cURL and Bash, capture variable and value from string [closed] - regex

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Using bash, grep, split, awk, or sed, I would like to capture
ASPSESSIONIDSUSTQBQS=AAHNFMBAGABAILMKCDGIIMFJ
from
Set-Cookie: ASPSESSIONIDSUSTQBQS=AAHNFMBAGABAILMKCDGIIMFJ; secure; path=/
'ASPSESSIONID' remains always the same + 8 random characters (SUSTQBQS).
Also this variable may not always be located in the second columns or right after 'Set-Cookie: '
Can anyone please help ?

Answer 1

With awk:
awk -F"[ ;]" '{print $2}' FileName
Set the field seperator as space and ;. Then print the 2nd field.

Answer 2

The basic regex structure is the same in various programs.
It may be explained in words as: The text between the colon/space(: ) and the semicolon (;). Which, in regex parlance is:
: ([^;]*);
And could be assigned to a var:
RE=': ([^;]*);'
Then, we could use it in
bash
while read l; do
[[ $l =~ $RE ]] && echo "${BASH_REMATCH[1]}";
done <file
gawk
gawk -v RE="$RE" '$0 ~ RE { print gensub(".*"RE".*","\\1",1); }' file
sed
sed -rn 's/^.*'"$RE"'.*$/\1/p' file # using -r avoids the several `\`

Answer 3

Try this sed command
sed 's/[^:]\+..\([^;]\+\).*/\1/' FileName
Explanation:
[^:]\+ -- Remove the charecters until :
.. -- Remove two characters
\([^;]\+\) -- Capture the group until ; found
.* -- Remove the all character after capture the group
\1 -- Finally print the captured group
Output :
ASPSESSIONIDSUSTQBQS=AAHNFMBAGABAILMKCDGIIMFJ