bash中的shell腳本（sed）問題

Question

我正在嘗試在bash中運行以下腳本：

#! /bin/Bash

cp '../Text_Files_Backups/'*.txt .

sed -i '1,/Ref/d' *.txt

##Deletes all lines from the begining of file up to and including the line that includes the text 'Ref'
##      

sed -i -b '/^.$/,$d' *.txt
##Deletes all blank lines and text following and including the first blank line

sed -i 's/\([(a-zA-Z) ]\)\([(1-9)][(0-9)][ ][ ]\)/\1\n\2/g' *.txt
##Searches document for any instance of a letter character immediately followed by a 2 digit number ##immediately followed by 2 blank spaces
##  <or>
##a blank space immediately followed by a 2 digit number immediately followed by 2 blank spaces
##      and inserts a new line immediately prior to the 2 digit number

exit

每行都經過了單獨的測試，並可以正常運行，除非放到腳本中。

第一個文件似乎很好。 接下來的4個文件為空。 那么接下來的兩個文件是好的。 在我需要對其運行的550個文件中，這似乎保持隨機的間隔。

有任何想法嗎？

謝謝。

Answer 1

sed -i -b '/^.$/,$d' *.txt
##Deletes all blank lines and text following and including the first blank line

你可能是說

sed -i -b '/^$/,$d' *.txt

更深入

sed -i -b '/^[[:blank:]]*$/,$d' *.txt

這將包括那些只有空格的行。

在測試中，此命令

(echo a; echo b; echo; echo b; echo; echo; echo c; echo d) | sed '/^$/,$d'

展會

a
b

當這個命令

(echo a; echo b; echo; echo b; echo; echo; echo c; echo d) | sed '/^.$/,$d'

什么也沒顯示。