I want to grep some part of the sentence, for example: /hana/new/register. In this I need to grep the first element between the / characters, so here I want to get hana.
How can I do that in shell?
You can use sed and capture whatever is between the first /.../ using a character class and back reference. For example:
echo '/samarth/new/register' | sed 's/^\/\([^/]*\).*$/\1/'
samarth
Where the sed command is the basic substitute command with the form sed 's/find/replace/' where you find everything after the first / (escaped as \/) and anchored to the beginning with ^. You use a capture group \(...\) to capture the character class [^/]* (everything not a /) and in the replace side of the substitute, you use the backreference \1 to put what you first captured in as the replace.
To get the first word on a slash-separated line, we can use cut:
$ echo '/samarth/new/register then i want to grep samarth' | cut -d/ -f 2
samarth
$ echo '/hana/new/register' | cut -d/ -f 2
hana
Or, we can use awk:
$ echo '/samarth/new/register then i want to grep samarth' | awk -F/ '{print $2}'
samarth
$ echo '/hana/new/register' | awk -F/ '{print $2}'
hana
Related
How can I remove all text after a character, in this case a colon (":"), in bash? Can I remove the colon, too? I have no idea how to.
In Bash (and ksh, zsh, dash, etc.), you can use parameter expansion with % which will remove characters from the end of the string or # which will remove characters from the beginning of the string. If you use a single one of those characters, the smallest matching string will be removed. If you double the character, the longest will be removed.
$ a='hello:world'
$ b=${a%:*}
$ echo "$b"
hello
$ a='hello:world:of:tomorrow'
$ echo "${a%:*}"
hello:world:of
$ echo "${a%%:*}"
hello
$ echo "${a#*:}"
world:of:tomorrow
$ echo "${a##*:}"
tomorrow
An example might have been useful, but if I understood you correctly, this would work:
echo "Hello: world" | cut -f1 -d":"
This will convert Hello: world into Hello.
$ echo 'hello:world:again' |sed 's/:.*//'
hello
I know some solutions:
# Our mock data:
A=user:mail:password
With awk and pipe:
$ echo $A | awk -v FS=':' '{print $1}'
user
Via bash variables:
$ echo ${A%%:*}
user
With pipe and sed:
$ echo $A | sed 's#:.*##g'
user
With pipe and grep:
$ echo $A | egrep -o '^[^:]+'
user
With pipe and cut:
$ echo $A | cut -f1 -d\:
user
egrep -o '^[^:]*:'
trim off everything after the last instance of ":"
grep -o '^.*:' fileListingPathsAndFiles.txt
and if you wanted to drop that last ":"
grep -o '^.*:' file.txt | sed 's/:$//'
#kp123: you'd want to replace : with / (where the sed colon should be \/)
Let's say you have a path with a file in this format:
/dirA/dirB/dirC/filename.file
Now you only want the path which includes four "/". Type
$ echo "/dirA/dirB/dirC/filename.file" | cut -f1-4 -d"/"
and your output will be
/dirA/dirB/dirC
The advantage of using cut is that you can also cut out the uppest directory as well as the file (in this example), so if you type
$ echo "/dirA/dirB/dirC/filename.file" | cut -f1-3 -d"/"
your output would be
/dirA/dirB
Though you can do the same from the other side of the string, it would not make that much sense in this case as typing
$ echo "/dirA/dirB/dirC/filename.file" | cut -f2-4 -d"/"
results in
dirA/dirB/dirC
In some other cases the last case might also be helpful. Mind that there is no "/" at the beginning of the last output.
I want to delete three words with a special character on a line such as
Input:
\cf4 \cb6 1749,1789 \cb3 \
Output:
1749,1789
I have tried a couple sed and grep statements but so far none have worked, mainly due to the character \.
My unsuccessful attempt:
sed -i 's/ [.\c ] //g' inputfile.ext >output file.ext
Awk accepts a regex Field Separator (in this case, comma or space):
$ awk -F'[ ,]' '$0 = $3 "." $4' <<< '\cf4 \cb6 1749,1789 \cb3 \'
1749.1789
-F'[ ,]' - Use a single character from the set space/comma as Field Separator
$0 = $3 "." $4 - If we can set the entire line $0 to Field 3 $4 followed by a literal period "." followed by Field 4 $4, do the default behavior (print entire line)
Replace <<< 'input' with file if every line of that file has the same delimeters (spaces/comma) and number of fields. If your input file is more complex than the sample you shared, please edit your question to show actual input.
The backslash is a special meta-character that confuses bash.
We treat it like any other meta-character, by escaping it, with--you guessed it--a backslash!
But first, we need to grep this pattern out of our file
grep '\\... \\... [0-9]+,[0-9]+ \\... \\' our_file # Close enough!
Now, just sed out those pesky backslashes
| sed -e 's/\\//g' # Don't forget the g, otherwise it'll only strip out 1 backlash
Now, finally, sed out the clusters of 2 alpha followed by a number and a space!
| sed -e 's/[a-z][a-z][0-9] //g'
And, finally....
grep '\\... \\... [0-9]+,[0-9]+ \\... \\' our_file | sed -e 's/\\//g' | sed -e 's/[a-z][a-z][0-9] //g'
Output:
1749,1789
My guess is you are having trouble because you have backslashes in input and can't figure out how to get backslashes into your regex. Since backslashes are escape characters to shell and regex you end up having to type four backslashes to get one into your regex.
Ben Van Camp already posted an answer that uses single quotes to make the escaping a little easier; however I shall now post an answer that simply avoids the problem altogether.
grep -o '[0-9]*,[0-9]*' | tr , .
