I'm trying to pipe the output of grep to sed so it will only edit specific files. I don't want sed to edit something without changing it. (Changing the modified date.)
I'm searching with grep and writing with sed. That's it
The thing I am trying to change is a dash, not the normal type, a special type. "-" is normal. "–" isn't normal
The code I currently have:
sed -i 's/– foobar/- foobar/g' * ; perl-rename 's/– foobar/- foobar/' *'– foobar'*
Sorry about the trouble, I'm inexperienced.
Are you sure about what you want to achieve? Let me explain you:
grep "string_in_file" <filelist> | sed <sed_script>
This is first showing the "string_in_file", preceeded by the filename.
If you launch a sed on this, then it will just show you the result of that sed-script on screen, but it will not change the files itself. In order to do this, you need the following:
grep -l "string_in_file" <filelist> | sed <sed_script_on_file>
The grep -l shows you some filenames, and the new sed_script_on_file needs to be a script, reading the file, and altering it.
Thank you all for helping, I'm sorry about not being fast in responding
After a bit of fiddling with the command, I got it:
grep -l 'old' * | xargs -d '\n' sed -i 's/old/new/'
This should only touch files that contain old and leave all other files.
This might be what you're trying to do if your file names don't contain newlines:
grep -l -- 'old' * | xargs sed -i 's/old/new/'
Related
I'm using this command here of grep and sed,
grep -w 'HN8 .* ATP' topol.top | sed -i 's_0[.]09_0.054_' topol.top
In my head, I thought the command is fetch lines that have HN8 and ATP, in those lines, replace 0.09 with 0.054. However, when this happens, Lines with HN7 or HN6 etc, with 0.09 are also getting swapped to 0.054.
Whys that? so I can understand how to just look for the values I require changing.
sed can either process named files or stdin, but not both. And the -i option can only be used when processing named files. So when you specify filename arguments, stdin is not used.
You should use the regexp as the address part of the sed command.
sed -i '/HN8 .* ATP/s_0[.]09_0.054_' topol.top
Okay so I have a textfile containing multiple strings, example of this -
Hello123
Halo123
Gracias
Thank you
...
I want grep to use these strings to find lines with matching strings/keywords from other files within a directory
example of text files being grepped -
123-example-Halo123
321-example-Gracias-com-no
321-example-match
so in this instance the output should be
123-example-Halo123
321-example-Gracias-com-no
With GNU grep:
grep -f file1 file2
-f FILE: Obtain patterns from FILE, one per line.
Output:
123-example-Halo123
321-example-Gracias-com-no
You should probably look at the manpage for grep to get a better understanding of what options are supported by the grep utility. However, there a number of ways to achieve what you're trying to accomplish. Here's one approach:
grep -e "Hello123" -e "Halo123" -e "Gracias" -e "Thank you" list_of_files_to_search
However, since your search strings are already in a separate file, you would probably want to use this approach:
grep -f patternFile list_of_files_to_search
I can think of two possible solutions for your question:
Use multiple regular expressions - a regular expression for each word you want to find, for example:
grep -e Hello123 -e Halo123 file_to_search.txt
Use a single regular expression with an "or" operator. Using Perl regular expressions, it will look like the following:
grep -P "Hello123|Halo123" file_to_search.txt
EDIT:
As you mentioned in your comment, you want to use a list of words to find from a file and search in a full directory.
You can manipulate the words-to-find file to look like -e flags concatenation:
cat words_to_find.txt | sed 's/^/-e "/;s/$/"/' | tr '\n' ' '
This will return something like -e "Hello123" -e "Halo123" -e "Gracias" -e" Thank you", which you can then pass to grep using xargs:
cat words_to_find.txt | sed 's/^/-e "/;s/$/"/' | tr '\n' ' ' | dir_to_search/*
As you can see, the last command also searches in all of the files in the directory.
SECOND EDIT: as PesaThe mentioned, the following command would do this in a much more simple and elegant way:
grep -f words_to_find.txt dir_to_search/*
I need some guidance manipulating a text file that is the result of a diff. I only want those results listed after the > delimiter (which are file names) and then I will add a path to the file name for further work.
I am not dealing with large files.
I am hoping to do it all in place.
Essentially I want to take something like this
96a97,98
> SCR-33333.sql
> SCR-33333-WEB.sql
and create an action like
cp /add/this/path/SCR-33333.sql /to/somewhere/else
Can anyone please give me a quick example I can run with?
