如何删除文本文件中第一个空格后每行中的所有字符？

Question

I have a large log file from which I need to extract file names.

The file looks like this:

/path/to/loremIpsumDolor.sit /more/text/here/notAlways/theSame/here
/path/to/anotherFile.ext /more/text/here/differentText/here
.... about 10 million times

I need to extract the file names like this:

loremIpsumDolor.sit
anotherFile.ext

I figure my first strategy is to find/replace all /path/to/ with ''. But I'm stuck how to remove all characters after the space.

Can you help?

Answer 1

sed 's/ .*//' file

It doesn't take any more. The transformed output appears on standard output, of course.

Answer 2

传递它cut ：

cut '-d ' -f1 yourfile

Answer 3

In theory, you could also use awk to grab the filename from each line as:

awk '{ print $1 }' input_file.log

That, of course, assumes that there are no spaces in any of the filenames. awk defaults to looking for whitespace as the field delimiters, so the above snippet would take the first "field" from your log file (your filename) for each line, and output it.

Answer 4

a bash-only solution:

while read path otherstuff; do
    echo ${path##*/}
done < filename