bash 使用 VBoxManage 获取正在运行的虚拟机列表
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Get a list of running VMs using VBoxManage
提问by capdragon
I want to loop through my running VM's and return only what is between quotes.
我想遍历我正在运行的 VM 并仅返回引号之间的内容。
So this command:
所以这个命令：
VBoxManage list runningvms
returns:
返回：
"UbuntuServer" {7ef01f8d-a7d5-4405-af42-94d85f999dff}
And I only want it to return:
我只希望它返回：
UbuntuServer
This is what i have so far (fail):
这是我到目前为止（失败）：
#!/bin/bash
for machine in `cat VBoxManage list runningvms`; do
echo "$machine"
done
exit
回答by Mat
Warning: all of this is is risky if your VM names have shell glob characters in them, or contain spaces.
警告：如果您的 VM 名称中包含 shell glob 字符或包含空格，则所有这些都是有风险的。
You can do something like this if there is only one running VM:
如果只有一个正在运行的 VM，您可以执行以下操作：
read machine stuff <<< $(VBoxManage list runningvms)
echo "$machine"
Alternative with bash arrays (same condition):
bash 数组的替代方案（相同条件）：
vbm=($(VBoxManage list runningvms))
echo "${vbm[0]}"
If that program returns more than one line, a more classic approach would be:
如果该程序返回多行，则更经典的方法是：
for machine in $(VBoxManage list runningvms|cut -d" " -f 1); do
echo "$machine"
done
回答by jfg956
VBoxManage list runningvms | cut -d '"' -f 2 | while read machine; do
echo "$machine"
done
回答by Omriko
for one-liner fans:
对于单线风扇：
VBoxManage list runningvms | cut -d" " -f 1 | grep -oP "(?<=\").*(?=\")"
回答by Charles Duffy
To validate each line as you read it, the safe way to do this is to write a regular expression and use BASH_REMATCHto extract match groups from it.
要在阅读时验证每一行，执行此操作的安全方法是编写一个正则表达式并用于BASH_REMATCH从中提取匹配组。
With the following code:
使用以下代码：
re='^"(.*)" [{]([0-9a-f-]+)[}]$'
while read -r line; do
if [[ $line =~ $re ]]; then
name=${BASH_REMATCH[1]}; uuid=${BASH_REMATCH[2]}
echo "Found VM with name $name and uuid $uuid" >&2
else
echo "ERROR: Could not parse line: $line" >&2
fi
done < <(VBoxManage list runningvms)
...and the following mock implementation of VBoxManage (to allow folks without VirtualBox to reproduce the test):
...以及 VBoxManage 的以下模拟实现（允许没有 VirtualBox 的人重现测试）：
VBoxManage() { printf '%s\n' '"UbuntuServer" {7ef01f8d-a7d5-4405-af42-94d85f999dff}'; }
...output is as follows:
...输出如下：
Found VM with name UbuntuServer and uuid 7ef01f8d-a7d5-4405-af42-94d85f999dff
Note advantages of this approach:
请注意这种方法的优点：
- It doesn't make unfounded assumptions, such as excluding virtual machines with whitespace or quote literals in their names from support.
- It detects any line that doesn't match the expected pattern, rather than behaving in an unanticipated way in the presence of such values.
- It still behaves correctly with data that doesmatch the pattern but has unanticipated values. (For instance, a virtual machine named
*won't have that name silently replaced with the name of a file in the current directory). - It doesn't involve tools external to the shell such as
sed,cut, &c., but relies purely on shell-builtin functionality -- see BashFAQ #1documenting the use ofwhile read, and the bash-hackers' wiki on regular expression matchingdocumenting[[ $string =~ $re ]].
- 它不会做出毫无根据的假设，例如从支持中排除名称中包含空格或引号文字的虚拟机。
- 它检测任何与预期模式不匹配的行，而不是在存在此类值的情况下以意外方式运行。
- 它仍然与数据正确的行为是不匹配的模式，但有意外的值。（例如，一个名为的虚拟机
*不会用当前目录中的文件名静默替换该名称）。 - 它不涉及 shell 外部的工具，例如
sed,cut, &c.，而是纯粹依赖于 shell 内置功能——参见BashFAQ #1记录 的使用while read，以及bash-hackers' wiki 上的正则表达式匹配记录[[ $string =~ $re ]]。
回答by Shizzmo
VBoxManage list runningvms | sed 's/"//g;s/ .*//'
To loop through:
循环：
for machine in `VBoxManage list runningvms | sed 's/"//g;s/ .*//'` ; do
echo $machine
done
This will break if your machine has spaces in its name.
如果您的机器名称中有空格，这将中断。
