仅在使用sed或awk将其为空白的情况下删除模式之后的行

Question

I want to delete a blank line only if this one is after the line of my pattern using sed or awk for example if I have

G

O TO P999-ERREUR

END-IF.

the pattern in this case is G I want to have this output

 G
 O TO P999-ERREUR

 END-IF.

Answer 1

This will do the trick:

$ awk -v n=-2 'NR==n+1 && !NF{next} /G/ {n=NR}1' file
G
O TO P999-ERREUR

END-IF.

Explanation:

-v n=-2    # Set n=-2 before the script is run to avoid not printing the first line
NR == n+1  # If the current line number is equal to the matching line + 1
&& !NF     # And the line is empty 
{next}     # Skip the line (don't print it)
/G/        # The regular expression to match
{n = NR}   # Save the current line number in the variable n
1          # Truthy value used a shorthand to print every (non skipped) line

Answer 2

Using sed

sed '/GG/{N;s/\n$//}' file

If it sees GG, gets the next line, removes the newline between them if the next line is empty.

---

Note this will only remove one blank line after, and the line must be blank ie not spaces or tabs.

Answer 3

Using ex (edit in-place):

ex +'/G/j' -cwq foo.txt 

or print to the standard output (from file or stdin):

ex -s +'/GG/j|%p|q!' file_or_/dev/stdin

where:

• /GG/j - joins the next line when the pattern is found
• %p - prints the buffer
• q! - quits

For conditional checking (if there is a blank line), try:

ex -s +'%s/^\(G\)\n/\1/' +'%p|q!' file_or_/dev/stdin

Answer 4

This might work for you (GNU sed):

sed -r 'N;s/(G.*)\n\s*$/\1/;P;D' file

Keep a moving window of two lines throughout the length of the file and remove a newline (and any whitespace) if it follows the intended pattern.