Sed - 替换包含括号、逗号和未指定数量的空格的文本字符串

Question

Assume an alphanumeric text string that contains a section comprising a keyword, parentheses, and commas as well as a line break and an unspecified number of whitespaces immediately following some or all of the commas. How do I replace such a section from the text string with a simple comma in bash (preferentially using sed )?

Example:

$ cat have.txt
foo (keyword(00001..00002),keyword(00003..00004),
   keyword(00005..00006),keyword(00007..00008)) foo 
$ cat want.txt 
foo (keyword(00001..00002,00003..00004,00005..00006,00007..00008)) foo

Attempt:

$ sed 's/),keyword(/,/g' have.txt
foo (keyword(00001..00002,00003..00004),
   keyword(00005..00006,00007..00008)) foo

(And, yes, I know that whitespaces can be captured via [[:space:]] .)

Answer 1

With GNU sed:

sed -z 's/),\s*keyword(/,/g' file

With -z , you will be able to match linebreaks, the \s* will match zero or more whitespace including those linebreaks.

To actually modify the file use

sed -z -i 's/),\s*keyword(/,/g' file

Answer 2

cat >> edrep.txt << EOF

2s/   //
%s/(//g
%s/)//g
%s/keyword//g
1s/00001/\(keyword\(00001/
2s/8/8))/
1,2j
wq
EOF

ed -s file.txt < edrep.txt

I know this one has already been answered, but I felt like posting this solution anyway. People here will probably consider it hackish; but the point is that you can climb around text files like a monkey with ed, and I don't know of any other program which lets you do that to the same degree.