bash：将 '\n' 分隔的字符串转换为数组
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/24628076/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
bash: convert '\n' delimited strings into array
提问by ramgorur
I have this script --
我有这个脚本——
nmapout=`sudo nmap -sP 10.0.0.0/24`
names=`echo "$nmapout" | grep "MAC" | grep -o '(.\+)'`
echo "$names"
now the $namesvariable contains strings delimited with newlines --
现在$names变量包含用换行符分隔的字符串——
>_
(Netgear)
(Hon Hai Precision Ind. Co.)
(Apple)
I tried to do the array conversion with the sub-string approach --
我尝试使用子字符串方法进行数组转换——
names=(${names//\n/ })
echo "${names[@]}"
But the problem is that I can't access them by indexing (i.e. ${names[$i]etc), if I run this loop --
但问题是我无法通过索引（即${names[$i]等）访问它们，如果我运行这个循环——
for (( i=0; i<${#names[@]}; i++ ))
do
echo "$i: ${names[$i]"
# do some processing with ${names[$i]}
done
I get this output --
我得到这个输出——
>_
0: (Netgear)
1: (Hon
2: Hai
but what I want is --
但我想要的是——
>_
0: (Netgear)
1: (Hon Hai Precision Ind. Co.)
2: (Apple)
I could not figure out a good way to do this, please note that the second string has spaces in it.
我想不出一个好的方法来做到这一点，请注意第二个字符串中有空格。
Any idea ?
任何的想法 ？
回答by Sanket Parmar
Set IFS. Shell uses IFSvariable to determine what the field separators are. By default IFSis set to the space character. Change it to newline.
设置IFS。Shell 使用IFS变量来确定字段分隔符是什么。默认IFS设置为空格字符。将其更改为换行符。
#!/bin/bash
names="Netgear
Hon Hai Precision Ind. Co.
Apple"
SAVEIFS=$IFS # Save current IFS
IFS=$'\n' # Change IFS to new line
names=($names) # split to array $names
IFS=$SAVEIFS # Restore IFS
for (( i=0; i<${#names[@]}; i++ ))
do
echo "$i: ${names[$i]}"
done
Output
输出
0: Netgear
1: Hon Hai Precision Ind. Co.
2: Apple
回答by firegurafiku
Let me contribute to Sanket Parmar's answer. If you can extract string splitting and processing into a separate function, there is no need to save and restore $IFS— use localinstead:
让我为Sanket Parmar 的回答做出贡献。如果您可以将字符串拆分和处理提取到单独的函数中，则无需保存和恢复$IFS—local而是使用：
#!/bin/bash
function print_with_line_numbers {
local IFS=$'\n'
local lines=()
local i
for (( i=0; i<${#lines[@]}; i++ )) ; do
echo "$i: ${lines[$i]}"
done
}
names="Netgear
Hon Hai Precision Ind. Co.
Apple"
print_with_line_numbers "$names"
See also:
也可以看看：
回答by Ale
Bash also has a readarraybuiltin command, easily searchable in the man page. It uses newline (\n) as the default delimiter, and MAPFILEas the default array, so one can do just like so:
Bash 还有一个readarray内置命令，可以在手册页中轻松搜索。它使用换行符 ( \n) 作为默认分隔符和MAPFILE默认数组，因此可以这样做：
names="Netgear
Hon Hai Precision Ind. Co.
Apple"
readarray -t <<<$names
printf "0: ${MAPFILE[0]}\n1: ${MAPFILE[1]}\n2: ${MAPFILE[2]}\n"
The -toption removes the delimiter ('\n'), so that it can be explicitly added in printf. The output is:
该-t选项删除分隔符 ( '\n')，以便它可以显式添加到printf. 输出是：
0: Netgear
1: Hon Hai Precision Ind. Co.
2: Apple
回答by Hari Bharathi
As others said, IFS will help you.IFS=$'\n' read -ra array <<< "$names"if your variable has string with spaces, put it between double quotes.
Now you can easily take all values in a array by ${array[@]}
正如其他人所说，IFS 会帮助你。IFS=$'\n' read -ra array <<< "$names"如果您的变量有带空格的字符串，请将其放在双引号之间。现在您可以通过以下方式轻松获取数组中的所有值${array[@]}
