bash 变量中的令牌数
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number of tokens in bash variable
提问by flybywire
how can I know the number of tokens in a bash variable (whitespace-separated tokens) - or at least, wether it is one or there are more.
我如何知道 bash 变量中的标记数（以空格分隔的标记）——或者至少，它是一个还是更多。
回答by guns
The $# expansion will tell you the number of elements in a variable / array. If you're working with a bash version greater than 2.05 or so you can:
$# 扩展将告诉您变量/数组中的元素数。如果您使用的 bash 版本高于 2.05 左右，您可以：
VAR='some string with words'
VAR=( $VAR )
echo ${#VAR[@]}
This effectively splits the string into an array along whitespace (which is the default delimiter), and then counts the members of the array.
这有效地将字符串沿空格（这是默认分隔符）拆分为一个数组，然后计算数组的成员。
EDIT:
编辑：
Of course, this recasts the variable as an array. If you don't want that, use a different variable name or recast the variable back into a string:
当然，这会将变量重新转换为数组。如果您不想这样，请使用不同的变量名称或将变量重新转换为字符串：
VAR="${VAR[*]}"
回答by Jo So
I can't understand why people are using those overcomplicated bashisms all the time. There's almost always a straight-forward, no-bashism solution.
我不明白为什么人们一直在使用那些过于复杂的 bashism。几乎总是有一个直截了当的、无偏见的解决方案。
howmany() { echo $#; }
myvar="I am your var"
howmany $myvar
This uses the tokenizer built-in to the shell, so there's no discrepancy.
这使用 shell 内置的标记器，因此没有差异。
Here's one related gotcha:
这是一个相关的问题：
myvar='*'
echo $myvar
echo "$myvar"
set -f
echo $myvar
echo "$myvar"
Note that the solution from @guns using bash array has the same gotcha.
请注意，@guns 使用 bash 数组的解决方案具有相同的问题。
The following is a (supposedly) super-robust version to work around the gotcha:
以下是一个（据说）超级强大的版本，可以解决这个问题：
howmany() ( set -f; set -- ; echo $# )
If we want to avoid the subshell, things start to get ugly
如果我们想避免使用子外壳，事情就会开始变得丑陋
howmany() {
case $- in *f*) set -- ;; *) set -f; set -- ; set +f;; esac
echo $#
}
These two must be used WITH quotes, e.g. howmany "one two three"returns 3
这两个必须使用 WITH 引号，例如howmany "one two three"返回3
回答by Mykola Golubyev
set VAR='hello world'
echo $VAR | wc -w
here is how you can check.
这是您可以检查的方法。
if [ `echo $VAR | wc -w` -gt 1 ]
then
echo "Hello"
fi
回答by ata
To count:
计算：
sentence="This is a sentence, please count the words in me."
words="${sentence//[^\ ]} "
echo ${#words}
To check:
去检查：
sentence1="Two words"
sentence2="One"
[[ "$sentence1" =~ [\ ] ]] && echo "sentence1 has more than one word"
[[ "$sentence2" =~ [\ ] ]] && echo "sentence2 has more than one word"
回答by mklement0
For a robust, portableshsolution, see @JoSo's functionsusing set -f.
有关强大、可移植的sh解决方案，请参阅@JoSo 的使用set -f.
(Simple bash-only solution for answering (only) the "Is there at least 1 whitespace?" question; note: will also match leading and trailing whitespace, unlike the awksolution below:
（用于回答（仅）“是否有至少 1 个空格？”问题的简单 bash-only 解决方案；注意：也将匹配前导和尾随空格，与以下awk解决方案不同：
[[ $v =~ [[:space:]] ]] && echo "$v has at least 1 whitespace char."
)
)
Here's a robust awk-based bash solution (less efficient due to invocation of an external utility, but probably won't matter in many real-world scenarios):
这是一个awk基于强大的 bash 解决方案（由于调用外部实用程序而效率较低，但在许多实际场景中可能无关紧要）：
# Functions - pass in a quoted variable reference as the only argument.
# Takes advantage of `awk` splitting each input line into individual tokens by
# whitespace; `NF` represents the number of tokens.
# `-v RS=$''` ensures that even multiline input is treated as a single input
# string.
countTokens() { awk -v RS=$'' '{print NF}' <<<""; }
hasMultipleTokens() { awk -v RS=$'' '{if(NF>1) ec=0; else ec=1; exit ec}' <<<""; }
# Example: Note the use of glob `*` to demonstrate that it is not
# accidentally expanded.
v='I am *'
echo "$v has $(countTokens "$v") token(s)."
if hasMultipleTokens "$v"; then
echo "$v has multiple tokens."
else
echo "$v has just 1 token."
fi
回答by Eugene Morozov
Simple method:
简单方法：
$ VAR="a b c d"
$ set $VAR
$ echo $#
4
