Bash：需要去除字符串周围的单引号
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Bash: need to strip single quotes around a string
提问by Maddy
Following is the code snippet from my bash script:
以下是我的 bash 脚本中的代码片段：
....
....
for ((i=0; i<${#abc[@]}; i++))
do
xyz=${abc[i]}
....
....
When the value of 'xyz' is substituted in the script, the value has single quotes around it:
当在脚本中替换 'xyz' 的值时，该值周围有单引号：
'"6b76cdae-a4a8-4e88-989d-1581ae2d5b98"'
Why are the single quotes added and how do I strip them?
为什么要添加单引号以及如何去除它们？
Thanks!
谢谢！
采纳答案by nino_mezza
You can replace
你可以更换
xyz=${abc[i]}
xyz=${abc[i]}
with
和
eval xyz=${abc[i]}
评估 xyz=${abc[i]}
And here is an illustrating example what happens:
这是一个说明会发生什么的示例：
$ foo="'"'"bar"'"'"
$ echo $foo
'"bar"'
$ eval foo=$foo
$ echo $foo
"bar"
$ eval foo=$foo
$ echo $foo
bar
So, what happens with the holy evalis that the assignment gets split into two parts:
所以，神圣的事情eval是分配被分成两部分：
- Evaluate
$abc - Assign to
$xyz
- 评估
$abc - 分配给
$xyz
instead of just Step 2.
而不仅仅是第 2 步。
Maybe in your case you should consider to already change the assignment of abcbut i don't know because of ignorance ;)
也许在您的情况下，您应该考虑已经更改分配，abc但由于无知，我不知道；)
回答by Cyrus
You can try to remove allsingle quotes in array abc with Parameter Expansion:
您可以尝试使用参数扩展删除数组 abc中的所有单引号：
abc=(${abc[@]//\'/})
You can try to remove allsingle quotes in string xyz with Parameter Expansion:
您可以尝试使用参数扩展删除字符串 xyz中的所有单引号：
xyz=${xyz//\'/}
