问题
I have a file named test.txt which has:
abc.cde.ccd.eed.12345.5678.txt
abcd.cdde.ccdd.eaed.12346.5688.txt
aabc.cade.cacd.eaed.13345.5078.txt
abzc.cdae.ccda.eaed.29345.1678.txt
abac.cdae.cacd.eead.18145.2678.txt
aabc.cdve.cncd.ened.19945.2345.txt
If I want to remove everything beyond the first . like:
cde.ccd.eed.12345.5678.txt
cdde.ccdd.eaed.12346.5688.txt
cade.cacd.eaed.13345.5078.txt
cdae.ccda.eaed.29345.1678.txt
cdae.cacd.eead.18145.2678.txt
cdve.cncd.ened.19945.2345.txt
Then I will do
for i in `cat test.txt`; do echo ${i#*.}; done
but If I want to remove everything after the last . like:
abc.cde.ccd.eed.12345.5678
abcd.cdde.ccdd.eaed.12346.5688
aabc.cade.cacd.eaed.13345.5078
abzc.cdae.ccda.eaed.29345.1678
abac.cdae.cacd.eead.18145.2678
aabc.cdve.cncd.ened.19945.2345
what should I do?
回答1:
With awk:
awk 'BEGIN{FS=OFS="."} NF--' file
In case there are no empty lines, this works. It sets input and output field separators to the dot .. Then, decreases the number of fields in one, so that the last one is kept out. Then it performs the default awk action: {print $0}, that is, print the line.
With sed:
sed 's/\.[^.]*$//' file
This catches the last block of . + text + end of line and replaces it with nothing. That is, it removes it.
With rev and cut:
rev file | cut -d'.' -f2- | rev
rev reverses the line, so that cut can print from the 2nd word to the end. Then, rev back to get the correct output.
With bash:
while ISF= read -r line
do
echo "${line%.*}"
done < file
This perform a string operation consisting in replacing the shortest match of .* from the end of the variable $line content.
With grep:
grep -Po '.*(?=\.)' file
Look-ahead to print just what is before the last dot.
All of them return:
abc.cde.ccd.eed.12345.5678
abcd.cdde.ccdd.eaed.12346.5688
aabc.cade.cacd.eaed.13345.5078
abzc.cdae.ccda.eaed.29345.1678
abac.cdae.cacd.eead.18145.2678
aabc.cdve.cncd.ened.19945.2345
来源：https://stackoverflow.com/questions/26633623/remove-all-text-from-last-dot-in-bash