Bash 创建变量然后为其赋值
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Bash create variable then assign value to it
提问by mike
For this problem I have two values, curdirand curlevel, which change throughout my script. I want to know if it's possible to create a variable and then use that value as the name for another value. For example
对于这个问题，我有两个值curdir和 curlevel，它们在我的脚本中都发生了变化。我想知道是否可以创建一个变量，然后将该值用作另一个值的名称。例如
temp="dir_${curdir}_${curlevel}"
$temp=$name_of_directory **<----Is there a legitimate way to do this?**
so if initially curdir=1and curlevel=0then
$(temp)=directory_oneis equal to
因此，如果最初curdir=1和curlevel=0然后
$(temp)=directory_one是等于
dir_1_0=directory_one
then later if curdir=2and curlevel=4, I can reset temp and then have
然后稍后如果curdir=2和curlevel=4，我可以重置温度然后有
$(temp)=another_directory
is the same as
是相同的
dir_2_4=another_directory
so I could make a call such as
所以我可以拨打电话，例如
cd $(temp)
which will move me to different directories when I need to
当我需要时，这会将我移动到不同的目录
采纳答案by Brian
I think what you want is to use eval. Like so:
我认为你想要的是使用eval. 像这样：
$ foo=bar
$ bar=baz
$ eval qux=$$foo
$ echo $qux
baz
So what you could do is something like
所以你可以做的是
eval temp=$$"dir_${curdir}_${curlevel}"
cd $temp
回答by Jonathan Leffler
The trick for this is to use eval- several times.
诀窍是使用eval- 多次。
curdir=1
curlevel=0
temp='dir_${curdir}_${curlevel}' # Note single quotes!
x=$(eval echo $temp)
eval $x=$PWD
cd /tmp
curdir=2
curlevel=4
x=$(eval echo $temp)
eval $x=$PWD
echo $dir_1_0
echo $dir_2_4
The output of sh -x script:
的输出sh -x script：
+ curdir=1
+ curlevel=0
+ temp='dir_${curdir}_${curlevel}'
++ eval echo 'dir_${curdir}_${curlevel}'
+++ echo dir_1_0
+ x=dir_1_0
+ eval dir_1_0=/Users/jleffler/tmp/soq
++ dir_1_0=/Users/jleffler/tmp/soq
+ cd /tmp
+ curdir=2
+ curlevel=4
++ eval echo 'dir_${curdir}_${curlevel}'
+++ echo dir_2_4
+ x=dir_2_4
+ eval dir_2_4=/tmp
++ dir_2_4=/tmp
+ echo /Users/jleffler/tmp/soq
/Users/jleffler/tmp/soq
+ echo /tmp
/tmp
The output of sh script:
的输出sh script：
/Users/jleffler/tmp/soq
/tmp
Converted to a function:
转换为函数：
change_dir()
{
temp='dir_${curdir}_${curlevel}' # Note single quotes!
x=$(eval echo $temp)
eval $x=$PWD
cd
}
curdir=1
curlevel=0
change_dir /tmp
curdir=2
curlevel=4
change_dir $HOME
echo $dir_1_0
echo $dir_2_4
pwd
Output:
输出：
/Users/jleffler/tmp/soq
/tmp
/Users/jleffler
The recorded names are the names of the directory being left, not the one you arrive at.
记录的名称是留下的目录的名称，而不是您到达的名称。
回答by Paused until further notice.
The secure way to do this is to use indirection, associative arrays (Bash 4), functions or declare:
这样做的安全方法是使用间接、关联数组（Bash 4）、函数或declare：
Use declare:
使用declare：
declare $temp=$name_of_directory
Use indirection:
使用间接：
bar=42
foo=bar
echo ${!foo}
IFS= read -r $foo <<< 101
echo ${!foo}
Please take note of the security implications of eval.
请大家注意的的安全隐患eval。
