从 bash 脚本运行时出现 Rsync 语法错误
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Rsync syntax error when run from bash script
提问by KroniK907
I have been working on a backup script that uses rsync to do an incremental backup.
我一直在研究使用 rsync 进行增量备份的备份脚本。
I have tested the following rsync command manually, and it runs and completes a backup without error:
我手动测试了以下 rsync 命令，它运行并完成备份没有错误：
rsync -aAXv --delete --progress --link-dest=/backup/Uyuk/Uyuk-backup-part1/2014-02-24/ /mnt/backup/ /backup/Uyuk/Uyuk-backup-part1/2014-02-25/
however when I run that same command in my backup script it gives me the following error:
但是，当我在备份脚本中运行相同的命令时，它给了我以下错误：
rsync: -aAXv --delete --progress --link-dest=/backup/Uyuk/Uyuk-backup-part1/2014-02-24/ /mnt/backup/ /backup/Uyuk/Uyuk-backup-part1/2014-02-25/: unknown option
rsync error: syntax or usage error (code 1) at main.c(1422) [client=3.0.6]
I ran bash -x on my script to figure out exactly what is sent to the console and here is what was printed:
我在我的脚本上运行 bash -x 以确定发送到控制台的确切内容，这是打印的内容：
+ rsync '-aAXv --delete --progress --link-dest=/backup/Uyuk/Uyuk-backup-part1/2014-02-24/ /mnt/backup/ /backup/Uyuk/Uyuk-backup-part1/2014-02-25/'
Does anyone see what is wrong? I cant find anything that would cause the syntax error.
有没有人看到有什么问题？我找不到任何会导致语法错误的东西。
EDIT: Here is the actual code I have in the script, and this is a pretty large script so yes some variables are not defined here, but you get the idea.
编辑：这是我在脚本中的实际代码，这是一个非常大的脚本，所以是的，这里没有定义一些变量，但你明白了。
mkdir -p "/backup/$HOST/$NAME/$TODAY"
#source directory
SRC="$MNT"
#link directory
LNK="/backup/$HOST/$NAME/$LAST/"
#target directory
TRG="/backup/$HOST/$NAME/$TODAY/"
#rsync options
OPT1="-aAXv --delete --progress --link-dest=$LNK"
#run the rsync command
echo "rsync $OPT1 $SRC $TRG"
rsync "$OPT1 $SRC $TRG" > /var/log/backup/backup.rsync.log 2>&1
回答by chepner
You are passing your option list as a single argument, when it needs to be passed as a listof arguments. In general, you should use an array in bashto hold your arguments, in case any of them contain whitespace. Try the following:
你是通过你的选项列表作为一个参数，当需要作为传递它列表的参数。通常，您应该使用数组 inbash来保存您的参数，以防它们中的任何一个包含空格。请尝试以下操作：
mkdir -p "/backup/$HOST/$NAME/$TODAY"
#source directory
SRC="$MNT"
#link directory
LNK="/backup/$HOST/$NAME/$LAST/"
#target directory
TRG="/backup/$HOST/$NAME/$TODAY/"
#rsync options
OPTS=( "-aAXv" "--delete" "--progress" "--link-dest=$LNK" )
#run the rsync command
echo "rsync $OPT1 $SRC $TRG"
rsync "${OPTS[@]}" "$SRC" "$TRG" > /var/log/backup/backup.rsync.log 2>&1
An array expansion ${OPTS[@]}, when quoted, is treated specially as a sequence of arguments, each of which is quoted individually to preserve any whitespace or special characters in the individual elements. If arr=("a b" c d), then echo "${arr[@]}"is the same as
${OPTS[@]}引用时，数组扩展被特别视为一系列参数，每个参数都单独引用以保留单个元素中的任何空格或特殊字符。如果arr=("a b" c d)，则echo "${arr[@]}"与
echo "a b" "c" "d"
rather than
而不是
echo "a b c d"
This will not work in a shell that doesn't support arrays, but then, arrays were invented because there wasn't a safe way (that is, without using eval) to handle this use case without them.
这在不支持数组的 shell 中不起作用，但是后来发明了数组，因为没有安全的方法（即，不使用eval）来处理没有它们的用例。
回答by izilotti
The approach suggested by @chepner didn't work on my Mac OS X (10.9.4), but evaldid.
@chepner 建议的方法不适用于我的 Mac OS X (10.9.4)，但适用eval。
eval rsync "$OPT1 $SRC $TRG"
回答by twalberg
This:
这个：
rsync "$OPT1 $SRC $TRG"
passes all your intended arguments lumped together as one argument, which rsyncdoesn't know how to deal with.
将所有您想要的参数作为一个参数一起传递，它rsync不知道如何处理。
Try this instead:
试试这个：
rsync ${OPT1} ${SRC} ${TRG}
