Get version of Podspec via command line (bash, zsh) [duplicate] - azure-devops

Given a file, for example:
potato: 1234
apple: 5678
potato: 5432
grape: 4567
banana: 5432
sushi: 56789
I'd like to grep for all lines that start with potato: but only pipe the numbers that follow potato:. So in the above example, the output would be:
1234
5432
How can I do that?

grep 'potato:' file.txt | sed 's/^.*: //'
grep looks for any line that contains the string potato:, then, for each of these lines, sed replaces (s/// - substitute) any character (.*) from the beginning of the line (^) until the last occurrence of the sequence : (colon followed by space) with the empty string (s/...// - substitute the first part with the second part, which is empty).
or
grep 'potato:' file.txt | cut -d\ -f2
For each line that contains potato:, cut will split the line into multiple fields delimited by space (-d\ - d = delimiter, \ = escaped space character, something like -d" " would have also worked) and print the second field of each such line (-f2).
or
grep 'potato:' file.txt | awk '{print $2}'
For each line that contains potato:, awk will print the second field (print $2) which is delimited by default by spaces.
or
grep 'potato:' file.txt | perl -e 'for(<>){s/^.*: //;print}'
All lines that contain potato: are sent to an inline (-e) Perl script that takes all lines from stdin, then, for each of these lines, does the same substitution as in the first example above, then prints it.
or
awk '{if(/potato:/) print $2}' < file.txt
The file is sent via stdin (< file.txt sends the contents of the file via stdin to the command on the left) to an awk script that, for each line that contains potato: (if(/potato:/) returns true if the regular expression /potato:/ matches the current line), prints the second field, as described above.
or
perl -e 'for(<>){/potato:/ && s/^.*: // && print}' < file.txt
The file is sent via stdin (< file.txt, see above) to a Perl script that works similarly to the one above, but this time it also makes sure each line contains the string potato: (/potato:/ is a regular expression that matches if the current line contains potato:, and, if it does (&&), then proceeds to apply the regular expression described above and prints the result).

Or use regex assertions: grep -oP '(?<=potato: ).*' file.txt

grep -Po 'potato:\s\K.*' file
-P to use Perl regular expression
-o to output only the match
\s to match the space after potato:
\K to omit the match
.* to match rest of the string(s)

sed -n 's/^potato:[[:space:]]*//p' file.txt
One can think of Grep as a restricted Sed, or of Sed as a generalized Grep. In this case, Sed is one good, lightweight tool that does what you want -- though, of course, there exist several other reasonable ways to do it, too.

This will print everything after each match, on that same line only:
perl -lne 'print $1 if /^potato:\s*(.*)/' file.txt
This will do the same, except it will also print all subsequent lines:
perl -lne 'if ($found){print} elsif (/^potato:\s*(.*)/){print $1; $found++}' file.txt
These command-line options are used:
-n loop around each line of the input file
-l removes newlines before processing, and adds them back in afterwards
-e execute the perl code

You can use grep, as the other answers state. But you don't need grep, awk, sed, perl, cut, or any external tool. You can do it with pure bash.
Try this (semicolons are there to allow you to put it all on one line):
$ while read line;
do
if [[ "${line%%:\ *}" == "potato" ]];
then
echo ${line##*:\ };
fi;
done< file.txt
## tells bash to delete the longest match of ": " in $line from the front.
$ while read line; do echo ${line##*:\ }; done< file.txt
1234
5678
5432
4567
5432
56789
or if you wanted the key rather than the value, %% tells bash to delete the longest match of ": " in $line from the end.
$ while read line; do echo ${line%%:\ *}; done< file.txt
potato
apple
potato
grape
banana
sushi
The substring to split on is ":\ " because the space character must be escaped with the backslash.
You can find more like these at the linux documentation project.

Modern BASH has support for regular expressions:
while read -r line; do
if [[ $line =~ ^potato:\ ([0-9]+) ]]; then
echo "${BASH_REMATCH[1]}"
fi
done

grep potato file | grep -o "[0-9].*"

Related

Use sed to replace every character by itself followed by $n times a char?

