我有一个备份脚本(bash)。下面显示了其中的一部分。此脚本每 14 天轮换一次备份。如果我想将其更改为 30 天,我必须写出 30 个这样的 if-then 块。我确信这可以用一个漂亮的 for 循环来代替。它会是什么?
# step 1: delete the oldest snapshot, if it exists:
if [ -d $BACKUP_DIR/daily.14 ] ; then \
$RM -rf $BACKUP_DIR/daily.14 ; \
fi ;
# step 2: shift the middle snapshot(s) back by one, if they exist
if [ -d $BACKUP_DIR/daily.13 ] ; then \
$MV $BACKUP_DIR/daily.13 $BACKUP_DIR/daily.14 ; \
fi;
if [ -d $BACKUP_DIR/daily.12 ] ; then \
$MV $BACKUP_DIR/daily.12 $BACKUP_DIR/daily.13 ; \
fi;
if [ -d $BACKUP_DIR/daily.11 ] ; then \
$MV $BACKUP_DIR/daily.11 $BACKUP_DIR/daily.12 ; \
fi;
if [ -d $BACKUP_DIR/daily.10 ] ; then \
$MV $BACKUP_DIR/daily.10 $BACKUP_DIR/daily.11 ; \
fi;
if [ -d $BACKUP_DIR/daily.9 ] ; then \
$MV $BACKUP_DIR/daily.9 $BACKUP_DIR/daily.10 ; \
fi;
if [ -d $BACKUP_DIR/daily.8 ] ; then \
$MV $BACKUP_DIR/daily.8 $BACKUP_DIR/daily.9 ; \
fi;
if [ -d $BACKUP_DIR/daily.7 ] ; then \
$MV $BACKUP_DIR/daily.7 $BACKUP_DIR/daily.8 ; \
fi;
if [ -d $BACKUP_DIR/daily.6 ] ; then \
$MV $BACKUP_DIR/daily.6 $BACKUP_DIR/daily.7 ; \
fi;
if [ -d $BACKUP_DIR/daily.5 ] ; then \
$MV $BACKUP_DIR/daily.5 $BACKUP_DIR/daily.6 ; \
fi;
if [ -d $BACKUP_DIR/daily.4 ] ; then \
$MV $BACKUP_DIR/daily.4 $BACKUP_DIR/daily.5 ; \
fi;
if [ -d $BACKUP_DIR/daily.3 ] ; then \
$MV $BACKUP_DIR/daily.3 $BACKUP_DIR/daily.4 ; \
fi;
if [ -d $BACKUP_DIR/daily.2 ] ; then \
$MV $BACKUP_DIR/daily.2 $BACKUP_DIR/daily.3 ; \
fi;
if [ -d $BACKUP_DIR/daily.1 ] ; then \
$MV $BACKUP_DIR/daily.1 $BACKUP_DIR/daily.2 ; \
fi;
# step 3: make a hard-link-only (except for dirs) copy of the latest snapshot, if that exists
if [ -d $BACKUP_DIR/daily.0 ] ; then \
$CP -al $BACKUP_DIR/daily.0 $BACKUP_DIR/daily.1 ; \
fi;
答案1
#!/bin/bash
MAX=29
for i in `seq 1 $MAX`
do
echo " check file exists directory.$i"
echo "mv directory.$i to directory" `expr $i - 1`
done
这可能是一个很好的起点
答案2
for i in {7..1};
do
#echo "$i"
if [ -d $BACKUP_DIR/daily.${i} ]
then
z=$(($i+1));
#echo "$z"
echo "moving daily.$i to daily.$z"
$MV $BACKUP_DIR/daily.$i $BACKUP_DIR/daily.$z ;
fi;
done