ftok函数的shell脚本实现
2007-11-29 22:53 来源: sss0213.cublog.cn 作者:sss0213 网友评论 0 条 浏览次数 36
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-
- let key=0
- function ftok() {
- pathname=$1;
- proj_id=$2;
-
- str_st_ino=`stat --format='%i' "${pathname}" 2>/dev/null`;
- str_st_dev=`stat --format='%d' "${pathname}" 2>/dev/null`;
- if [ "x${str_st_ino}" = "x" -o "x${str_st_dev}" = "x" ] ; then
- return 1;
- fi
-
- let st_ino=${str_st_ino}
- let st_dev=${str_st_dev}
-
-
- let key1=${st_ino}\&16
- let key2=${st_dev}\&16
- let key2=${key2}\<\<16
- let key3=${proj_id}\&16
- let key3=${key3}\<\<24
- let key=${key1}\|${key2}
- let key=${key}\|${key3}
- }
-
- function echohelp(){
- echo "ftok generator"
- echo "Usage:ftok pathname projid"
- exit 5
- }
-
- if [ $
- echohelp
- fi
-
- sPathName=$1
- let nProjectID=$2
-
- if [ "${sPathName:0:1}" != "/" ] ; then
- sPathName=${PWD}/${sPathName}
- fi
-
- if ! test -f ${sPathName} ; then
- echo "No File Found![${sPathName}]"
- exit 4
- fi
-
- ftok "${sPathName}" "${nProjectID}"
- echo ${key}
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