I'm trying to run the command below to replace every char in DECEMBER by itself followed by $n question marks. I tried both escaping {$n} like so {$n} and leaving it as is. Yet my output just keeps being D?{$n}E?{$n}... Is it just not possible to do this with a sed?
How should i got about this.
echo 'DECEMBER' > a.txt
sed -i "s%\(.\)%\1\(?\){$n}%g" a.txt
cat a.txt
This might work for you (GNU sed):
n=5
sed -E ':a;s/[^\n]/&\n/g;x;s/^/x/;/x{'"$n"'}/{z;x;y/\n/?/;b};x;ba' file
Append a newline to each non-newline character in a line $n times then replace all newlines by the intended character ?.
N.B. The newline is chosen as the initial substitute character as it is not possible for it to be within a line (sed uses newlines to separate lines) and if the final substitution character already exists within the current line, the substitutions are correct.
Range (also, interval or limiting quantifiers), like {3} / {3,} / {3,6}, are part of regex, and not replacement patterns.
You can use
sed -i "s/./&$(for i in {1..7}; do echo -n '?'; done)/g" a.txt
See the online demo:
#!/bin/bash
sed "s/./&$(for i in {1..7}; do echo -n '?'; done)/g" <<< "DECEMBER"
# => D???????E???????C???????E???????M???????B???????E???????R???????
Here, . matches any char, and & in the replacement pattern puts it back and $(for i in {1..7}; do echo -n '?'; done) adds seven question marks right after it.
This one-liner should do the trick:
sed 's/./&'$(printf '%*s' "$n" '' | tr ' ' '?')'/g' a.txt
with the assumption that $n expands to a positive integer and the command is executed in a POSIX shell.
Efficiently using any awk in any shell on every Unix box after setting n=2:
$ awk -v n="$n" '
BEGIN {
new = sprintf("%*s",n,"")
gsub(/./,"?",new)
}
{
gsub(/./,"&"new)
print
}
' a.txt
D??E??C??E??M??B??E??R??
To make the changes "inplace" use GNU awk with -i inplace just like GNU sed has -i.
Caveat - if the character you want to use in the replacement text is & then you'd need to use gsub(/./,"\\\\\\&",new) in the BEGIN section to make it is treated as literal instead of a backreference metachar. You'd have that issue and more (e.g. handling \1 or /) with any sed solution and any solution that uses double quotes around the script would have more issues with handling $s and the solutions that have a shell script expanding unquoted would have even more issues with globbing chars.

Using a single sed call to split and grep

This is mostly by curiosity, I am trying to have the same behavior as:
echo -e "test1:test2:test3"| sed 's/:/\n/g' | grep 1
in a single sed command.
I already tried
echo -e "test1:test2:test3"| sed -e "s/:/\n/g" -n "/1/p"
But I get the following error:
sed: can't read /1/p: No such file or directory
Any idea on how to fix this and combine different types of commands into a single sed call?
Of course this is overly simplified compared to the real usecase, and I know I can get around by using multiple calls, again this is just out of curiosity.
EDIT: I am mostly interested in the sed tool, I already know how to do it using other tools, or even combinations of those.
EDIT2: Here is a more realistic script, closer to what I am trying to achieve:
arch=linux64
base=https://chromedriver.storage.googleapis.com
split="<Contents>"
curl $base \
| sed -e 's/<Contents>/<Contents>\n/g' \
| grep $arch \
| sed -e 's/^<Key>\(.*\)\/chromedriver.*/\1/' \
| sort -V > out
What I would like to simplify is the curl line, turning it into something like:
curl $base \
| sed 's/<Contents>/<Contents>\n/g' -n '/1/p' -e 's/^<Key>\(.*\)\/chromedriver.*/\1/' \
| sort -V > out
Here are some alternatives, awk and sed based:
sed -E "s/(.*:)?([^:]*1[^:]*).*/\2/" <<< "test1:test2:test3"
awk -v RS=":" '/1/' <<< "test1:test2:test3"
# or also
awk 'BEGIN{RS=":"} /1/' <<< "test1:test2:test3"
Or, using your logic, you would need to pipe a second sed command:
sed "s/:/\n/g" <<< "test1:test2:test3" | sed -n "/1/p"
See this online demo. The awk solution looks cleanest.
Details
In sed solution, (.*:)?([^:]*1[^:]*).* pattern matches an optional sequence of any 0+ chars and a :, then captures into Group 2 any 0 or more chars other than :, 1, again 0 or more chars other than :, and then just matches the rest of the line. The replacement just keeps Group 2 contents.
In awk solution, the record separator is set to : and then /1/ regex is used to only return the record having 1 in it.
This might work for you (GNU sed):
sed 's/:/\n/;/^[^\n]*1/P;D' file
Replace each : and if the first line in the pattern space contains 1 print it.
Repeat.
An alternative:
sed -Ez 's/:/\n/g;s/^[^1]*$//mg;s/\n+/\n/;s/^\n//' file
This slurps the whole file into memory and replaces all colons by newlines. All lines that do not contain 1 are removed and surplus newlines deleted.
An alternative to the really ugly sed is: grep -o '\w*2\w*'
$ printf "test1:test2:test3\nbob3:bob2:fred2\n" | grep -o '\w*2\w*'
test2
bob2
fred2
grep -o: only matching
Or: grep -o '[^:]*2[^:]*'
echo -e "test1:test2:test3" | sed -En 's/:/\n/g;/^[^\n]*2[^\n]*(\n|$)/P;//!D'
sed -n doesn't print unless told to
sed -E allows using parens to match (\n|$) which is newline or the end of the pattern space
P prints the pattern buffer up to the first newline.
D trims the pattern buffer up to the first newline
[^\n] is a character class that matches anything except a newline
// is sed shorthand for repeating a match
//! is then matching everything that didn't match previously
So, after you split into newlines, you want to make sure the 2 character is between the start of the pattern buffer ^ and the first newline.
And, if there is not the character you are looking for, you want to D delete up to the first newline.
At that point, it works for one line of input, with one string containing the character you're looking for.
To expand to several matches within a line, you have to ta, conditionally branch back to label :a:
$ printf "test1:test2:test3\nbob3:bob2:fred2\n" | \
sed -En ':a s/:/\n/g;/^[^\n]*2[^\n]*(\n|$)/P;D;ta'
test2
bob2
fred2
This is simply NOT a job for sed. With GNU awk for multi-char RS:
$ echo "test1:test2:test3:test4:test5:test6"| awk -v RS='[:\n]' '/1/'
test1
$ echo "test1:test2:test3:test4:test5:test6"| awk -v RS='[:\n]' 'NR%2'
test1
test3
test5
$ echo "test1:test2:test3:test4:test5:test6"| awk -v RS='[:\n]' '!(NR%2)'
test2
test4
test6
$ echo "foo1:bar1:foo2:bar2:foo3:bar3" | awk -v RS='[:\n]' '/foo/ || /2/'
foo1
foo2
bar2
foo3
With any awk you'd just have to strip the \n from the final record before operating on it:
$ echo "test1:test2:test3:test4:test5:test6"| awk -v RS=':' '{sub(/\n$/,"")} /1/'
test1

