${var}、“$var"和“${var}"之间有什么区别?在 Bash shell 中? [英] What is the difference between ${var}, "$var", and "${var}" in the Bash shell?
问题描述
标题所说的:将变量封装在 {}、"" 或 "{}" 中是什么意思?我无法在网上找到任何关于此的解释 - 除了使用符号外，我无法参考它们，这不会产生任何结果.
这是一个例子:
声明 -a 组groups+=("CN=exampleexample,OU=exampleexample,OU=exampleexample,DC=example,DC=com")groups+=("CN=示例，OU=示例，OU=示例，DC=示例，DC=com")这个:
for group in "${groups[@]}";做回声$组完毕事实证明与此大不相同:
for group in $groups;做回声$组完毕还有这个:
for group in ${groups};做回声$组完毕只有第一个完成了我想要的:遍历数组中的每个元素.我不太清楚 $groups、"$groups"、${groups} 和 "${ 之间的区别组}".如果有人能解释一下，我将不胜感激.
作为一个额外的问题 - 有没有人知道引用这些封装的公认方式?
Braces ($var vs. ${var})
在大多数情况下，$var 和 ${var} 是相同的:
var=foo回声 $var# 富回声 ${var}# 富大括号仅用于解决表达式中的歧义:
var=foo回声 $varbar# 不打印任何内容，因为没有变量 'varbar'回声 ${var}bar# foobar引用($var vs. "$var" vs. "${var}")
当您在变量周围添加双引号时，您会告诉 shell 将其视为单个单词，即使它包含空格:
var="foo bar"因为我在$var"中；do # 扩展为 'for i in "foo bar";做...'echo $i # 所以只运行一次循环完毕# foo 栏将该行为与以下内容进行对比:
var="foo bar"因为我在 $var 中；do # 扩展为 'for i in foo bar;做...'echo $i # 所以循环运行两次，每个参数一次完毕# 富# 酒吧与 $var 和 ${var} 一样，大括号仅用于消除歧义，例如:
var="foo bar"因为我在$varbar"中；do # 扩展为 'for i in "";做...'因为没有echo $i # 变量名为'varbar'，所以循环运行一次done # 不打印任何内容(实际上是 "")var="foo 酒吧"对于我在${var}bar"中；do # 在 "foo barbar" 中扩展为 'for i;做...'echo $i # 所以运行一次循环完毕# foo 酒吧注意上面第二个例子中的 "${var}bar" 也可以写成 "${var}"bar，在这种情况下你不需要不再需要大括号，即 "$var"bar.但是，如果您的字符串中有很多引号，这些替代形式可能会变得难以阅读(因此难以维护).此页面很好地介绍了 Bash 中的引用.>
数组($var vs. $var[@] vs. ${var[@]})
现在为您的阵列.根据bash手册:
<块引用>引用一个没有下标的数组变量相当于引用一个下标为0的数组.
换句话说，如果你没有用 [] 提供索引，你会得到数组的第一个元素:
foo=(a b c)回声 $foo# 一种完全一样
foo=(a b c)回声 ${foo}# 一种要获取数组的所有元素，您需要使用 @ 作为索引，例如${foo[@]}.数组需要大括号，因为没有它们，shell 将首先扩展 $foo 部分，给出数组的第一个元素，后跟文字 [@]:
foo=(a b c)回声 ${foo[@]}# a b c回声 $foo[@]# 一种[@]本页很好地介绍了 Bash 中的数组.
引用重新访问(${foo[@]} vs. "${foo[@]}")
你没有问这个问题，但这是一个很好的细微差别，了解一下.如果数组中的元素可能包含空格，则需要使用双引号，以便将每个元素视为单独的单词:"
foo=(第一个"第二个")对于我在${foo[@]}"中；do # 在第一个"第二个"中扩展为 'for i;做...'echo $i # 所以循环运行两次完毕# 首先# 第二将此与没有双引号的行为进行对比:
foo=(第一个"第二个")对于我在 ${foo[@]};do # 在第一个第二个中扩展为 'for i;做...'echo $i # 所以循环运行四次！完毕# 这# 第一的# 这# 第二What the title says: what does it mean to encapsulate a variable in {}, "", or "{}"? I haven't been able to find any explanations online about this - I haven't been able to refer to them except for using the symbols, which doesn't yield anything.
