Question

我试图在bash中运行以下脚本：

#! /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

每一行都经过单独测试，并按照应有的方式运行，除非将它们组合成一个脚本。

第一个文件似乎没问题。接下来的4个文件是空白的。然后接下来的两个文件是好的。这在我需要运行它的550个文件中保持看似随机的间隔。

任何想法？

感谢。

Answer 1

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

你可能意味着

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

更进一步

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

其中包括那些只有空格的行。

在测试中，这个命令

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

显示

a
b

这个命令

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

什么都不显示。

bash中shell脚本（sed）的问题

1 个答案: