Extracting n characters from a test line

Question

How can I extract the 6 characters after Z in the following text line and put it into a file using grep, egrep, sed or awk?

B1_Site4_5aT4ZNHN691AQSB6B65_KYEC_SLT_2013-11-24-00-30_935985e7_100m_PASS1.tar

Answer 1

file="B1_Site4_5aT4ZNHN691AQSB6B65_KYEC_SLT_2013-11-24-00-30_935985e7_100m_PASS1.tar"

with GNU grep when built with PCRE support (and with zsh or recent versions of ksh93 or bash for <<<):

grep -oP '(?<=Z).{6}' <<< "$file" > file

with ksh93, bash or recent versions of zsh:

tmp=${file#*Z}              # remove chars up to the first Z
echo "${tmp:0:6}" > file

Just for fun, awk

awk -F Z '{print substr($2, 1, 6)}' <<< "$file"

Answer 2

expr B1_Site4_5aT4ZNHN691AQSB6B65_KYEC_SLT_2013-11-24-00-30_935985e7_100m_PASS1.tar : '.*Z\(.\{6\}\)' > file

Or just with the shell

string=B1_Site4_5aT4ZNHN691AQSB6B65_KYEC_SLT_2013-11-24-00-30_935985e7_100m_PASS1.tar
printf '%.6s\n' "${string#*Z}" > file

(the first one will consider the last Z followed by 6 characters, the second one the first; behaviour will vary if their not ZXXXXXX in there).

Answer 3

>>echo \
B1_Site4_5aT4ZNHN691AQSB6B65_KYEC_SLT_2013-11-24-00-30_935985e7_100m_PASS1.tar |
sed 's/.*Z\(.\{6\}\).*/\1/'
>>NHN691

Adding a >file will put it in a file.