访问bash阵列而无需落后逗号

Question

这是一个简单的问题，但我找不到答案。我有一系列的IP，我想将文件分配到，并且不想每次执行单独的SCP命令。我为我设计了这个狂欢功能：

function scp_Targets () {
    loopControl=0
    declare -a targets=("200.150.100.2", "200.150.100.3", "200.150.100.4")
    arraySize=${#targets[@]}
    while [ $loopControl -lt $arraySize ]
    do
        echo "hello, loopControl is $loopControl, targetValue is ${targets[$loopControl]}"
        scp $1 root@${targets[$loopControl]}:$2
        if [ $? -eq 0 ]
        then
                echo "Transferred $1 to $2 on target at ${targets[$loopControl]}"
        fi
        ((loopControl++))
    done
}

吐出

hello, loopControl is 0, targetValue is 200.150.100.2,
ssh: Could not resolve hostname 200.150.100.2,: Name or service not known
lost connection
hello, loopControl is 1, targetValue is 200.150.100.3,
ssh: Could not resolve hostname 200.150.100.3,: Name or service not known
lost connection
hello, loopControl is 2, targetValue is 200.150.100.4
root@200.150.100.4's password: 
script.sh                                                                                                                                                              
100%  326     0.3KB/s   00:00    
Transferred script.sh to /usr/bin on target at 200.150.100.4

我想要什么

hello, loopControl is 0, targetValue is 200.150.100.2
root@200.150.100.2's password: 
script.sh                                                                                                                                                              
100%  326     0.3KB/s   00:00    
Transferred script.sh to /usr/bin on target at 200.150.100.2
... (same for the other two IPs)

表明我访问数组包括一个尾随逗号，这是我如何访问数组的副作用吗？如何从值中获得逗号？我知道我可以进行长度检查，然后删除最后一个字符，但似乎应该有更明显的方法。

Answer 1

您可以使用此简单 bash参数扩展：

来修剪所有逗号,

$ declare -a targets=("200.150.100.2", "200.150.100.3", "200.150.100.4")
$ new_targets="${targets[@]%,}"
$ printf '%sn' "${new_targets[@]}"
200.150.100.2 200.150.100.3 200.150.100.4