I'd like to cut a string from a number of position until a specific character "/" :
It would cut this line :
Export text H8X7IS5G.FIC NB regs COLOLO 4138/4138
To this one :
4138
What i tried is to use cut -c with the position and the character but of course it doesn't work :
cut -c 57-'/'
If you want to stick with cut then this might be what you want:
echo 'Export text H8X7IS5G.FIC NB regs COLOLO 4138/4138' |
cut -c41- | cut -d/ -f1
There are many other ways to accomplish this task. If you have a grep which supports perl-compatible regular expressions, for instance, then I'd suggest something along this line:
grep -Po '.{40}\K[^/]*'
Or, a sed one-liner:
sed 's/.\{40\}//; s|/.*||'
Or, using pure bash
[[ $line =~ .{40}([^/]*) ]] && printf '%s\n' "${BASH_REMATCH[1]}"
Assuming you're trying to process a single variable at a time (rather than a stream with hundreds or thousands of lines), you don't need cut for this at all.
input='Export text H8X7IS5G.FIC NB regs COLOLO 4138/4138'
result=${input:40}
echo "${result%%/*}"
...emits 4138.
Both ${var:start:len} (and its shorter synonym ${var:start}) and ${var%%PATTERN} are examples of parameter expansion syntax; the former takes only a subset of a string starting at a given position; the latter trims the longest possible match of PATTERN (${var%PATTERN} trims the shortest possible match of PATTERN instead).
These and other string manipulations in bash are also documented in BashFAQ #100.
Related
How do I cut everything starting with a given multi-character string using a common shell command?
e.g., given:
foo+=bar
I want:
foo
i.e., cut everything starting with +=
cut doesn't work because it only takes a single-character delimiter, not a multi-character string:
$ echo 'foo+=bar' | cut -d '+=' -f 1
cut: bad delimiter
If I can't use cut, I would consider using perl instead, or if there's another shell command that is more commonly installed.
cut only allows single character delimiter.
You may use bash string manipulation:
s='foo+=bar'
echo "${s%%+=*}"
foo
or use more powerful awk:
awk -F '\\+=' '{print $1}' <<< "$s"
foo
'\\+=' is a regex that matches + followed by = character.
You can use 'sed' command to do this:
string='foo+=bar'
echo ${string} | sed 's/+=.*//g'
foo
or if you're using Bash shell, then use the below parameter expansion (recommended) since it doesn't create unnecessary pipeline and another sed process and so is efficient:
echo ${string%%\+\=*}
or
echo ${string%%[+][=]*}
I want to convert the text in a bash variable i.e. HttpStatus.NOT_FOUND into status().isNotFound() and I had accomplished this by using sed:
result=HttpStatus.NOT_FOUND
result=$(echo $result | cut -d'.' -f2- | sed -r 's/(^|_)([A-Z])/\L\2/g' | sed -E 's/([[:lower:]])|([[:upper:]])/\U\1\L\2/g')
echo "status().is$result()"
Output:
status().isNotFound()
As you can see here I'm using 2 sed commands.
Is there a way to achieve the same result using 1 sed or any other simpler way?
Since it involves a lot of new text insertion in the replacement part, the sed command can be written in detail as below. Just pass the variable content over a pipe without using cut
result=HttpStatus.NOT_FOUND
echo "$result" |
sed -E 's/^.*(Status)\.([[:upper:]])([[:upper:]]+)_([[:upper:]])([[:upper:]]+)$/\L\1().is\u\2\L\3\u\4\L\5()/g'
The idea is add the case conversion functions of GNU sed on the captured groups. So we capture
(Status) in \1 in which we just lowercase the entire string and then append a ().is to the result
The next captured group, \2 would be first uppercase character following the . which would be N and the rest of the string OT in \3. We retain the second as such and do lower case of the third group.
The same sequence as above is repeated for the next word FOUND in \4 and \5.
The \L, \u are case conversion operators available in GNU sed.
