I'm using bash and I have a file that is in 3 parts of text. The first part, then a blank line, then the 2nd part then another blank line, then the file 3 part of text. I need to output this to a new file that contains only the first 2 parts without the blank line in between. I've been playing with sed and awk, but can't quite figure it out.
我正在使用bash，我有一个文件的3个部分。第一部分，然后是空白行，然后是第二部分，然后是另一个空行，然后是文件的3部分文本。我需要将其输出到一个新文件，该文件只包含前两个部分，中间没有空白行。我一直在玩sed和awk，但无法弄明白。
1 个解决方案
#1
2
Most simply with awk:
最简单的用awk：
awk -v RS= 'NR <= 2' filename
With an empty record separator RS, awk splits the file into records at empty lines. With the selection NR <= 2, only the first two are printed (delimited by the default output record separator, which is a newline).
使用空记录分隔符RS，awk将文件拆分为空行记录。选择NR <= 2时，仅打印前两个（由默认输出记录分隔符分隔，这是一个换行符）。
If the file is very large, it might be prudent to amend this to
如果文件非常大，修改它可能是谨慎的
awk -v RS= '1; NR == 2 { exit }' filename
This stops processing the file after the second record and prints all until then.
这将停止在第二个记录之后处理文件并在此之前打印所有文件。
Addendum: Obligatory crazy sed solution (not recommended for use, written for fun):
附录：强制性的疯狂sed解决方案（不推荐使用，为了好玩而写）：
sed -n '/^$/ { x; /./q; H; d; }; p' filename
#1
2
Most simply with awk:
最简单的用awk：
awk -v RS= 'NR <= 2' filename
With an empty record separator RS, awk splits the file into records at empty lines. With the selection NR <= 2, only the first two are printed (delimited by the default output record separator, which is a newline).
使用空记录分隔符RS，awk将文件拆分为空行记录。选择NR <= 2时，仅打印前两个（由默认输出记录分隔符分隔，这是一个换行符）。
If the file is very large, it might be prudent to amend this to
如果文件非常大，修改它可能是谨慎的
awk -v RS= '1; NR == 2 { exit }' filename
This stops processing the file after the second record and prints all until then.
这将停止在第二个记录之后处理文件并在此之前打印所有文件。
Addendum: Obligatory crazy sed solution (not recommended for use, written for fun):
附录：强制性的疯狂sed解决方案（不推荐使用，为了好玩而写）：
sed -n '/^$/ { x; /./q; H; d; }; p' filename