问题
I want to count the number of matches there is on one single line (or all lines as there always will be only one line).
I want to count not just one match per line as in
echo "123 123 123" | grep -c -E "123" # Result: 1
Better example:
echo "1 1 2 2 2 5" | grep -c -E '([^ ])( \1){1}' # Result: 1, expected: 2 or 3
回答1:
You could use grep -o then pipe through wc -l:
$ echo "123 123 123" | grep -o 123 | wc -l
3
回答2:
Maybe you should convert spaces to newlines first:
$ echo "1 1 2 2 2 5" | tr ' ' $'\n' | grep -c 2
3
回答3:
Maybe below:
echo "123 123 123" | sed "s/123 /123\n/g" | wc -l
( maybe ugly, but my bash fu is not that great )
回答4:
Why not use awk?
You could use awk '{print gsub(your_regex,"&")}'
to print the number of matches on each line, or
awk '{c+=gsub(your_regex,"&")}END{print c}'
to print the total number of matches. Note that relative speed may vary depending on which awk implementation is used, and which input is given.
回答5:
This might work for you:
sed -n -e ':a' -e 's/123//p' -e 'ta' file | sed -n '$='
GNU sed could be written:
sed -n ':;s/123//p;t' file | sed -n '$='
来源：https://stackoverflow.com/questions/6181324/counting-regex-pattern-matches-in-one-line-using-sed-or-grep