If you are looking to modify only the part beyond the . to CamelCase, then you can use sed as
result=HttpStatus.NOT_FOUND
result=$(echo "$result" |
sed -E 's/^.*\.([[:upper:]])([[:upper:]]+)_([[:upper:]])([[:upper:]]+)/\u\1\L\2\u\3\L\4/g')
echo "status().is$result()"
This might work for you (GNU sed):
<<<"$result" sed -r 's/.*(Status)\.(.*)_(.*)/\L\1().is\u\2\u\3()/'
Use pattern matching/grouping/back references. The majority of the RHS is lowercase, so use the \L metacharacter to convert from Status... to lowercase and uppercase just the start of words using \u which converts only the next character to uppercase.
N.B. \L and likewise \U converts all following characters to lowercase/uppercase until \E or \U/\L, \l and \u only interrupt this for the next character.
Since you are using GNU sed (-r switch), here's another sed solution,
just a little bit more concise, and locale safe:
$ result=HttpStatus.NOT_FOUND
$ echo "$result" | sed -r 's/^.*([A-Z][a-z]*)\.([a-zA-Z])([a-zA-Z]*)_([a-zA-Z])([a-zA-Z]*)/\L\1().is\u\2\L\3\U\4\L\5()/'
status().isNotFound()
An even more concise way of sed is:
echo "$result" | sed -r 's/^.*([A-Z][a-z]*)\.([a-zA-Z]*)_([a-zA-Z]*)/\L\1().is\u\2\u\3()/'
They both are case insensitive for the second part, for example .nOt_fOuNd also works here.
And an GNU awk solution:
echo "$result" | awk 'function cap(str){return (toupper(substr(str,1,1)) tolower(substr(str,2)))}match($0, /([A-Z][a-z]*)\.([a-zA-Z]*)_([a-zA-Z]*)/, m){print tolower(m[1]) ".is" cap(m[2]) cap(m[3]) "()"}'
You can use the sed option "-e" to concatenate multible expressions.
This question already has answers here:
How to grep for contents after pattern?
(8 answers)
Closed 5 years ago.
I'm trying to read values from a text file.
I have test1.txt which looks like
sub1 1 2 3
sub8 4 5 6
I want to obtain values '1 2 3' when I specify 'sub1'.
The closest I get is:
subj="sub1"
grep "$subj" test1.txt
But the answer is:
sub8 4 5 6
I've read that grep gives you the next line to the match, so I've tried to change the text file to the following:
test2.txt looks like:
sub1
1 2 3
sub8
4 5 6
However, when I type
grep "$subj" test2.txt
The answer is:
sub1
It should be something super simple but I've tried awk, seg, grep,egrep, cat and none is working...I've also read some posts somehow related but none was really helpful
Awk works: awk '$1 == "'"$subj"'" { print $2, $3, $4 }' test1.txt
The command outputs fields two, three, and four for all lines in test1.txt where the first field is $subj (i.e.: the contents of the variable named subj).
With your original text file format:
target=sub1
while IFS=$' \t\n' read -r key values; do
if [[ $key = "$target" ]]; then
echo "Found values: $values"
fi
done <test1.txt
This requires no external tools, using only functionality built into bash itself. See BashFAQ #1.
As has come up during debugging in comments, if you have a traditional Apple-format text file (CR newlines only), then you might want something more like:
target=sub1
while IFS=$' \t\n' read -r -d $'\r' key values || [[ $key ]]; do
if [[ $key = "$target" ]]; then
echo "Found values: $values"
fi
done <test1.txt
Alternately, using awk (for a standard UNIX text file):
target="sub1"
awk -v target="$target" '$1 == target { $1 = ""; print; }' <test1.txt
...or, for a file with CR-only newlines:
target="sub1"
tr '\r' '\n' <test1.txt | awk -v target="$target" '$1 == target { $1 = ""; print; }'
This version will be slower if the text file being read is small (since awk, like any other external tool, takes time to start up); but faster if it's large (since awk's operation is much faster than that of bash's built-ins once it's done starting up).
grep "sub1" test1.txt | cut -c6-
or
grep -A 1 "sub1" test2.txt | tail -n 1
You doing it right, but it seems like test1.txt has a wrong value in it.
with grep foo you get all lines with foo in it. use grep -m1 foo to find the first line with foo in it only.
then you can use cut -d" " -f2- to get all the values behind foo, while seperated by empty spaces.
