Grep Text between pattern and space - sed

I have data in below format.
sometext NAME=abc TIME_TAKEN
sometext NAME=xyz{123} REQUEST
I am trying to grep the data between Name and space . I want output in below format
abc
xyz{123}
I tried using sed along with cat like below but it is not working .
cat test.txt | sed 's/NAME=\(.*\)"\+[[:space:]]\+"/\1/g'
awk 'sub(/.*NAME= +/,"") && sub(/ +.*/,"")' test.txt
Regards

Match everything before and after the assignment:
sed -ne 's/.* NAME=\(.*\) .*/\1/p' test.txt
Beware! It will extract the last NAME=... from each line.

This might work for you (GNU sed):
sed -En 's/.*\<NAME=(\S*).*/\1/p' file
Turn off implicit printing -n and turn on extended regexp -E.
Match a word NAME, followed by = and then any number of non-white spaced characters and return the any number of non-white spaced characters.
N.B. The replacement must replace the entire line, hence the .* at the start and end of the regexp.

Try the following command
cat test.txt | sed 's#.*NAME=\(.* \).*#\1#'
Explanation:
cat test.txt | sed 's# .*NAME= \(.* \) .* #\1#'
cat fileName | sed 's# All upto NAME= extract(upto any space) skip all

Related

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

search a string which contains "/" and replace using sed

How to search a pattern and remove the line using sed which contains special characters like "ranasnfs2:/SA_kits/prod"
I tried using a variable to hold the complete string and then recall the variable in sed command but it is not working.
echo $a
ranasnfs2:/SA_kits/prod
sed -i '/"$a"/d' test.txt
cat test.txt | grep -i SA
/SA_kits -rw,suid,soft,retry=4 ranasnfs2:/SA_kits/prod
You need to escape the slash character.
Use this for deleting lines which contain a /:
sed '/\//d' file

tcsh & sed: no output

I’m trying to replace the 3rd column of a file for itself plus the value of column 2 (without any space). I get the proper value for variable c and a but then sed doesn't give any output. Any clue?
#!/bin/tcsh
setenv c `cat lig_mod.pdb | awk '{print $3}'`
echo $c
setenv a `cat lig_mod.pdb | awk '{print $3=$3$2}'`
echo $a
sed -i "" 's/^'"${c}"'$/^'"${a}"'$/g' lig_mod.pdb
Even though awk is usually better for columns parsing this one-liner sed can work for you as well:
sed -i 's/ \(\w*\) \(\w*\) / \1 \2\1 /1' lig_mod.pdb
the '/1' at the end denote the instance number you desire to change which for the 2nd and 3rd columns is the first, but you could use it for any adjacent columns.

Sed uppercase lines if they starting with an uppercase character

I want the lines starting with one uppercase character to be uppercased, other lines should be not touched.
So this input:
cat myfile
a
b
Cc
should result in this output:
a
b
CC
I tried this command, but this not matches if i use grouping:
cat myfile | sed -r 's/\([A-Z]+.*\)/\U\1/g'
What am i doing wrong?
When you use the -r option, you must not put \ before parentheses used for grouping. So it should be:
sed -r 's/^([A-Z].*)/\U\1/' myfile
Also, notice that you need ^ to match the beginning of the line. The g modifier isn't needed, since you're matching the entire line.
cat myfile | sed 's/^\([A-Z].*\)$/\U\1/'
\U for uppercase conversion is a GNU sed extension.
Alternative for platforms where that is not available (e.g., macOS, with its BSD awk implementation):
awk '/^[A-Z]/ { print toupper($0); next } 1'
sed '/^[A-Z].*[a-z]/ s/.*/\U\1/' YourFile
only on line that are not compliant
This might work for you (GNU sed):
sed 's/^[[:upper:]].*/\U&/' file

How to use a sed one-liner to parse "rec:id=1&name=zz&age=21" into "1 zz 21"?

I can chain multiple sed substitutions and a awk operation to achieve this, but is there a single sed substitution that can do it?
Also is there any other tool that is more suitable for this parsing task?
You could try:
sed -r 's!rec:id=(.*?)&name=(.*?)&age=(.*?)!\1 \2 \3!' input_file
If you don't know the rec:id etc in advance but you know there's three, you could try:
sed -r 's![^=]+=(.*?)&[^=]+=(.*?)&[^=]+=(.*?)!\1 \2 \3!' input_file
If you don't know how many &name=value pairs you're after in advance but want to output all the values, you could try something like:
grep -P -o '(?<==)([^&]*)(?=&|$)' | xargs
where the -P means 'perl regex', the regex says "find the string followed by an & (or end of string) and preceded by and equals sign", the -o means to print just the matches (ie the 1, zz, and 21) each on their own line, and the | xargs moves these from their own line to one line and space separated (ie 1\nzz\n21 to 1 zz 21).
This might work for you:
echo "rec:id=1&name=zz&age=21" | sed 's/[^=]*=\([^&]*\)/\1 /g'
1 zz 21
However this leaves an extra space at the end, to solve this use:
echo "rec:id=1&name=zz&age=21"|sed 's/[^=]*=\([^&]*\)/\1 /g:;s/ $//'
1 zz 21
How about parsing the values directly into variables?
inbound="rec:id=1&name=zz&age=21"
eval $(echo $inbound | cut -c5- | tr \& "\n")
echo "Name:$name, ID:$id, Age:$age"
Or even better, though slightly more arcane:
inbound="rec:id=1&name=zz&age=21"
IFS=\& eval $(cut -c5- <<< $inbound)
echo "Name:$name, ID:$id, Age:$age"