使用 sed 从文件中的匹配行中提取几个匹配的字符串

Question

I have a file with strings similar to this:

abcd u'current_count': u'2', u'total_count': u'3', u'order_id': u'90'

I have to find current_count and total_count for each line of file. I am trying below command but its not working. Please help.

grep current_count file | sed "s/.*\('current_count': u'\d+'\).*/\1/"

It is outputting the whole line but I want something like this:

'current_count': u'3', 'total_count': u'3'

Answer 1

It's printing the whole line because the pattern in the s command doesn't match, so no substitution happens.

sed regexes don't support \\d for digits, or x+ for xx* . GNU sed has a -r option to enable extended-regex support so + will be a meta-character, but \\d still doesn't work. GNU sed also allows \\+ as a meta-character in basic regex mode, but that's not POSIX standard.

So anyway, this will work:

echo -e "foo\nabcd u'current_count': u'2', u'total_count': u'3', u'order_id': u'90'" |
sed -nr "s/.*('current_count': u'[0-9]+').*/\1/p"
# output:  'current_count': u'2'

Notice that I skip the grep by using sed -ns///p . I could also have used /current_count/ as an address:

sed  -r -e '/current_count/!d' -e "s/.*('current_count': u'[0-9]+').*/\1/"

Or with just grep printing only the matching part of the pattern, instead of the whole line:

grep -E -o "'current_count': u'[[:digit:]]+'

(or egrep instead of grep -E). I forget if grep -o is POSIX-required behaviour.

Answer 2

For me this looks like some sort of serialized Python data. Basically I would try to find out the origin of that data and parse it properly.

However, while being hackish, sed can also being used here:

sed "s/.*current_count': [a-z]'\([0-9]\+\).*/\1/" input.txt
sed "s/.*total_count': [a-z]'\([0-9]\+\).*/\1/" input.txt