I only want to select the first occurrence of a pattern on the line, e.g. the '53.0' instead I get:
$ echo "Core 0: +53.0°C (high = +80.0°C, crit = +100.0°C)" |
sed -n 's/.*+\([0-9.]*\).*/\1/p'
which, for me, prints: 100.0 when I expected: 53.0.
I am using Ubuntu 18.04.3 LTS.
What am I missing?
Could you please try following.
echo "Core 0: +53.0°C (high = +80.0°C, crit = +100.0°C)" |
sed 's/[^+]*+\([^ ]*\).*/\1/'
OR in case you don't want ° in output then try following.
echo "Core 0: +53.0°C (high = +80.0°C, crit = +100.0°C)" |
sed 's/[^+]*+\([^°]*\).*/\1/'
Explanation of above code:
I have used sed's capability of saving matched regex into temp buffer memory and given it a reference as \1 to print it later.
Following explanation is only for explaining purposes to run code use above code please.
sed ' ##Starting sed program here.
s ##Using s option for substitution method by sed on Lines of Input_file.
/[^+]*+ ##Using regex to have till 1st occurrence of + here.
\([^ ]*\) ##Using memory buffer saving capability of sed to save matched regex in it.
##Have used regex to match everything till occurrence of space. Which should catch 1st temperature.
.* ##Mentioning .* to match all rest line.
/\1/' ##Now substituting whole line with 1st buffer matched value.
Explanation of why OP's code not working: Since OP is using .*+ in his/her attempt which is a GREEDY match so hence it is matching the last occurrence of + thus it is giving last temperature value.
Note: Added link provided by Sundeep sir in comments for Greedy match understanding.
Quite simple to do with awk (no need for complex regex)
echo "Core 0: +53.1°C (high = +80.0°C, crit = +100.0°C)" | awk '{print $3+0}'
53.1
Print the third field $3, then the +0 is used to get only the digits, it trims of °C
