linux获取文件的路径名及其文件名的方法

分类： linux 2009-11-07 09:39 1330人阅读 评论(2) 收藏 举报

1、简单测试实例

 for i in `find . -type f -name "*.c"`
do
  echo $i
  basename $i    获取*.c文件名
  dirname $i        获取*.c对应的目录名
done

2、实际应用

diff Linux源码，并将有不一样的源码整理在一起，要求：文件名以new和old区分。

(1)、简单测试

rm -rf ~/Desktop/20091107

cp_source()
{
#for i in `find . -type d -print`
for i in `find $1 -type f -name "*.s"`
do
  #echo $i
  #basename $i
  dirname $i

  mkdir -p ~/Desktop/20091107/`dirname $i`
  cp $i ~/Desktop/20091107/`dirname $i`/`basename $i`_"$2"
done
}

cd linux-0.11 && cp_source "." "new"
cd ../linux-0.11 && cp_source "." "old"

 

(2)、实际实现(以diff -Nur linux-2.6.30 linux-2.6.31.5为例)

 

rm -rf ~/Desktop/20091107

#cat linux-patch.txt | grep 'diff -Nur ' | sed 's/^.*linux-2.6.30////g'| sed 's/ linux-2.6.31.5.*$//g' > ~/Desktop/diff.txt

old_source_noprefix=`cat linux-patch.txt | grep 'diff -Nur ' | sed 's/^.*linux-2.6.30////g'| sed 's/ linux-2.6.31.5.*$//g'`
new_source_noprefix=`cat linux-patch.txt | grep 'diff -Nur ' | sed 's/^.*linux-2.6.31.5////g'`

old_source_haveprefix=`cat linux-patch.txt | grep 'diff -Nur ' | sed 's/^.*linux-2.6.30/linux-2.6.30/g'| sed 's/ linux-2.6.31.5.*$//g'`
new_source_haveprefix=`cat linux-patch.txt | grep 'diff -Nur ' | sed 's/^.*linux-2.6.31.5/linux-2.6.31.5/g'`

cp_source()
{
#for i in `find . -type d -print`
for i in $1
do
  #echo $i

  #basename $i
  #dirname $i
 
  #Used "diff -Nur", some file didn't exist!
  mkdir -p ~/Desktop/20091107/`dirname $i` && cp ./$2/$i ~/Desktop/20091107/`dirname $i`/`basename $i`_"$3" 2>/dev/null
done
}

cp_source "$old_source_noprefix" "linux-2.6.30" "old"
cp_source "$new_source_noprefix" "linux-2.6.31.5" "new"