Here's an example:
declare -a groups
groups+=("CN=exampleexample,OU=exampleexample,OU=exampleexample,DC=example,DC=com")
groups+=("CN=example example,OU=example example,OU=example example,DC=example,DC=com")
This:
for group in "${groups[@]}"; do
echo $group
done
Proves to be much different than this:
for group in $groups; do
echo $group
done
and this:
for group in ${groups}; do
echo $group
done
Only the first one accomplishes what I want: to iterate through each element in the array. I'm not really clear on the differences between $groups, "$groups", ${groups} and "${groups}". If anyone could explain it, I would appreciate it.
As an extra question - does anyone know the accepted way to refer to these encapsulations?
Braces ($var vs. ${var})
In most cases, $var and ${var} are the same:
var=foo
echo $var
# foo
echo ${var}
# foo
The braces are only needed to resolve ambiguity in expressions:
var=foo
echo $varbar
# Prints nothing because there is no variable 'varbar'
echo ${var}bar
# foobar
Quotes ($var vs. "$var" vs. "${var}")
When you add double quotes around a variable, you tell the shell to treat it as a single word, even if it contains whitespaces:
var="foo bar"
for i in "$var"; do # Expands to 'for i in "foo bar"; do...'
echo $i # so only runs the loop once
done
# foo bar
Contrast that behavior with the following:
var="foo bar"
for i in $var; do # Expands to 'for i in foo bar; do...'
echo $i # so runs the loop twice, once for each argument
done
# foo
# bar
As with $var vs. ${var}, the braces are only needed for disambiguation, for example:
var="foo bar"
for i in "$varbar"; do # Expands to 'for i in ""; do...' since there is no
echo $i # variable named 'varbar', so loop runs once and
done # prints nothing (actually "")
var="foo bar"
for i in "${var}bar"; do # Expands to 'for i in "foo barbar"; do...'
echo $i # so runs the loop once
done
# foo barbar
Note that "${var}bar" in the second example above could also be written "${var}"bar, in which case you don't need the braces anymore, i.e. "$var"bar. However, if you have a lot of quotes in your string these alternative forms can get hard to read (and therefore hard to maintain). This page provides a good introduction to quoting in Bash.
Arrays ($var vs. $var[@] vs. ${var[@]})
Now for your array. According to the bash manual:
Referencing an array variable without a subscript is equivalent to referencing the array with a subscript of 0.
In other words, if you don't supply an index with [], you get the first element of the array:
foo=(a b c)
echo $foo
# a
Which is exactly the same as
foo=(a b c)
echo ${foo}
# a
To get all the elements of an array, you need to use @ as the index, e.g. ${foo[@]}. The braces are required with arrays because without them, the shell would expand the $foo part first, giving the first element of the array followed by a literal [@]:
foo=(a b c)
echo ${foo[@]}
# a b c
echo $foo[@]
# a[@]
This page is a good introduction to arrays in Bash.
Quotes revisited (${foo[@]} vs. "${foo[@]}")
You didn't ask about this but it's a subtle difference that's good to know about. If the elements in your array could contain whitespace, you need to use double quotes so that each element is treated as a separate "word:"
foo=("the first" "the second")
for i in "${foo[@]}"; do # Expands to 'for i in "the first" "the second"; do...'
echo $i # so the loop runs twice
done
# the first
# the second
Contrast this with the behavior without double quotes:
foo=("the first" "the second")
for i in ${foo[@]}; do # Expands to 'for i in the first the second; do...'
echo $i # so the loop runs four times!
done
# the
# first
# the
# second
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