找到具有特定字符串的行，然后在 bash 中删除其末尾的换行符

Question

I'm trying to prepare a file for a report. What I have is like this:

foo 
bar bar oof 
bar oof 
foo 
bar bar

I'm trying to get an output like this:

foo bar bar oof
bar off
foo bar bar

I wanted to search for a string, in this case 'foo', and within the line where the string is found I have to remove the newline.

I did search but I can only find solutions where 'foo' is also removed. How can I do this?

Answer 1

Using awk :

awk -v search='foo' '$0 ~ search{printf $0; next}1' infile

You may use printf $0 OFS like below, if your field doesn't have leading space before newline char

awk -v search='foo' '$0 ~ search{printf $0 OFS; next}1' infile

Test Results:

$ cat infile
foo 
bar bar oof 
bar oof 
foo 
bar bar

$ awk -v search='foo' '$0 ~ search{printf $0; next}1' infile
foo bar bar oof 
bar oof 
foo bar bar

Explanation:

• -v search='foo' - set variable search
• $0 ~ search - if lines/record/row contains regexp/pattern/string mentioned in variable
• {printf $0; next} {printf $0; next} - print current record without record separator and go to next line
• }1 1 at the end does default operation that is print current record/row.

Answer 2

You can do this quite easily with sed , for example:

$ sed '/^foo$/N;s/\n/ /' file
foo bar bar oof
bar oof
foo bar bar

Explanation

1. /^foo$/ find lines containing only foo

2. N read/append next line of input into pattern space.

3. s/\\n/ / substitute the '\\n' with a space .