I am having following syntax for one of my file.Could you please anyone explain me what is this command doing
path = /document/values.txt
where we have different username specified e.g username1 = john,username2=marry
cat ${path} | grep -e username1 | cut -d'=' -f2`
my question here is cat command is reading from the file value of username1 but why why we need to use cut command?
Cat is printing the file. The file has username1=something in one of the lines. The cut command splits this and prints out the second argument.
your command was not written well. the cat is useless.
you can do:
grep -e pattern "$path"|cut ...
you can of course do it with single process with awk if you like. anyway the line in your question smells not good.
awk example:
awk -F'=' '/pattern/{print $2}' inputFile
cut -d'=' -f2`
This cut uses -d'=' that means you use '=' as 'field delimiter' and -f2 will take only de second field.
So in this case you want only the value after the "=" .
Related
Basically, I have a file formatted like
ABC:123
And I would like to flip the strings around the delimiter, so it would look like this
123:ABC
I would prefer to do this with bash/linux tools.
Thanks for any help!
That's reasonably easy with internal bash commands, assuming two fields, as per the following transcript:
pax:~$ x='abc:123'
pax:~$ echo "${x#*:}:${x%:*}"
123:abc
The first substitution ${x#*:} removes everything from the start up to the colon. The second, ${x%:*}, removes everything from the colon to the end.
Then you just re-join them with the colon in-between.
It doesn't matter for your particular data but % and # use the shortest possible pattern. The %% and ## variants will give you the longest possible pattern (greedy).
As an aside, this is ideal if you doing it for one string at a time since you don't need to kick up an external process to do the work for you. But, if you're processing an entire file, there are better ways to do it, such as with awk:
pax:~$ printf "abc:123\ndef:456\nghi:789\n" | awk -F: '{print $2 FS $1}'
123:abc
456:def
789:ghi
#!/bin/sh -x
var1=$(echo -e 'ABC:123' | cut -d':' -f1)
var2=$(echo -e 'ABC:123' | cut -d':' -f2)
echo -e "${var2}":"${var1}"
I use cut to split the string into two parts, and store both of those parts as variables.
From there, it's possible to use echo to re-arrange the variables as you see fit.
Using sed.
sed -E 's/(.*):(.*)/\2:\1/' file.txt
Using paste and cut with process substitution.
paste -d: <(cut -d : -f2 file.txt) <(cut -d : -f1 file.txt)
A slower/slowest shell solution on large set of data/files.
while IFS=: read -r left rigth; do printf '%s:%s\n' "$rigth" "$left"; done < file.txt
Extract the value for OWNER in the following:
{{USERID 9898}}{{OWNER Wayne, Daniel}}{{EMAIL danielwayne#blah.com}}
To get this string I am using grep on a text file. In all other cases only one value is contained on each line, so they are not an issue.
My problem is removing the text after OWNER but before the }} brackets, leaving me with only the string 'Wayne, Daniel'.
So far I have began looking into writing a for loop to go through the text a character at a time, but I feel there is a more elegant solution then my limited knowledge of unix.
With grep
> cat file
{{USERID 9898}}{{OWNER Wayne, Daniel}}{{EMAIL danielwayne#blah.com}}
> grep -Po '(?<=OWNER )[\w, ]*' file
Wayne, Daniel
Try cat file.txt | perl -n -e'/OWNER ([^\}]+)/ && print $1'
You can use this awk:
awk -F '{{|}}' '{sub(/OWNER +/, "", $4); print $4}' file
Wayne, Daniel
Try this. I use cut
INPUT="{{USERID 9898}}{{OWNER Wayne, Daniel}}{{EMAIL danielwayne#blah.com}}"
SUBSTRING=`echo $INPUT| cut -d' ' -f3`
SUBSTRING2=`echo $INPUT| cut -d',' -f2`
SUBSTRING2=`echo $SUBSTRING2| cut -d'}' -f1`
echo $SUBSTRING$SUBSTRING2
maybe is not the most elegant way but works.
I have 2mill lines of content and all lines look like this:
--username:orderID:email:country
I already added a -- prefix to all usernames.
What I need now is to get ONLY the usernames from the file. I think its possible with grep file starting with "--" ending with ":", but I have absolutely no idea.
So output should be:
usernameThank you all for the help.
THIS WORKED:
cut -d: -f1
Even without adding the prefix, you should be able to get the usernames with cut:
cut -d: -f1
-d says what the delimiter is, -f says which field(s) to return.
Try this:
cat YOUR_FILE | sed "s/:/\n/g" | grep "\-\-"
I need to get a list of matches with grep including filename and line number but without the match string
I know that grep -Hl will give only file names and grep -Hno will give filename with only matching string. But those not ideal for me. I need to get a list without match but with line no. For this grep -Hln doesn't work. I tried with grep -Hn 'pattern' | cut -d " " -f 1 But it doesn't cut the filename and line no properly.
awk can do that in single command:
awk '/pattern/ {print FILENAME ":" NR}' *.txt
You were pointing it well with cut, only that you need the : field separator. Also, I think you need the first and second group. Hence, use:
grep -Hn 'pattern' files* | cut -d: -f1,2
Sample
$ grep -Hn a a*
a:3:are
a:10:bar
a:11:that
a23:1:hiya
$ grep -Hn a a* | cut -d: -f1,2
a:3
a:10
a:11
a23:1
I guess you want this, just line numbers:
grep -nh PATTERN /path/to/file | cut -d: -f1
example output:
12
23
234
...
Unfortunately you'll need to use cut here. There is no way to do it with pure grep.
Try
grep -RHn Studio 'pattern' | awk -F: '{print $1 , ":", $2}'
I have a number of log files in a directory. I am trying to write a script to search all the log files for a string and echo the name of the files and the line number that the string is found.
I figure I will probably have to use 2 grep's - piping the output of one into the other since the -l option only returns the name of the file and nothing about the line numbers. Any insight in how I can successfully achieve this would be much appreciated.
Many thanks,
Alex
$ grep -Hn root /etc/passwd
/etc/passwd:1:root:x:0:0:root:/root:/bin/bash
combining -H and -n does what you expect.
If you want to echo the required informations without the string :
$ grep -Hn root /etc/passwd | cut -d: -f1,2
/etc/passwd:1
or with awk :
$ awk -F: '/root/{print "file=" ARGV[1] "\nline=" NR}' /etc/passwd
file=/etc/passwd
line=1
if you want to create shell variables :
$ awk -F: '/root/{print "file=" ARGV[1] "\nline=" NR}' /etc/passwd | bash
$ echo $line
1
$ echo $file
/etc/passwd
Use -H. If you are using a grep that does not have -H, specify two filenames. For example:
grep -n pattern file /dev/null
My version of grep kept returning text from the matching line, which I wasn't sure if you were after... You can also pipe the output to an awk command to have it ONLY print the file name and line number
grep -Hn "text" . | awk -F: '{print $1 ":" $2}'