I was searching for a change that included "foreach" so I used this Mercurial command:
$ hg grep -r "user(mjh) & public() & date(-30)" --diff -i foreach
and it does return the hits where "foreach" was added and removed.
However, I'd like to know the actual commit hashes too. If I add a template:
$ hg grep ... -T '{date|shortdate}\n{node|short}\n{desc|firstline}\n\n'
then I get the commit hash and description as expected, but then I don't see the changed files listed.
Is there a template to capture the output of hg grep? The {files} template lists the files associated with a commit, but that's not the actual grep output. Is there an iterable template keyword available for the grep results?
Please, re-read carefully hg help grep -v (-v is important option), note the following part (new and unexpected for me also)
The following keywords are supported in addition to the common
template
keywords and functions. See also 'hg help templates'.
change String. Character denoting insertion "+" or removal "-".
Available if "--diff" is specified.
lineno Integer. Line number of the match.
path String. Repository-absolute path of the file.
texts List of text chunks.
After it you'll be able to repeat (so-so, because some details will differ slightly) default output of grep in you template
>hg grep --diff -i -r 1166 to_try
>hg grep --diff -i -r 1166 -T "{path}:{rev}:{change}:{texts}\n" to_try
hggit/compat.py:1166:-: for args in parameters_to_try:
hggit/compat.py:1166:+: for (args, kwargs) in parameters_to_try:
and after replacing {rev} by {node|short}
>hg grep --diff -i -r 1166 -T "{path}:{node|short}:{change}:{texts}\n" to_try
hggit/compat.py:f6cef55e6aeb:-: for args in parameters_to_try:
hggit/compat.py:f6cef55e6aeb:+: for (args, kwargs) in parameters_to_try:
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 "${PWD#/home/foo/}"
tmp
Use grep -Po 'hrq/\K.*', for example:
grep -Po 'hrq/\K.*' <<< '/home/hrq/my-project/'
my-project/
Here, grep uses the following options:
-P : Use Perl regexes.
-o : Print the matches only (1 match per line), not the entire lines.
\K : Cause the regex engine to "keep" everything it had matched prior to the \K and not include it in the match. Specifically, ignore the preceding part of the regex when printing the match.
SEE ALSO:
grep manual
perlre - Perl regular expressions
I am running grep -r to look for the context of a word in multiple files.
I am using -r to do it recursively, -i to ignore case and -C to get lines below and above:
grep -r -i -C 10 --group-separator="==========" "29/04/2020" "$dir" >> output.txt
In my output, however, I get the filenames before the match, like:
../data/filename1.txt- (other text)
../data/filename1.txt- 29/04/2020 is the date for etc
../data/filename1.txt- (other text)
==========
../data/different_filename.txt- (other text)
../data/different_filename.txt- something in 29/04/2020
../data/different_filename.txt- (other text)
I would like as output just:
(other text)
29/04/2020 is the date for etc
(other text)
==========
(other text)
something in 29/04/2020
(other text)
Do you know how I could alter the grep -r command to exclude the filepaths?
Use grep -h, as described in man grep:
-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.
SEE ALSO:
Get grep to not output file name
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 want to remove an entire line of text from all files in a given directory. I know I can use grep -v foo filename to do this one file at a time. And I know I can use grep -r foo to search recursively through a directory. How do I combine these commands to remove a given line of text from all files in a directory?
The UNIX command to find files is named find, not grep. Forget you ever heard of grep -r as it's just a bad idea, here's the right way to find files and perform some action on them:
find . -type f -print | xargs sed -i '/badline/d'
Try something like:
grep -vlre 'foo' . | xargs sed -i 's/pattern/replacement/g'
Broken down:
grep:
-v 'Inverse match'
-l 'Show filename'
-r 'Search recursively'
-e 'Extended pattern search'
xargs: For each entry perform
sed -i: replace inline
I think this would work:
grep -ilre 'Foo' . | xargs sed -i 'extension' 'Foo/d'
Where 'extension' refers to the addition to the file name. It will make a copy of the original file with the extension you designated and the modified file will have the original filename. I added -i in case you require it to be case insensitive.
modified file1 becomes "file1"
original file1 becomes "file1extension"
invalid command code ., despite escaping periods, using sed
One of the responses suggests that the newer version of sed's -i option in OSX is slightly different so you need to add an extension. The file is being interpreted as a command, which is why you are seeing that error.