Bash 数组声明和附加
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Bash array declaration and appendation
提问by A.Jac
I'm trying to declare and append to an array in a bash script, after searching i resulted in this code.
我试图在 bash 脚本中声明并附加到一个数组，搜索后我得到了这段代码。
list=()
list+="string"
But when i echo this out it results in nothing. I have also tried appending to the array like this
但是当我回应这一点时，它什么也没有。我也试过像这样附加到数组
list[$[${#list[@]}+1]]="string"
I don't understand why this is not working, anyone have any suggestions?
我不明白为什么这不起作用，有人有什么建议吗？
EDIT:The problem is list is appended to inside a while loop.
编辑：问题是 list 附加到 while 循环内。
list=()
git ls-remote origin 'refs/heads/*' | while read sha ref; do
list[${#list[@]}+1]="$ref"
done
declare -p list
see stackoverflow.com/q/16854280/1126841
见 stackoverflow.com/q/16854280/1126841
回答by Ali Okan Yüksel
You can append new string to your array like this:
您可以像这样将新字符串附加到数组中：
#!/bin/bash
mylist=("number one")
#append "number two" to array
mylist=("${mylist[@]}" "number two")
# print each string in a loop
for mystr in "${mylist[@]}"; do echo "$mystr"; done
For more information you can check http://wiki.bash-hackers.org/syntax/arrays
有关更多信息，您可以查看http://wiki.bash-hackers.org/syntax/arrays
回答by Jdamian
Ali Okan Yüksel has written down an answer for the first method you mentioned about adding items in an array.
Ali Okan Yüksel 已经为您提到的关于在数组中添加项目的第一种方法写下了答案。
list+=( newitem another_new_item ··· )
The right way of the second method you mentioned is:
您提到的第二种方法的正确方法是：
list[${#list[@]}]="string"
Assuming that a non-sparsearray has Nitems and because bash array indexes starts from 0, the last item in the array is N-1th. Therefore the next item must be added in the Nposition (${#list[@]}); not necessarily in N+1as you wrote.
假设一个非稀疏数组有N项并且因为 bash 数组索引从 开始0，数组中的最后一项是N-1th。因此必须在N位置 ( ${#list[@]}) 中添加下一项；不一定N+1像你写的那样。
Instead, if a sparsearray is used, it is very useful the bash parameter expansionwhich provides the indexes of the array:
相反，如果使用稀疏数组，则提供数组索引的bash 参数扩展非常有用：
${!list[@]}
For instance,
例如，
$ list[0]=3
$ list[12]=32
$ echo ${#list[@]}
2
$ echo ${!list[@]}
0 12
