shell 编程 取文件绝对路径

wwwjy2000

hansfly

pwd | sed 's/\/[^\/]*$//'

apsong

原帖由 leyyer 于 2007-8-28 18:12 发表
bash ?

a=/home/me/a.out
echo ${a%/*}

强哦……找到了相关解释(man bash)，顶一下:

       ${parameter#word}
       ${parameter##word}
              The  word  is  expanded to produce a pattern just as in pathname
              expansion.  If the pattern matches the beginning of the value of
              parameter,  then  the  result  of  the expansion is the expanded
              value of parameter with the shortest matching pattern (the ``#''
              case) or the longest matching pattern (the ``##'' case) deleted.
              If parameter is @ or *, the pattern removal operation is applied
              to  each  positional parameter in turn, and the expansion is the
              resultant list.  If parameter is an array  variable  subscripted
              with  @  or  *, the pattern removal operation is applied to each
              member of the array in turn, and the expansion is the  resultant
              list.

       ${parameter%word}
       ${parameter%%word}
              The  word  is  expanded to produce a pattern just as in pathname
              expansion.  If the pattern matches a  trailing  portion  of  the
              expanded value of parameter, then the result of the expansion is
              the expanded value of parameter with the shortest matching  pat-
              tern  (the  ``%''  case)  or  the  longest matching pattern (the
              ``%%'' case) deleted.  If parameter  is  @  or  *,  the  pattern
              removal  operation  is  applied  to each positional parameter in
              turn, and the expansion is the resultant list.  If parameter  is
              an  array  variable subscripted with @ or *, the pattern removal
              operation is applied to each member of the array  in  turn,  and
              the expansion is the resultant list.

       ${parameter/pattern/string}
       ${parameter//pattern/string}
              The pattern is expanded to produce a pattern just as in pathname
              expansion.  Parameter is expanded and the longest match of  pat-
              tern  against  its  value is replaced with string.  In the first
              form, only the first match is replaced.  The second form  causes
              all  matches  of pattern to be replaced with string.  If pattern
              begins with #, it must match at the beginning  of  the  expanded
              value  of parameter.  If pattern begins with %, it must match at
              the end of the expanded value of parameter.  If string is  null,
              matches  of  pattern are deleted and the / following pattern may
              be omitted.  If parameter is @ or *, the substitution  operation
              is  applied to each positional parameter in turn, and the expan-
              sion is the resultant list.  If parameter is an  array  variable
              subscripted  with  @ or *, the substitution operation is applied
              to each member of the array in turn, and the  expansion  is  the
              resultant list.

guyulengyi

zwylinux

很简单的问题嘛，而且楼上的几个兄弟已经给了几个方法了

jarhead1

path="$(pwd -P)"
path="$(echo "${path%/*}")"
#弱引用是为了应付含有空格的目录的情况

[ 本帖最后由 jarhead1 于 2007-8-29 12:52 编辑 ]

codekiller

没有dirname的情况下，2楼的答案是最佳答案。不同意的请举手。

fogsky

path1=/a/b/c
path2=`echo $path1 | awk '{split($0,array,"/");print "/"array[2]"/"array[3]}'`
echo $path2

[ 本帖最后由 fogsky 于 2007-8-29 14:38 编辑 ]

honckly

还是2楼比较正 解。

coffee_45

dirs  可以，与pwd类似。