I tried without the ='' portion, or tried adding \'\' to that part.. or remove the ='' altogether to simplify. nothing.
find . -maxdepth 5 -mindepth 4 -type f -name ‘*.php’ -print | xargs grep "\$slides=’’" *
I'm already running it under the directory under which I want to recursively search.
Also - I have the filter to look for *.php only but I still get a bunch of directory names in the return with a warning that says grep: [dir_name]: Is a directory
Clearly I am missing something here as far as syntax of grep command goes, or how the filter works here. I use grep more in PHP so this is quite a transition for me!
So you were almost right. The problem looks to have been the grep part of the command
grep "\$slides=''" *
Namely the * was the issue. From the bash manual
After word splitting, unless the -f option has been set (see The Set
Builtin), Bash scans each word for the characters ‘*’, ‘?’, and ‘[’.
If one of these characters appears, and is not quoted, then the word
is regarded as a pattern, and replaced with an alphabetically sorted
list of filenames matching the pattern
When you piped the found files with find into xargs and attempted to grep them with *, grep would have interpreted this as you wanting to find the string $slides='' in a list of filenames/directories returned by the glob *, and you cannot grep directories without supplying the -r flag to grep, so it returned an error.
Instead, what you wanted to do is pipe the found files with find into xargs so it can add the list of filenames to the grep command, as that's what xargs does. From the xargs manual
xargs reads items from the standard input, delimited by blanks (which
can be protected with double or single quotes or a backslash) or
newlines, and executes the command (default is /bin/echo) one or more
times with any initial- arguments followed by items read from
standard input. Blank lines on the standard input are ignored.
Making the correct command
find . -maxdepth 5 -mindepth 4 -type f -name '*.php' -print0 | xargs -0 grep "\$slides=''"
Using the -print0 flag in find, and the -0 flag in xargs, to use NUL as the delimiter, in case any filenames contained newlines.
If you want to use shell_exec from your PHP code, it is a program execution function which allows you to run a command like 'ls -al' in the operating system shell and get the result returned into a variable. Querystrings are not commands you can use in this way.
Do you mean running PHP from the command line so that it runs from the shell, not from the web server:
php -r 'echo "hello world\n";'
If you run PHP 4.3 and above, you can use the PHP Command Line Interface (CLI) which can also execute scripts stored in files. Have a look at the syntax and examples at: http://php.net/features.commandline
If I have two strings, for example "class" and "btn", what is the linux command that would allow me to search for these two strings in the entire directory.
To be more specific, lets say I have directory that contains few folders with bunch of .php files. My goal is to be able to search throughout those .php files so that it prints out only files that contain "class" and "btn" in one line. Hopefully this clarifies things better.
Thanks,
I normally use the following to search for strings inside my source codes. It searches for string and shows the exact line number where that text appears. Very helpful for searching string in source code files. You can always pipes the output to another grep and filter outputs.
grep -rn "text_to_search" directory_name/
example:
$ grep -rn "angular" menuapp
$ grep -rn "angular" menuapp | grep some_other_string
output would be:
menuapp/public/javascripts/angular.min.js:251://# sourceMappingURL=angular.min.js.map
menuapp/public/javascripts/app.js:1:var app = angular.module("menuApp", []);
grep -r /path/to/directory 'class|btn'
grep is used to search a string in a file. With the -r flag, it searches recursively all files in a directory.
Or, alternatively using the find command to "identify" the files to be searched instead of using grep in recursive mode:
find /path/to/your/directory -type f -exec grep "text_to_search" {} \+;
I have several documents hosted on a cloud instance. I want to extract all words conforming to a specific pattern into a .txt file. This is the pattern:
ABC123A
ABC123B
ABC765A
and so one. Essentially the words start with a specific character string 'ABC', have a fixed number of numerals, and end with a letter. This is my code:
grep -oh ABC[0-9].*[a-zA-Z]$ > /home/user/abcLetterMatches.txt
When I execute the query, it runs for several hours without generating any output. I have over 1100 documents. However, when I run this query:
grep -r ABC[0-9].*[a-zA-Z]$ > /home/user/abcLetterMatches.txt
the list of files with the strings is generated in a matter for seconds.
What do I need to correct in my query? Also, what is causing the delay?
UPDATE 1
Based on the answers, it's evident that the command is missing the file name on which it needs to be executed. I want to run the code on multiple document files (>1000)
The documents I want searched are in multiple sub-directories within a directory. What is a good way to search through them? Doing
grep -roh ABC[0-9].*[a-zA-Z]$ > /home/user/abcLetterMatches.txt
only returns the file names.
UPDATE 2
If I use the updated code from the answer below:
find . -exec grep -oh "ABC[0-9].*[a-zA-Z]$" >> ~/abcLetterMatches.txt {} \;
I get a no file or directory error
UPDATE 3
The pattern can be anywhere in the line.
You can use this regexp :
~/ grep -E "^ABC[0-9]{3}[A-Z]$" docs > filename
ABC123A
ABC123B
ABC765A
There is no delay, grep is just waiting for the input you didn't give it (and therefore it waits on standard input, by default). You can correct your command by supplying argument with filename:
grep -oh "ABC[0-9].*[a-zA-Z]$" file.txt > /home/user/abcLetterMatches.txt
Source (man grep):
SYNOPSIS
grep [OPTIONS] PATTERN [FILE...]
