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fnan 写了：

ChenFengyuan 写了：to fnan,你的bash也太长了。。。。。原来是要有+1的效果。。。。
交给perl好了。写得又短又漂亮。

确是还能突破，谢谢提醒。不过能短多少呢？无需太执着吧。
ls -1|while read n;do z=${n/#*[!0-9]};mv $n ${n/%*([0-9])}$(printf ".%0${#z}d" $((${z/#*(0)}+1)));done
rename -v 's/(?<![.\d])(\d++)$/sprintf ".%0".length($1)."d",$1+1/e' *

ls
a.12mkv000103 a.mkv000 a.mkv000000 a.mkv00000000004 a.mkv00002 a.mkv002 a.mkv004
ls -1|while read n;do z=${n/#*[!0-9]};mv $n ${n/%*([0-9])}$(printf ".%0${#z}d" $((${z/#*(0)}+1)));done
ls
a.12mkv.000104 a.mkv.00000000005 a.mkv.000001 a.mkv.00003 a.mkv.001 a.mkv.003 a.mkv.005
ls -1|while read n;do z=${n/#*[!0-9]};mv $n ${n/%*([0-9])}$(printf ".%0${#z}d" $((${z/#*(0)}+1)));done
ls
a.12mkv..000105 a.mkv..00000000006 a.mkv..000002 a.mkv..00004 a.mkv..002 a.mkv..004 a.mkv..006

ls
a.12mkv000103 a.mkv000 a.mkv000000 a.mkv00000000004 a.mkv00002 a.mkv002 a.mkv004
rename -v 's/(?<![.\d])(\d++)$/sprintf ".%0".length($1)."d",$1+1/e' *
ls
a.12mkv.000104 a.mkv.00000000005 a.mkv.000001 a.mkv.00003 a.mkv.001 a.mkv.003 a.mkv.005
rename -v 's/(?<![.\d])(\d++)$/sprintf ".%0".length($1)."d",$1+1/e' *
ls
a.12mkv.000104 a.mkv.00000000005 a.mkv.000001 a.mkv.00003 a.mkv.001 a.mkv.003 a.mkv.005

多个点就不工作了，改一下吧。

我就是我2 写了：

代码： 全选

for i in a.mkv*;do mv -v $i ${i/mkv/mkv.};done

會有什麽問題嗎？或者

代码： 全选

for i in a.mkv*;do mv -v $i ${i//mkv/mkv.};done

文件名會含有兩個mkv?
或者使用變量來替換mkv來實現靈活性？