In the end the command would look like this ...
$ subj="sub1"
$ grep -m1 "$subj" test1.txt | cut -d" " -f2-
But this doenst explain why you could not find sub1 in the first place.
Did you read the proper file ?
There's a bunch of ways to do this (and shorter/more efficient answers than what I'm giving you), but I'm assuming you're a beginner at bash, and therefore I'll give you something that's easy to understand:
egrep "^$subj\>" file.txt | sed "s/^\S*\>\s*//"
or
egrep "^$subj\>" file.txt | sed "s/^[^[:blank:]]*\>[[:blank:]]*//"
The first part, egrep, will search for you subject at the beginning of the line in file.txt (that's what the ^ symbol does in the grep string). It also is looking for a whole word (the \> is looking for an end of word boundary -- that way sub1 doesn't match sub12 in the file.) Notice you have to use egrep to get the \>, as grep by default doesn't recognize that escape sequence. Once done finding the lines, egrep then passes it's output to sed, which will strip the first word and trailing whitespace off of each line. Again, the ^ symbol in the sed command, specifies it should only match at the beginning of the line. The \S* tells it to read as many non-whitespace characters as it can. Then the \s* tells sed to gobble up as many whitespace as it can. sed then replaces everything it matched with nothing, leaving the other stuff behind.
BTW, there's a help page in Stack overflow that tells you how to format your questions (I'm guessing that was the reason you got a downvote).
-------------- EDIT ---------
As pointed out, if you are on a Mac or something like that you have to use [:alnum:] instead of \S, and [:blank:] instead of \s in your sed expression (as these are portable to all platforms)
awk '/sub1/{ print $2,$3,$4 }' file
1 2 3
What happens? After regexp /sub1/ the three following fields are printed.
Any drawbacks? It affects the space.
Sed also works: sed -n -e 's/^'"$subj"' *//p' file1.txt
It outputs all lines matching $subj at the beginning of a line after having removed the matching word and the spaces following. If TABs are used the spaces should be replaced by something like [[:space:]].
I get a string that can have from zero to multiple leading and trailing spaces.
I'm trying to get rid of them without lot of hackery but my code looks huge.
How to do this in a clean way?
as easy as:
$ src=" some text "
$ dst=" $(echo $src)"
$ echo ":$dst:"
: some text:
$(echo $src) will get rid of all around spaces.
than you simply add how much spaces you need before it.
How are you calling out the string? If it's an echo you can just put
Echo "<2 spaces>". "string";
if it's a normal string you just put 2 spaces between the first qoute and the string.
"<2spaces> string here"
One way using GNU sed:
sed 's/^[ \t]*/ /; s/[ \t]*$//' file.txt
You can apply this to a bash variable like this:
echo "$string" | sed 's/^[ \t]*/ /; s/[ \t]*$//'
And save it like this:
variable=$(echo "$string" | sed 's/^[ \t]*/ /; s/[ \t]*$//')
Explanation:
The first substitution will remove all leading whitespace and replace it with two spaces.
The second substitution will simply remove all lagging whitespace from a line.
The simplest is probably to use an external process.
value=$(echo "$value" | sed 's/^ *\(.*[^ ]\) *$/ \1/')
If you need to transform an empty string into two spaces, you'll need to modify the regex; and if you're not on Linux, your sed dialect may differ slightly. For maximum portability, switch to awk or Perl, or do it all in Bash. That gets a bit more complex, but for a start, trailing=${value##*[! ]} contains any trailing spaces, and you can trim them off with ${value%$trailing}, and similarly for leading spaces. See the section on variable substitution in the Bash manual for details.
You can use a regular expression to match everything between the leading and trailing spaces. The matched text is found in the BASH_REMATCH array (the text matching the first parentheses group is in element 1).
spcs='\ *'
text='.*[^ ]'
[[ $src =~ ^$spcs($text)$spcs$ ]]
dst=" ${BASH_REMATCH[1]}"
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2`
outputs something like this
"Abc Inc";
What I want to do is I want to remove the trailing ";" as well. How can i do that? I am a beginner to bash. Any thoughts or suggestions would be helpful.
