如何在shell中找到数组的长度？

Question

如何在unix shell中找到数组长度？

Answer 1

$ a=(1 2 3 4)
$ echo ${#a[@]}
4

Answer 2

假设bash:

~> declare -a foo
~> foo[0]="foo"
~> foo[1]="bar"
~> foo[2]="baz"
~> echo ${#foo[*]}
3

因此,${#ARRAY[*]}扩展到数组的长度ARRAY.

Answer 3

从Bash 手册:

$ {#参数}

参数展开值的字符长度被替换.如果参数是' '或'@',则替换的值是位置参数的数量.如果parameter是由' '或'@' 下标的数组名,则替换的值是数组中元素的数量.如果参数是由负数下标的索引数组名称,则该数字被解释为相对于大于参数的最大索引的数字,因此负数索引从数组的末尾开始计数,索引-1引用最后一个元件.

字符串,数组和关联数组的长度

string="0123456789"                   # create a string of 10 characters
array=(0 1 2 3 4 5 6 7 8 9)           # create an indexed array of 10 elements
declare -A hash
hash=([one]=1 [two]=2 [three]=3)      # create an associative array of 3 elements
echo "string length is: ${#string}"   # length of string
echo "array length is: ${#array[@]}"  # length of array using @ as the index
echo "array length is: ${#array[*]}"  # length of array using * as the index
echo "hash length is: ${#hash[@]}"    # length of array using @ as the index
echo "hash length is: ${#hash[*]}"    # length of array using * as the index

输出:

string length is: 10
array length is: 10
array length is: 10
hash length is: 3
hash length is: 3

处理$@参数数组:

set arg1 arg2 "arg 3"
args_copy=("$@")
echo "number of args is: $#"
echo "number of args is: ${#@}"
echo "args_copy length is: ${#args_copy[@]}"

输出:

number of args is: 3
number of args is: 3
args_copy length is: 3

Answer 4

在tcsh或csh中:

~> set a = ( 1 2 3 4 5 )
~> echo $#a
5

Answer 5

在鱼壳中，可以通过以下方式找到数组的长度：

$ set a 1 2 3 4
$ count $a
4

Answer 6

这对我来说效果很好：

arglen=$#
argparam=$*
if [ $arglen -eq '3' ];
then
    echo Valid Number of arguments
    echo "Arguments are $*"
else
    echo only four arguments are allowed
fi