Tried grep -xf test $option in a bash script, searching for test file as pattern in option file. Despite that in some cases test matches exactly in option, I take
grep: Invalid range end
How can I avoid it? Something wrong with escaping quotes maybe? -e and -E options didn't work.
Related
I have a test.txt file with links for example:
google.com?test=
google.com?hello=
and this code
xargs -0 -n1 -a FUZZvul.txt -d '\n' -P 20 -I % curl -ks1L '%/?=DarkLotus' | grep -a 'DarkLotus'
When I type a specific word, such as DarkLotus, in the terminal, it checks the links in the file and it brings me the word which is reflected in the links i provided in the test file
There is no problem here, the problem is that I have many links, and when the result appears in the terminal, I do not know which site reflected the DarkLotus word.
How can i do it?
Try -n option. It shows the line number of file with the matched line.
Best Regards,
Haridas.
I'm not sure what you are up to there, but can you invert it? grep by default prints matching lines. The problem here is you are piping the input from the stdout of the previous commands into grep, and that can lack context at grep. Since you have a file to work with:
$ grep 'DarkLotus' FUZZvul.txt
If your intention is to also follow the link then it might be easier to write a bash script:
#!/bin/bash
for line in `grep 'DarkLotus FUZZvul.txt`
do
link=# extract link from line
echo ${link}
curl -ks1L ${link}
done
Then you could make your script accept user input:
#/bin/bash
word="${0}"
for line in `grep ${word} FUZZvul.txt`
...
and then
$ my_link_getter "DarkLotus"
https://google?somearg=DarkLotus
...
And then you could make the txt file a parameter.
etc.
I am trying to use grep with the pwd command.
So, if i enter pwd, it shows me something like:
/home/hrq/my-project/
But, for purposes of a script i am making, i need to use it with grep, so it only prints what is after hrq/, so i need to hide my home folder always (the /home/hrq/) excerpt, and show only what is onwards (like, in this case, only my-project).
Is it possible?
I tried something like
pwd | grep -ov 'home', since i saw that the "-v" flag would be equivalent to the NOT operator, and combine it with the "-o" only matching flag. But it didn't work.
Given:
$ pwd
/home/foo/tmp
$ echo "$PWD"
/home/foo/tmp
Depending on what it is you really want to do, either of these is probably what you really should be using rather than trying to use grep:
$ basename "$PWD"
tmp
$ echo "${PWD#/home/foo/}"
tmp
Use grep -Po 'hrq/\K.*', for example:
grep -Po 'hrq/\K.*' <<< '/home/hrq/my-project/'
my-project/
Here, grep uses the following options:
-P : Use Perl regexes.
-o : Print the matches only (1 match per line), not the entire lines.
\K : Cause the regex engine to "keep" everything it had matched prior to the \K and not include it in the match. Specifically, ignore the preceding part of the regex when printing the match.
SEE ALSO:
grep manual
perlre - Perl regular expressions
In the tail output i have following string...Using grep command how can i search for the string "contentState\":\"STOPPED\". I have to search for the whole string "contentState\":\"STOPPED\" rather
than searching for STOPPED or contentState only.
I tried following command: But it is not working.
grep -e ""contentState\":\"STOPPED\" /opt/logs/out.log | tail -1
{\"eventType\":\"appAction\",\"action\":\"CONTENT_STATE_CHANGE\",\"evt\":{\"contentState\":\"STOPPED\"}}}
To search for "contentState\":\"STOPPED\", you need to (a) put the whole string in single-quotes to protect it, and (b) escape (double) the backslashes. Thus:
grep -e '"contentState\\":\\"STOPPED\\"' /opt/logs/out.log
Without the protection provided the outer single-quotes, the unescaped double-quotes would be subject to the shell's quote removal and grep would never see them.
Example
Consider this test file:
$ cat log
good "contentState\":\"STOPPED\"
bad contentState\":\"STOPPED\"
bad "contentState\":\"STOPPED"
Let's run our command:
$ grep -e '"contentState\\":\\"STOPPED\\"' log
good "contentState\":\"STOPPED\"
As we can see, the good line is returned and only the good line.
I sometimes want to grep for a function to see examples of how it is used in context, eg. what sort of parameters it is called with. When I am doing this, the name of the file the match appears in becomes useless clutter. Is there any way to instruct grep to not include it? (Or a grep alternative that solves the same problem?)
You can tell grep not to indicate the filename in the output with the option -h:
-h, --no-filename
Suppress the prefixing of file names on output. This is the
default when there is only one file (or only standard input) to
search.
Test
$ echo "hello" > f1
$ echo "hello man" > f2
$ grep "hello" f*
f1:hello
f2:hello man
$ grep -h "hello" f*
hello
hello man
I'll use an example to illustrate my problem. Suppose we have the file name 'file.txt' that contains the following string:
AooYoZooYZoAoooooYZ
I'd like to use grep to find all substrings that begin with 'A' and end with 'YZ' but do not contain 'YZ' in between the 'A' and 'YZ'. The desired output would be:
AooYoZooYZ
AoooooYZ
My best guess is to do the following:
$grep -E -o 'A[^(YZ)]*YZ' file.txt
But the output is only:
AoooooYZ
I'd like the parentheses to hold their meaning for the YZ but I read in the GNU grep manual (http://www.gnu.org/software/grep/manual/grep.html) that:
"Most meta-characters lose their special meaning inside bracket expressions." I've also tried:
$grep -E -o 'A.*YZ file.txt
But this outputs the entire line:
AooYoZooYZoAoooooYZ
Is there a way to override this or another way of solving my problem?
Maybe you can use non-greedy match which can be used in Perl regexp
echo 'AooYoZooYZoAoooooYZ' | grep -P -o 'A.*?YZ'
However, note that the manual for GNU grep says that -P option is highly experimental.