从数组中删除最后一个元素 [英] Remove the last element from an array

问题描述

我想删除数组中的最后一个条目，并且我希望数组在我使用 ${#array[@]} 时显示它少了 1 个条目.这是我正在使用的当前行:

unset GreppedURLs[${#GreppedURLs[@]} -1]

请纠正我并告诉我正确的方法.

解决方案

你的答案(几乎)是正确的 对于非稀疏索引数组¹:

unset 'arr[${#arr[@]}-1]'

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Bash 4.3 或更高版本添加了这个新语法来做同样的事情:

 取消设置 arr[-1]

(注意单引号:它们可以防止路径名扩展).

演示:

arr=( a b c )回声 ${#arr[@]}

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3

for a in "${arr[@]}";做 echo "$a";完毕

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a乙C

unset 'arr[${#arr[@]}-1]'for a in "${arr[@]}";做 echo "$a";完毕

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a乙

重点

echo ${#arr[@]}

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2

(GNU bash, 版本 4.2.8(1)-release (x86_64-pc-linux-gnu))

<小时>

¹ @Wil 提供了一个很好的答案，适用于各种数组

I want to remove the last entry in my array, and I want the array to show me that it has 1 less entry when I am using the ${#array[@]}. This is the current line I am using:

unset GreppedURLs[${#GreppedURLs[@]} -1]

Please correct me and show me the right way.

解决方案

The answer you have is (nearly) correct for non-sparse indexed arrays¹:

unset 'arr[${#arr[@]}-1]'

Bash 4.3 or higher added this new syntax to do the same:

 unset arr[-1]

(Note the single quotes: they prevent pathname expansion).

Demo:

arr=( a b c )
echo ${#arr[@]}

3

for a in "${arr[@]}"; do echo "$a"; done

a
b
c

unset 'arr[${#arr[@]}-1]'
for a in "${arr[@]}"; do echo "$a"; done

a
b

Punchline

echo ${#arr[@]}

2

(GNU bash, version 4.2.8(1)-release (x86_64-pc-linux-gnu))

---

¹ @Wil provided an excellent answer that works for all kinds of arrays

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