Well, you could try this, bearing in mind that it'll only work if filenames do not contains spaces...
diff this that | awk '/^>/{print "/add/this/path/" $2}' | xargs -i cp {} /to/somewhere/else
(note: this is a one-liner command. ignore wrapping caused by web browser.)
grep ">" dummy.txt | cut -f 2 -d ' ' | xargs -I{} cp /add/this/path/{} somewhere
where 'dummy.txt' is your diff file.
Trying to remove a line that contains a particular pattern in a text file. I have the following code which does not work
grep -v "$varName" config.txt
Can anyone tell me how I can make it work properly, I want to make it work using grep and not sed.
you can use sed, with in place -i
sed -i '/pattern/d' file
grep doesn't modify files. The best you can do if you insist on using grep and not sed is
grep -v "$varName" config.txt > $$ && mv $$ config.txt
Note that I'm using $$ as the temporary file name because it's the pid of your bash script, and therefore probably not a file name going to be used by some other bash script. I'd encourage using $$ in temp file names in bash, especially ones that might be run multiple times simultaneously.
try using -Ev
grep -Ev 'item0|item1|item2|item3'
That will delete lines containing item[0-3]. let me know if this helps
[Editorial insertion: Possible duplicate of the same poster's earlier question?]
Hi, I need to extract from the file:
first
second
third
using the grep command, the following line:
second
third
How should the grep command look like?
Instead of grep, you can use pcregrep which supports multiline patterns
pcregrep -M 'second\nthird' file
-M allows the pattern to match more than one line.
Your question abstract "bash grep newline", implies that you would want to match on the second\nthird sequence of characters - i.e. something containing newline within it.
Since the grep works on "lines" and these two are different lines, you would not be able to match it this way.
So, I'd split it into several tasks:
you match the line that contains "second" and output the line that has matched and the subsequent line:
grep -A 1 "second" testfile
you translate every other newline into the sequence that is guaranteed not to occur in the input. I think the simplest way to do that would be using perl:
perl -npe '$x=1-$x; s/\n/##UnUsedSequence##/ if $x;'
you do a grep on these lines, this time searching for string ##UnUsedSequence##third:
grep "##UnUsedSequence##third"
you unwrap the unused sequences back into the newlines, sed might be the simplest:
sed -e 's/##UnUsedSequence##/\n'
So the resulting pipe command to do what you want would look like:
grep -A 1 "second" testfile | perl -npe '$x=1-$x; s/\n/##UnUsedSequence##/ if $x;' | grep "##UnUsedSequence##third" | sed -e 's/##UnUsedSequence##/\n/'
Not the most elegant by far, but should work. I'm curious to know of better approaches, though - there should be some.
I don't think grep is the way to go on this.
If you just want to strip the first line from any file (to generalize your question), I would use sed instead.
sed '1d' INPUT_FILE_NAME
This will send the contents of the file to standard output with the first line deleted.
Then you can redirect the standard output to another file to capture the results.
sed '1d' INPUT_FILE_NAME > OUTPUT_FILE_NAME
That should do it.
If you have to use grep and just don't want to display the line with first on it, then try this:
grep -v first INPUT_FILE_NAME
By passing the -v switch, you are telling grep to show you everything but the expression that you are passing. In effect show me everything but the line(s) with first in them.
However, the downside is that a file with multiple first's in it will not show those other lines either and may not be the behavior that you are expecting.
To shunt the results into a new file, try this:
grep -v first INPUT_FILE_NAME > OUTPUT_FILE_NAME
Hope this helps.
I don't really understand what do you want to match. I would not use grep, but one of the following:
tail -2 file # to get last two lines
head -n +2 file # to get all but first line
sed -e '2,3p;d' file # to get lines from second to third
(not sure how standard it is, it works in GNU tools for sure)
So you just don't want the line containing "first"? -v inverts the grep results.
$ echo -e "first\nsecond\nthird\n" | grep -v first
second
third
Line? Or lines?
Try
grep -E -e '(second|third)' filename
Edit: grep is line oriented. you're going to have to use either Perl, sed or awk to perform the pattern match across lines.
BTW -E tell grep that the regexp is extended RE.
grep -A1 "second" | grep -B1 "third" works nicely, and if you have multiple matches it will even get rid of the original -- match delimiter
grep -E '(second|third)' /path/to/file
egrep -w 'second|third' /path/to/file
you could use
$ grep -1 third filename
this will print a string with match and one string before and after. Since "third" is in the last string you get last two strings.
I like notnoop's answer, but building on AndrewY's answer (which is better for those without pcregrep, but way too complicated), you can just do:
RESULT=`grep -A1 -s -m1 '^\s*second\s*$' file | grep -s -B1 -m1 '^\s*third\s*$'`
grep -v '^first' filename
Where the -v flag inverts the match.