Locks on to the comma and selects the digits on either side and outputs the number. Alternately if comma is not guaranteed we can do it this way:
egrep -o ' [0-9,]*|^[0-9,]*' | tr , . | tr -d ' '
Both of these assume there's only one usable number per line.
$ awk '{sub(/,/,".",$3); print $3}' file
1749.1789
$ sed 's/\([^ ]* \)\{2\}\([^ ]*\).*/\2/; s/,/./' file
1749.1789
I am working on shell script and new to it. I want to extract the string between double $$ characters, for example:
input:
$$extractabc$$
output
extractabc
I used grep and sed but not working out. Any suggestions are welcome!
You could do
awk -F"$" '{print $3}' file.txt
assuming the file contained input:$$extractabc$$ output:extractabc. awk splits your data into pieces using $ as a delimiter. First item will be input:, next will be empty, next will be extractabc.
You could use sed like so to get the same info.
sed -e 's/.*$$\(.*\)$$.*/\1/' file.txt
sed looks for information between $$s and outputs that. The goal is to type something like this .*$$(.*)$$.*. It's greedy but just stay with me.
looks for .* - i.e. any character zero or more times before $$
then the string should have $$
after $$ there'll be any character zero or more times
then the string should have another $$
and some more characters to follow
between the 2 $$ is (.*). String found between $$s is given a placeholder \1
sed finds such information and publishes it
Using grep PCRE (where available) and look-around:
$ echo '$$extractabc$$' | grep -oP "(?<=\\$\\$).*(?=\\$\\$)"
extractabc
echo '$$extractabc$$' | awk '{gsub(/\$\$/,"")}1'
extractabc
Here is an other variation:
echo "$$extractabc$$" | awk -F"$$" 'NF==3 {print $2}'
It does test of there are two set of $$ and only then prints whats between $$
Does also work for input like blabla$$some_data$$moreblabla
How about remove all the $ in the input?
$ echo '$$extractabc$$' | sed 's/\$//g'
extractabc
Same with tr
$ echo '$$extractabc$$' | tr -d '$'
extractabc
I have text data in this form:
^Well/Well[ADV]+ADV ^John/John[N]+N ^has/have[V]+V+3sg+PRES ^a/a[ART]
^quite/quite[ADV]+ADV ^different/different[ADJ]+ADJ ^not/not[PART]
^necessarily/necessarily[ADV]+ADV ^more/more[ADV]+ADV
^elaborated/elaborate[V]+V+PPART ^theology/theology[N]+N *edu$
And I want it to be processed to this form:
Well John have a quite different not necessarily more elaborate theology
Basically, I need every string between the starting character / and the ending character [.
Here is what I tried, but I just get empty files...
#!/bin/bash
for file in probe/*.txt
do sed '///,/[/d' $file > $file.aa
mv $file.aa $file
done
awk to the rescue!
$ awk -F/ -v RS=^ -v ORS=' ' '{print $1}' file
Well John has a quite different not necessarily more elaborated theology
Explanation set record separator (RS) to ^ to separate your logical groups, also set the field separator (FS) to / and print the first field as your requirement. Finally, setting the output field separator (OFS) to space (instead of the default new line) keeps the extracted fields on the same line.
With GNU grep and Perl compatible regular expressions (-P):
$ echo $(grep -Po '(?<=/)[^[]*' infile)
Well John have a quite different not necessarily more elaborate theology
-o retains just the matches, (?<=/) is a positive look-behind ("make sure there is a /, but don't include it in the match"), and [^[]* is "a sequence of characters other than [".
grep -Po prints one match per line; by using the output of grep as arguments to echo, we convert the newlines into spaces (could also be done by piping to tr '\n' ' ').
cat file|grep -oE "\/[^\[]*\[" |sed -e 's#^/##' -e 's/\[$//' | tr -s "\n" " "
I have a file as show below
1.2.3.4.ask
sanma.nam.sam
c.d.b.test
I want to remove the last field from each line, the delimiter is . and the number of fields are not constant.
Can anybody help me with an awk or sed to find out the solution. I can't use perl here.
Both these sed and awk solutions work independent of the number of fields.
Using sed:
$ sed -r 's/(.*)\..*/\1/' file
1.2.3.4
sanma.nam
c.d.b
Note: -r is the flag for extended regexp, it could be -E so check with man sed. If your version of sed doesn't have a flag for this then just escape the brackets:
sed 's/\(.*\)\..*/\1/' file
1.2.3.4
sanma.nam
c.d.b
The sed solution is doing a greedy match up to the last . and capturing everything before it, it replaces the whole line with only the matched part (n-1 fields). Use the -i option if you want the changes to be stored back to the files.
Using awk:
$ awk 'BEGIN{FS=OFS="."}{NF--; print}' file
1.2.3.4
sanma.nam
c.d.b
The awk solution just simply prints n-1 fields, to store the changes back to the file use redirection:
$ awk 'BEGIN{FS=OFS="."}{NF--; print}' file > tmp && mv tmp file
Reverse, cut, reverse back.
rev file | cut -d. -f2- | rev >newfile
Or, replace from last dot to end with nothing:
sed 's/\.[^.]*$//' file >newfile
The regex [^.] matches one character which is not dot (or newline). You need to exclude the dot because the repetition operator * is "greedy"; it will select the leftmost, longest possible match.
With cut on the reversed string
cat youFile | rev |cut -d "." -f 2- | rev
If you want to keep the "." use below:
awk '{gsub(/[^\.]*$/,"");print}' your_file