简单的 bash CSV 数组问题
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Simple bash CSV array issue
提问by Atomiklan
I'm trying to do something fairly simple, and I'm just coming up short. Here is an example variable I'm trying to work with:
我正在尝试做一些相当简单的事情，但我只是做空了。这是我正在尝试使用的示例变量：
20,80,443,53
The variable is just a string of ports separated by commas. I need to get those into an array.
该变量只是一串由逗号分隔的端口。我需要将它们放入一个数组中。
回答by iruvar
Set IFSto ,and use the read command with a here-string
设置IFS为,并使用带有 here-string 的 read 命令
IFS=, read -r -a arr <<<"20,80,443,53"
printf "%s\n" "${arr[@]}"
20
80
443
53
回答by Hai Vu
Here is one way:
这是一种方法：
#!/bin/bash
v="20,80,443,53"
IFS=, a=($v) # Split
echo ${a[0]} # Display
echo ${a[1]}
echo ${a[2]}
echo ${a[3]}
Update
更新
Thanks to gniourf_gniourf for pointing out that IFS was modified as the result of the assignment. Here is my quirkywork around. Now I see why others did things differently.
感谢 gniourf_gniourf 指出 IFS 被修改为赋值的结果。这是我古怪的工作。现在我明白了为什么其他人做事的方式不同。
v="20,80,443,53"
PREV_IFS="$IFS" # Save previous IFS
IFS=, a=($v)
IFS="$PREV_IFS" # Restore IFS
echo ${a[0]}
echo ${a[1]}
echo ${a[2]}
echo ${a[3]}
回答by glenn Hymanman
var="20,80,442,53"
IFS=, read -ra ary <<< "$var"
printf "%s\n" "${ary[@]}"
20
80
442
53
回答by user2448541
Ports=('20','80','443','53');
for e in "${lic4l[@]}"; do
echo $e
I hope this will help and print the ports in the given variable.
我希望这会有所帮助并打印给定变量中的端口。
回答by Eugene Wagner
With SED it becomes a one-liner:
使用 SED，它变成了单线：
a=($(sed 's/,/ /g' <<< "20,80,443,53"))
printf "%s\n" "${a[@]}"
20
80
443
53
