I wanted to search for this particular string secure="false" in my folder. But when I use a normal grep function grep -r "secure="false"" it gives me no results.
You need to escape your string: grep -r "secure=\"false\""
Related
I'm trying to use Grep to find a string with Tabs, Carriage Returns, & New Lines. Any other method would be helpful also.
grep -R "\x0A\x0D\x09<p><b>Site Info</b></p>\x0A\x0D\x09<blockquote>\x0A\x0D\x09\x09<p>\x0A\x0D\x09</blockquote>\x0A\x0D</blockquote>\x0A\x0D<blockquote>\x0A\x0D\x09<p><b>More Site Info</b></p>" *
From this answer
If using GNU grep, you can use the Perl-style regexp:
$ grep -P '\t' *
Also from here
Use Ctrl+V, Ctrl+M to enter a literal Carriage Return character into your grep string. So:
grep -IUr --color "^M"
will work - if the ^M there is a literal CR that you input as I suggested.
If you want the list of files, you want to add the -l option as well.
Quoting this answer:
Grep is not sufficient for this operation.
pcregrep, which is
found in most of the modern Linux systems can be used ...
Bash Example
$ pcregrep -M "try:\n fro.*\n.*except" file.py
returns
try:
from tifffile import imwrite
except (ModuleNotFoundError, ImportError):
I need some help with a grep command (in the Bash).
In my source files, I want to list all unique parameters of a function. Background: I want to search through all files, to see, which permissions ([perm("abc")] are used.
Example.txt:
if (x) perm("this"); else perm("that");
perm("what");
I'd like to have my grep output:
this
that
what
If I do my grep with this search expression
perm\(\"(.*?)\"\)
I'll get perm("this), perm("that"), etc. but I'd like to have just the permissions: this and that and what.
How can I do that?
Use a look-behind:
$ grep -Po '(?<=perm\(")[^"]*' file
this
that
what
This looks for all the text occurring after perm(" and until another " is found.
Note -P is used to allow this behaviour (it is a Perl regex) and -o to just print the matched item, instead of the whole line.
Here is a gnu awk version (due to multiple characters in RS)
awk -v RS='perm\\("' -F\" 'NR>1 {print $1}' file
this
that
what
Okay I have a file that contains numbers like this:
L21479
What I am trying to do is use grep (or a similar tool) to find all the strings in a file that have the format:
L#####
The # will be the number. SO an L followed by 5 numbers.
Is this even possible in grep? Should I load the file and perform regex?
You can do this with grep, for example with the following command:
grep -E -o 'L[0-9]{5}' name_of_file
For example, given a file with the text:
kasdhflkashl143112343214L232134614
3L1431413543454L2342L3523269ufoidu
gl9983ugsdu8768IUHI/(JHKJASHD/(888
The command above will output:
L23213
L14314
L35232
If it is just in a single file, you can do something along the lines of:
grep -e 'L[0-9]{5}' filename
If you need to search all files in a directory for these strings:
find . -type f | xargs grep -e 'L[0-9]{5}'
I need to find ALL files that have multiple keywords anywhere in the file (not necessarily on the same line), given a starting directory like ~/. Does "grep -ro" do this?
(I'm using Unix, Mac OSX 10.4)
You can use the -l option to get a list of filenames with matches, so it's just a matter of finding all of the files that have the first keyword and then filtering that list down to the files that also have the second keyword:
grep -rl first_keyword basedir | xargs grep -l second_keyword
To search just *.txt
find ~/. -name "*.txt" | xargs grep -l first_keyword | xargs grep -l second_keyword
Thanks Adam!
I want to grep for a function call 'init()' in all JavaScript files in a directory. How do I do this using grep?
Particularly, how do I escape parenthesis, ()?
It depends. If you use regular grep, you don't escape:
echo '(foo)' | grep '(fo*)'
You actually have to escape if you want to use the parentheses as grouping.
If you use extended regular expressions, you do escape:
echo '(foo)' | grep -E '\(fo*\)'
If you want to search for exactly the string "init()" then use fgrep "init()" or grep -F "init()".
Both of these will do fixed string matching, i.e. will treat the pattern as a plain string to search for and not as a regex. I believe it is also faster than doing a regex search.
$ echo "init()" | grep -Erin 'init\([^)]*\)'
1:init()
$ echo "init(test)" | grep -Erin 'init\([^)]*\)'
1:init(test)
$ echo "initwhat" | grep -Erin 'init\([^)]*\)'
Move to your root directory (if you are aware where the JavaScript files are). Then do the following.
grep 'init()' *.js