To perform the same grepping on several files recursively, combine it with find command:
find . -exec grep -oh "ABC[0-9].*[a-zA-Z]$" >> ~/abcLetterMatches.txt {} \;
This does what you ask for:
grep -hr '^ABC[0-9]\{3\}[A-Za-z]$'
-h to not get the filenames.
-r to search recursively. If no directory is given (as above) the current one is used. Otherwise just specify one as the last argument.
Quotes around the pattern to avoid accidental globbing, etc.
^ at the beginning of the pattern to — together with $ at the end — only match whole lines. (Not sure if this was a requirement, but the sample data suggests it.)
\{3\} to specify that there should be three digits.
No .* as that would match a whole lot of other things.
I'm trying to list all entries in a directory whose names contain ONLY upper-case letters. Directories need "/" appended.
#!/bin/bash
cd ~/testfiles/
ls | grep -r *.*
Since grep by default looks for upper-case letters only (right?), I'm just recursively searching through the directories under testfiles for all names who contain only upper-case letters.
Unfortunately this doesn't work.
As for appending directories, I'm not sure why I need to do this. Does anyone know where I can start with some detailed explanations on what I can do with grep? Furthermore how to tackle my problem?
No, grep does not only consider uppercase letters.
Your question I a bit unclear, for example:
from your usage of the -r option, it seems you want to search recursively, however you don't say so. For simplicity I assume you don't need to; consider looking into #twm's answer if you need recursion.
you want to look for uppercase (letters) only. Does that mean you don't want to accept any other (non letter) characters, but which are till valid for file names (like digits or dashes, dots, etc.)
since you don't say th it i not permissible to have only on file per line, I am assuming it is OK (thus using ls -1).
The naive solution would be:
ls -1 | grep "^[[:upper:]]\+$"
That is, print all lines containing only uppercase letters. In my TEMP directory that prints, for example:
ALLBIG
LCFEM
WPDNSE
This however would exclude files like README.TXT or FILE001, which depending on your requirements (see above) should most likely be included.
Thus, a better solution would be:
ls -1 | grep -v "[[:lower:]]\+"
That is, print all lines not containing an lowercase letter. In my TEMP directory that prints for example:
ALLBIG
ALLBIG-01.TXT
ALLBIG005.TXT
CRX_75DAF8CB7768
LCFEM
WPDNSE
~DFA0214428CD719AF6.TMP
Finally, to "properly mark" directories with a trailing '/', you could use the -F (or --classify) option.
ls -1F | grep -v "[[:lower:]]\+"
Again, example output:
ALLBIG
ALLBIG-01.TXT
ALLBIG005.TXT
CRX_75DAF8CB7768
LCFEM/
WPDNSE/
~DFA0214428CD719AF6.TMP
Note a different option would to be use find, if you can live with the different output (e.g. find ! -regex ".*[a-z].*"), but that will have a different output.
The exact regular expression depend on the output format of your ls command. Assuming that you do not use an alias for ls, you can try this:
ls -R | grep -o -w "[A-Z]*"
note that with -R in ls you will recursively list directories and files under the current directory. The grep option -o tells grep to only print the matched part of the text. The -w options tell grep to consider as match only for whole words. The "[A-Z]*" is a regexp to filter only upper-cased words.
Note that this regexp will print TEST.txt as well as TEXT.TXT. In other words, it will only consider names that are formed by letters.
It's ls which lists the files, not grep, so that is where you need to specify that you want "/" appended to directories. Use ls --classify to append "/" to directories.
grep is used to process the results from ls (or some other source, generally speaking) and only show lines that match the pattern you specify. It is not limited to uppercase characters. You can limit it to just upper case characters and "/" with grep -E '^[A-Z/]*$ or if you also want numbers, periods, etc. you could instead filter out lines that contain lowercase characters with grep -v -E [a-z].
As grep is not the program which lists the files, it is not where you want to perform the recursion. ls can list paths recursively if you use ls -R. However, you're just going to get the last component of the file paths that way.
You might want to consider using find to handle the recursion. This works for me:
find . -exec ls -d --classify {} \; | egrep -v '[a-z][^/]*/?$'
I should note, using ls --classify to append "/" to the end of directories may also append some other characters to other types of paths that it can classify. For instance, it may append "*" to the end of executable files. If that's not OK, but you're OK with listing directories and other paths separately, this could be worked around by running find twice - once for the directories and then again for other paths. This works for me:
find . -type d | egrep -v '[a-z][^/]*$' | sed -e 's#$#/#'
find . -not -type d | egrep -v '[a-z][^/]*$'
Out of many results returned by grepping a particular pattern, if I want to use all the results one after the other in my script, how can I go about it?For e.g. I grep for .der in a certificate folder which returns many results. I want to use each and every .der certificate listed from the grep command. How can I use one file after the other out of the grep result?
Are you actually grepping content, or just filenames? If it's file names, you'd be better off using the find command:
find /path/to/folder -name "*.der" -exec some other commands {} ";"
It should be quicker in general.
One way is to use grep -l. This ensures you only get every file once. -l is used to print the name of each file only, not the matches.
Then, you can loop on the results:
for file in `grep ....`
do
# work on $file
done
Also note that if you have spaces in your filenames, there is a ton of possible issues. See Looping through files with spaces in the names on the Unix&Linux stackexchange.
You can use the output as part of a for loop, something like:
for cert in $(grep '\.der' *) ; do
echo ${cert} # or something else
done
Of course, if those der things are actually files (and you're using ls | grep to get them), you can directly use the files:
for cert in *.der ; do
echo ${cert} # or something else
done
In both cases, you may need to watch out for arguments with embedded spaces.