This will remove the last character contained in your COMPANY_NAME var regardless if it is or not a semicolon:
echo "$COMPANY_NAME" | rev | cut -c 2- | rev
I'd use sed 's/;$//'. eg:
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2 | sed 's/;$//'`
foo="hello world"
echo ${foo%?}
hello worl
I'd use head --bytes -1, or head -c-1 for short.
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2 | head --bytes -1`
head outputs only the beginning of a stream or file. Typically it counts lines, but it can be made to count characters/bytes instead. head --bytes 10 will output the first ten characters, but head --bytes -10 will output everything except the last ten.
NB: you may have issues if the final character is multi-byte, but a semi-colon isn't
I'd recommend this solution over sed or cut because
It's exactly what head was designed to do, thus less command-line options and an easier-to-read command
It saves you having to think about regular expressions, which are cool/powerful but often overkill
It saves your machine having to think about regular expressions, so will be imperceptibly faster
I believe the cleanest way to strip a single character from a string with bash is:
echo ${COMPANY_NAME:: -1}
but I haven't been able to embed the grep piece within the curly braces, so your particular task becomes a two-liner:
COMPANY_NAME=$(grep "company_name" file.txt); COMPANY_NAME=${COMPANY_NAME:: -1}
This will strip any character, semicolon or not, but can get rid of the semicolon specifically, too.
To remove ALL semicolons, wherever they may fall:
echo ${COMPANY_NAME/;/}
To remove only a semicolon at the end:
echo ${COMPANY_NAME%;}
Or, to remove multiple semicolons from the end:
echo ${COMPANY_NAME%%;}
For great detail and more on this approach, The Linux Documentation Project covers a lot of ground at http://tldp.org/LDP/abs/html/string-manipulation.html
Using sed, if you don't know what the last character actually is:
$ grep company_name file.txt | cut -d '=' -f2 | sed 's/.$//'
"Abc Inc"
Don't abuse cats. Did you know that grep can read files, too?
The canonical approach would be this:
grep "company_name" file.txt | cut -d '=' -f 2 | sed -e 's/;$//'
the smarter approach would use a single perl or awk statement, which can do filter and different transformations at once. For example something like this:
COMPANY_NAME=$( perl -ne '/company_name=(.*);/ && print $1' file.txt )
don't have to chain so many tools. Just one awk command does the job
COMPANY_NAME=$(awk -F"=" '/company_name/{gsub(/;$/,"",$2) ;print $2}' file.txt)
In Bash using only one external utility:
IFS='= ' read -r discard COMPANY_NAME <<< $(grep "company_name" file.txt)
COMPANY_NAME=${COMPANY_NAME/%?}
Assuming the quotation marks are actually part of the output, couldn't you just use the -o switch to return everything between the quote marks?
COMPANY_NAME="\"ABC Inc\";" | echo $COMPANY_NAME | grep -o "\"*.*\""
you can strip the beginnings and ends of a string by N characters using this bash construct, as someone said already
$ fred=abcdefg.rpm
$ echo ${fred:1:-4}
bcdefg
HOWEVER, this is not supported in older versions of bash.. as I discovered just now writing a script for a Red hat EL6 install process. This is the sole reason for posting here.
A hacky way to achieve this is to use sed with extended regex like this:
$ fred=abcdefg.rpm
$ echo $fred | sed -re 's/^.(.*)....$/\1/g'
bcdefg
Some refinements to answer above. To remove more than one char you add multiple question marks. For example, to remove last two chars from variable $SRC_IP_MSG, you can use:
SRC_IP_MSG=${SRC_IP_MSG%??}
cat file.txt | grep "company_name" | cut -d '=' -f 2 | cut -d ';' -f 1
I am not finding that sed 's/;$//' works. It doesn't trim anything, though I'm wondering whether it's because the character I'm trying to trim off happens to be a "$". What does work for me is sed 's/.\{1\}$//'.