Question

我已经阅读了一些关于从文件名中提取部分内容的线程，但仍然无法解决我的问题。有一些名为aa - bb.txt，cc - dd.txt，ee - ff.txt等的文件。每个文件的最后一行都是这样的：

somewordbbaa

aa - bb.txt和cc - dd.txt的

是：

somewordddcc

然后在ee - ff.txt中：

somewordffee

我想编写一个shell脚本来删除相应文件最后一行中的bbaa，ddcc，ffee。我试过以下：

#!/bin/bash
for file in *.txt
do
  artist=`echo $file | awk -F'[ .]' '{print $1}'`
  name=`echo $file | awk -F'[ .]' '{print $3}'`
  echo $artist >> artist
  echo $name >> name
  sed -i "s/$name$artist//" $file
done

在我跑完之后，它扔了这个

sed: can't read aa: No such file or directory
sed: can't read -: No such file or directory
sed: can't read bb.txt: No such file or directory
sed: can't read cc: No such file or directory
sed: can't read -: No such file or directory
sed: can't read dd.txt: No such file or directory
sed: can't read ee: No such file or directory
sed: can't read -: No such file or directory
sed: can't read ff.txt: No such file or directory

我也试过这个

#!/bin/bash
ls *.txt | sed 's/\.txt//' > full_name
cut -f1 -d" " full_name > artist
cut -f3 -d" " full_name > name

for file in `ls -1 *.txt`, item1 in artist, item2 in name #Is this right?
do
  tail -n -1 $file | sed 's/$item2$item1//' > $file.last #just the last line
done

它只是显示了这个并且在按 Ctrl + c

之前没有反应

tail: cannot open `aa' for reading: No such file or directory

我认为bash将文件名的空白作为分隔符，将$file读为aa，-，bb.txt。

有人能给我一些建议吗？

Answer 1

由于您的文件名称中有空格，请尝试使用原始脚本，但更改此行：

sed -i "s/$name$artist//" $file

到此：

sed -i "s/$name$artist//" "$file"

从filename中提取部分作为变量

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