I had to convert parts of it to make it compatible to my old shell, as I got a syntax error but all in all it works perfectly! I even tried to trick it with random "$" or random braces "{", but it still only outputs the correct ones!
Code:
line='aaaa$}aaa${important}xxxxxxxx${important2}oo{o$}oo$oo${importantstring3}'
IFS=\$ read -a words <<< "$line"
regex='(\{[^}]+})'
for e in "${words[@]}"; do
if [[ $e =~ $regex ]]; then
echo "\$${BASH_REMATCH[0]}";
fi;
done
Thanks again, you made a very happy user
---------- Post updated at 08:25 AM ---------- Previous update was at 07:05 AM ----------
Though I am satisfied with the solution, as I assume it will not produce errors, I have found something where I could trick it. If I use this line:
Code:
line='aaaa$aa{yyy}aaaaaa${important}xxxx
It will print ${yyy} as matching. That is because it only uses the "$" as separator and indirectly allows random characters to follow afterwards. I still wonder if there isn't any regex which will cover this (sorry, I am not the best at expressions and think in pseudo code, but somehow it bugs me):
First one would need to determine that these 2 characters must always come first:
[\$][\{]
Then comes a term where everything is allowed, except these:
[everything allowed except \$,\{]
The previous term is read until the closing bracket comes:
[\}].
This is my naive thinking, but it seems the thought process is easier than the actual implementation.
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Discussion started by: rajeshwebspere
5 Replies
LEARN ABOUT PHP
trim
TRIM(3) 1 TRIM(3)trim - Strip whitespace (or other characters) from the beginning and end of a stringSYNOPSIS
string trim (string $str, [string $character_mask = " 0r B"])
DESCRIPTION
This function returns a string with whitespace stripped from the beginning and end of $str. Without the second parameter, trim(3) will
strip these characters:
o " " (ASCII
32 ( 0x20)), an ordinary space.
o " " (ASCII
9 ( 0x09)), a tab.
o "
" (ASCII
10 ( 0x0A)), a new line (line feed).
o "
" (ASCII
13 ( 0x0D)), a carriage return.
o "