I'm after a command that will return results based on a pattern match that starts with or ends with a the given pattern.
This is what i have so far.
"cat input.txt | grep "^in|in$"
My main problem is that i cant get the (or) to work but i can get them to work individually.
Thanks for your help in advance.
try this:
grep "^in\|in$" input.txt
by default, grep use BRE, you have to escape the |. Or use grep's -E or -P, in order to avoid escaping those char with special meaning.
P.S, the cat is no necessary.
This awk should work:
awk '/^start|end$/' file
It will print all lines starting with start or ending with end
cat file
nothing
start with this
or it does have an end
or the end is near
awk '/^start|end$/' file
start with this
or it does have an end
Have you thought of using egrep rather than grep? Using the following should work for what you're after:
egrep "^in|in$" input.txt
There's no need to have the cat at the start, the above will work fine.
Related
I have a shell script that contains the following line
PROC_ID=$(cat myfile.sh | grep running)
which, after you echo out the value would be 1234 or something like that.
What I want to do is find and replace instances of this line with a literal value
I want to replace it with PROC_ID=1234 instead of having the function call.
I've tried doing this in another shell script using sed but I can't get it to work
STR_TO_USE="PROC_ID=${1}"
STR_TO_REP='PROC_ID=$(cat myfile.sh | grep running)'
sed -i "s/$STR_TO_REP/$STR_TO_USE/g" sample.sh
but it complains stating sed: 1: "sample.sh": unterminated substitute pattern
How can I achieve this?
EDIT:
sample.sh should contain beforehand
#!/bin/bash
....
PROC_ID=$(cat myfile.sh | grep running)
echo $PROC_ID
....
After, it should contain
#!/bin/bash
....
PROC_ID=1234
echo $PROC_ID
....
The script I'm using as described above will be taking the in an arg from the command line, hence STR_TO_USE="PROC_ID=${1}"
Simply:
sed /^PROC_ID=/s/=.*/=1234/
Translation:
At line begining by PROC_ID=
replace = to end of line by =1234.
or more accurate
sed '/^[ \o11]*PROC_ID=.*myfile.*running/s/=.*/=1234/'
could be enough
([ \o11]* mean some spaces and or tabs could even prepand)
Well, first, I want to point out something obvious. This: $(cat myfile.sh | grep running) will at the very least NOT only contain the string 1234 but will certainly also contain the string running. But since you aren't asking for help with that, I'll leave it alone.
All you need in your above sed, is first to backslash the $.
STR_TO_REP='PROC_ID=\$(cat myfile.sh | grep running)'
This allows the sed command to be terminated.
I'm new to Unix in all its forms, so please go easy on me!
I have a bash script that will pipe an ls command with arbitrary filenames into sed, which will use an arbitrary replacement pattern on the files, and then this will be piped into awk for some processing. The catch is, awk needs to know both the original file name and the new one.
I've managed everything except getting the original file names into awk. For instance, let's say my files are test.* and my replacement pattern is 's:es:ar;', which would change every occurrence of "test" to "tart". For testing purposes I'm just using awk to print what it's receiving:
ls "$#" | sed "$pattern" | awk '{printf "0: %s\n1: %s\n2: %s\n", $0,$1,$2}'
where test.* is in $# and the pattern is stored in $pattern.
Clearly, this doesn't get me to where I want to be. The output is obviously
0: tart.c
1: tart.c
2:
If I could get sed to output "test.c tart.c", then I'd have two parameters for awk. I've played around with the pattern to no avail, even hardcoding "test.c" into the replacement. But of course that just gave me amateur results like "ttest.c art.c". Is it possible for sed to remember the input, then work it into the beginning of the output? Do I even have the right ideas? Thanks in advance!
Two ways to change the first t in a b in the duplicated field.
Duplicate (& replays the matched part), change first word and swap (remember 2 strings with a space in between):
echo test.c | sed -r 's/.*/& &/;s/t/b/;s/([^ ]*) (.*)/\2 \1/'
or with more magic (copy original value to buffer, make the change, insert value from buffer as the first line and replace eond of line with a space)
echo test.c | sed 'h;s/t/b/;x;G;s/\n/ /'
Use Perl instead of sed:
echo test.c | perl -lne 'print "$_ ", s/es/ar/r'
-l removes the newline from input and adds it after each print. The /r modifier to the substitution returns the modified string instead of changing the variable (Perl 5.14+ needed).
Old answer, not working for s/t/b/2 or s/.*/replaced/2:
You can duplicate the contents of the line with s/.*/& &/, then just tell sed that it should only apply the second substitution (this works at least in GNU sed):
echo test.c | sed 's/.*/& &/; s/es/ar/2'
$ echo 'foo' | awk '{old=$0; gsub(/o/,"e"); print old, $0}'
foo fee
I came across the following in our legacy bash code
sed -itmp <something> file.txt
Since I did not understand what it does clearly, I tried the following
Here is the content of file.txt before running sed
dummy={my.java.home}
dummy={my_java_home}
I run now
sed -itmp "s#{my.java.home}#${JAVA_HOME}#g" file.txt
After I run this, I get the following
dummy=/Library/Java/JavaVirtualMachines/jdk1.7.0_71.jdk/Contents/Home
dummy=/Library/Java/JavaVirtualMachines/jdk1.7.0_71.jdk/Contents/Home
I can see how sed replaces. But I fail to understand sed replaces both my_java_home and my.java.home in my original file although I asked it to change only my.java.home when issuing sed command above.
Thanks
Because dot in your regex matches any charcter not only the literal dot. So i suggest you to escape the dots present in your regex , so that it matches my.java.home string only.
sed -itmp "s#{my\.java\.home}#${JAVA_HOME}#g" file.txt
And you won't actually need a tmp parameters.
sed -i "s#{my\.java\.home}#${JAVA_HOME}#g" file.txt
I'm trying to find whole strings using grep. I am familiar with -w flag, but it gives me hard time since it refers a dot as a delimiter.
For example, I have a file named "a.txt" and a directory names a in some directory, this is what happens:
> ls | grep -w a
a
a.txt
What I want it to find is only "a" and that's it.
How can I do that?
If you want a single a on a line, use
grep '^a$'
If you only take whitespace as the delimiter, use
grep '\([[:space:]]\|^\)a\([[:space:]]\|$\)'
(i.e. whitespace or beginning of the line, a, whitespace or end of the line).
use the x optin of grep
ls | grep -x a
A simpler approach too would be:
grep '^[[:space:]]*a[[:space:]]*$'
Something more friendly with variables is by use of awk. It would not interpret input pattern as regex.
awk -v v="$var" '{ sub(/^[[:space:]]*/, ""); sub(/[[:space:]]*$/, ""); }; $0 == v;'
[Editorial insertion: Possible duplicate of the same poster's earlier question?]
Hi, I need to extract from the file:
first
second
third
using the grep command, the following line:
second
third
How should the grep command look like?
Instead of grep, you can use pcregrep which supports multiline patterns
pcregrep -M 'second\nthird' file
-M allows the pattern to match more than one line.
Your question abstract "bash grep newline", implies that you would want to match on the second\nthird sequence of characters - i.e. something containing newline within it.
Since the grep works on "lines" and these two are different lines, you would not be able to match it this way.
So, I'd split it into several tasks:
you match the line that contains "second" and output the line that has matched and the subsequent line:
grep -A 1 "second" testfile
you translate every other newline into the sequence that is guaranteed not to occur in the input. I think the simplest way to do that would be using perl:
perl -npe '$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.