Data transformation using sed

I have a file like:
A
B
C
D
E
F
G
H
I
J
K
L
and I want it to come out like
A,B,C,D
E,F,G,H
I'm assuming I'd use sed, but actually I'm not even sure if that's the best tool. I'm open to using anything commonly available on a Linux system.
In perl, I did it like this ... it works, but it's dirty and has a trailing comma. Was hoping for something simpler:
$ perl -ne 'if (/^(\w)\R/) {print "$1,";} else {print "\n";}' test
A,B,C,D,
E,F,G,H,
I,J,K,L,
Set the input record separator to paragraph mode (-00) and then split each record on any remaining whitespace:
$ perl -00 -ne 'print join("," => split), "\n"' test
Add -l to enable automatic newlines (but make sure it comes before -00, because we want $\ to be set to the value of $/ before modification):
$ perl -l -00 -ne 'print join("," => split)' test
Add -a to enable autosplit mode and implicitly split to #F:
$ perl -l -00 -ane 'print join("," => #F)' test
Swap out -n for -p for automatic printing:
$ perl -l -00 -ape '$_ = join("," => #F)' test
You could use
awk 'BEGIN {RS=""; FS="\n"; ORS="\n"; OFS=","} {$1=$1} 1' file
I see the gawk manual says this:
If RS
is set to the null string, then records are separated by blank lines. When RS is set to the null string, the newline character always acts as a field separator, in addition to whatever value FS may have.
So we don't actually need to specify FS to get the desired output:
awk 'BEGIN {RS=""; ORS="\n"; OFS=","} {$1=$1} 1' file
xargs could do it,
$ xargs -n4 < file | tr ' ' ','
A,B,C,D
E,F,G,H
I,J,K,L
Replacing newlines with sed is a bit complicated (see this question). It is easier to use tr for the newlines. The rest can be done by sed.
The following command assumes that yourFile does not contain any ,.
tr '\n' , < yourFile | sed 's/,*$/\n/;s/,,/\n/g'
The tr part converts all newlines to ,. The resulting string will have no newlines.
s/,*$/\n/ removes trailing commas and appends a newline (text files usually end with a newline).
s/,,/\n/g replaces ,, by a newline. Two consecutive commas appear only where your original file contained two consecutive newlines, that is where the sections are separated by an empty line.

