Read a part of directory name and get rid of carriage control usind sed or similar

In some script, it reads the directory name and get things like

66.9090_89.4450_168.0250_ABC3/

I need to extract the thing "ABC3" so I try

    sed -i -e "s/开发者_Go百科_/ /g" temp_direc 

so I get

66.9090 89.4450 168.0250 ABC3/

and then

read LABEL <<< $(awk '{print $4}' temp_direc)

and now I get

ABC3/

but I do not know how to remove the final "/". Furthermore, I need later to do this

echo $A" "$LABEL

being $A some string like 45.56

and I would like the output to be

45.56 ABC3

but I really get

 ABC3/6

so it seems the carriage control character is somehow embedded there. How could I get rid of this and get my desired output?

---

A pure sed solution would be:

sed 's#.*_\([^_/]*\)\(/\|$\)#\1#'

resp. with GNU sed:

sed -r 's#.*_([^_/]*)(/|$)#\1#'

I chose the # as a delimiter (instead of the usual /) because there's a / in the pattern which otherwise would have to be escaped.

---

Try simplier:

echo '66.9090_89.4450_168.0250_ABC3/' | cut -d"_" -f4 | cut -d"/" -f1

which yields ABC3

---

Not 100% sure what you're after here. But to get the ABC3 you have 90% of it already:

$echo '66.9090_89.4450_168.0250_ABC3/' | awk -F'_' '{print $4;}' | sed 's/\///'
ABC3

---

An awk only solution:

read LABEL <<< $( echo '66.9090_89.4450_168.0250_ABC3/' \
                  | awk -F '[_/]' '{printf("%s", $4);}' )

Opss ... I supposed that the field is ever the forth, the following line has not this limitation:

read LABEL <<< $( echo '66.9090_89.4450_168.0250_ABC3/' \
                  | awk -F '[_/]' '{printf("%s", $(NF -1));}' ) ; echo $LABEL

---

echo '66.9090_89.4450_168.0250_ABC3/' | sed -e 's/^.*_//' -e 's/.$//'

---

To get output like ABC3/6 indicates your $A ends with a carriage return. You can use something like dos2unix on your input file, or use tr to remove the carriage returns:

A=$'45.56\r'
B="66.9090_89.4450_168.0250_ABC3/"

a=$( echo "$A" | tr -d $'\r' )   # remove all carriage returns

b=${B##*_}   # remove up to (and including) the last underscore
b=${b%/}     # remove a trailing slash 

echo "$a $b"  # ==> "45.56 ABC3"