如何找到传递给 Bash 脚本的参数数量？
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How do I find the number of arguments passed to a Bash script?
提问by sabri
How do I find the number of arguments passed to a Bash script?
如何找到传递给 Bash 脚本的参数数量？
This is what I have currently:
这是我目前所拥有的：
#!/bin/bash
i=0
for var in "$@"
do
i=i+1
done
Are there other (better) ways of doing this?
是否有其他（更好的）方法可以做到这一点？
回答by zsalzbank
The number of arguments is $#
参数的数量是 $#
Search for it on this page to learn more: http://tldp.org/LDP/abs/html/internalvariables.html#ARGLIST
在此页面上搜索以了解更多信息：http: //tldp.org/LDP/abs/html/internalvariables.html#ARGLIST
回答by Paused until further notice.
#!/bin/bash
echo "The number of arguments is: $#"
a=${@}
echo "The total length of all arguments is: ${#a}: "
count=0
for var in "$@"
do
echo "The length of argument '$var' is: ${#var}"
(( count++ ))
(( accum += ${#var} ))
done
echo "The counted number of arguments is: $count"
echo "The accumulated length of all arguments is: $accum"
回答by Michael Brux
to add the original reference:
添加原始参考：
You can get the number of arguments from the special parameter $#. Value of 0 means "no arguments". $#is read-only.
您可以从特殊参数中获取参数的数量$#。值 0 表示“无参数”。$#是只读的。
When used in conjunction with shiftfor argument processing, the special parameter $#is decremented each time Bash Builtin shiftis executed.
当与shiftfor 参数处理结合使用时，$#每次shift执行Bash Builtin 时都会递减特殊参数。
see Bash Reference Manualin section 3.4.2 Special Parameters:
请参阅Bash 参考手册中的3.4.2 特殊参数部分：
"The shell treats several parameters specially. These parameters may only be referenced"
and in this section for keyword $#"Expands to the number of positional parameters in decimal."
“shell 对几个参数进行了特殊处理，这些参数只能被引用”
并在本节中为关键字$#“扩展到十进制位置参数的数量”。
回答by VIPIN KUMAR
Below is the easy one -
下面是简单的——
cat countvariable.sh
猫计数变量.sh
echo "$@" |awk '{for(i=0;i<=NF;i++); print i-1 }'
Output :
输出 ：
#./countvariable.sh 1 2 3 4 5 6
6
#./countvariable.sh 1 2 3 4 5 6 apple orange
8
回答by Jeff Gaer
that value is contained in the variable $#
该值包含在变量中 $#