sed or grep or awk to match very very long lines

more file
param1=" 1,deerfntjefnerjfntrjgntrjnvgrvgrtbvggfrjbntr*rfr4fv*frfftrjgtrignmtignmtyightygjn 2,3,4,5,6,7,8,
rfcmckmfdkckemdio8u548384omxc,mor0ckofcmineucfhcbdjcnedjcnywedpeodl40fcrcmkedmrikmckffmcrffmrfrifmtrifmrifvysdfn drfr4fdr4fmedmifmitfmifrtfrfrfrfnurfnurnfrunfrufnrufnrufnrufnruf"****
need to match the content of param1 as
sed -n "/$param1/p" file
but because the line length (very long line) I cant match the line
what’s the best way to match very long lines?
The problem you are facing is that param1 contains special characters which are being interpreted by sed. The asterisk ('*') is used to mean 'zero or more occurrences of the previous character', so when this character is interpreted by sed there is nothing left to match the literal asterisk you are looking for.
The following is a working bash script that should help:
#!/bin/bash
param1=' 1,deerfntjefnerjfntrjgntrjnvgrvgrtbvggfrjbntr\*rfr4fv\*frfftrjgtrignmtignmtyightygjn 2,3,4,5,6,7,8, rfcmckmfdkckemdio8u548384omxc,mor0ckofcmineucfhcbdjcnedjcnywedpeodl40fcrcmkedmrikmckffmcrffmrfrifmtrifmrifvysdfn'
cat <<EOF | sed "s/${param1}/Bubba/g"
1,deerfntjefnerjfntrjgntrjnvgrvgrtbvggfrjbntr*rfr4fv*frfftrjgtrignmtignmtyightygjn 2,3,4,5,6,7,8, rfcmckmfdkckemdio8u548384omxc,mor0ckofcmineucfhcbdjcnedjcnywedpeodl40fcrcmkedmrikmckffmcrffmrfrifmtrifmrifvysdfn
EOF
Maybe the problem is that your $param1 contains special characters? This works for me:
A="$(perl -e 'print "a" x 10000')"
echo $A | sed -n "/$A/p"
($A contains 10 000 a characters).
echo $A | grep -F $A
and
echo $A | grep -P $A
also works (second requires grep with built-in PCRE support. If you want pattern matching you should use either this or pcregrep. If you don't, use the fixed grep (grep -F)).
echo $A | grep $A
is too slow.

How to "grep" out specific line ranges of a file

There are often times I will grep -n whatever file to find what I am looking for. Say the output is:
1234: whatev 1
5555: whatev 2
6643: whatev 3
If I want to then just extract the lines between 1234 and 5555, is there a tool to do that? For static files I have a script that does wc -l of the file and then does the math to split it out with tail & head but that doesn't work out so well with log files that are constantly being written to.
Try using sed as mentioned on
http://linuxcommando.blogspot.com/2008/03/using-sed-to-extract-lines-in-text-file.html. For example use
sed '2,4!d' somefile.txt
to print from the second line to the fourth line of somefile.txt. (And don't forget to check http://www.grymoire.com/Unix/Sed.html, sed is a wonderful tool.)
The following command will do what you asked for "extract the lines between 1234 and 5555" in someFile.
sed -n '1234,5555p' someFile
If I understand correctly, you want to find a pattern between two line numbers. The awk one-liner could be
awk '/whatev/ && NR >= 1234 && NR <= 5555' file
You don't need to run grep followed by sed.
Perl one-liner:
perl -ne 'if (/whatev/ && $. >= 1234 && $. <= 5555) {print}' file
Line numbers are OK if you can guarantee the position of what you want. Over the years, my favorite flavor of this has been something like this:
sed "/First Line of Text/,/Last Line of Text/d" filename
which deletes all lines from the first matched line to the last match, including those lines.
Use sed -n with "p" instead of "d" to print those lines instead. Way more useful for me, as I usually don't know where those lines are.
Put this in a file and make it executable:
#!/usr/bin/env bash
start=`grep -n $1 < $3 | head -n1 | cut -d: -f1; exit ${PIPESTATUS[0]}`
if [ ${PIPESTATUS[0]} -ne 0 ]; then
echo "couldn't find start pattern!" 1>&2
exit 1
fi
stop=`tail -n +$start < $3 | grep -n $2 | head -n1 | cut -d: -f1; exit ${PIPESTATUS[1]}`
if [ ${PIPESTATUS[0]} -ne 0 ]; then
echo "couldn't find end pattern!" 1>&2
exit 1
fi
stop=$(( $stop + $start - 1))
sed "$start,$stop!d" < $3
Execute the file with arguments (NOTE that the script does not handle spaces in arguments!):
Starting grep pattern
Stopping grep pattern
File path
To use with your example, use arguments: 1234 5555 myfile.txt
Includes lines with starting and stopping pattern.
If I want to then just extract the lines between 1234 and 5555, is
there a tool to do that?
There is also ugrep, a GNU/BSD grep compatible tool but one that offers a -K option (or --range) with a range of line numbers to do just that:
ugrep -K1234,5555 -n '' somefile.log
You can use the usual GNU/BSD grep options and regex patterns (but it also offers a lot more such as -K.)
If you want lines instead of line ranges, you can do it with perl: eg. if you want to get line 1, 3 and 5 from a file, say /etc/passwd:
perl -e 'while(<>){if(++$l~~[1,3,5]){print}}' < /etc/passwd