通过忽略分隔符前后的空格来分割逗号分隔的字符串，并存储在shell脚本中的数组中

Question

我需要帮助在bash脚本中使用分隔符'，'拆分字符串，并忽略分隔符前后的空格。

My string looks like below and ',' is delimiter. 
mystring = < hostname1, hostname 2 , hostname value 3     > 
Notice that 
1. 'hostname1' has extra space in front
2. 'hostname 2' and 'hostname value 3' have extra spaces in front/back
I want to split above string and store in array like below:
mystring[0]:hostname1
mystring[1]:hostname 2
mystring[2]:hostname value 3
Please see below code and output:

HOSTNAME="hostname1 , hostname 2 , hostname value 3     "
IFS=',' read -ra hostnames <<< "$(HOSTNAME)"
for (( i=0; i<=2; i++ ))
do
echo hostname:[${hostnames[i]}] 
done

Output
hostname:[hostname1 ]
hostname:[ hostname 2 ]
hostname:[ hostname value 3 ]

Answer 1

尝试如下:

readarray -t hostnames <<< "$(awk -F, '{ for (i=1;i<=NF;i++) { gsub(/(^[[:space:]]+)|([[:space:]]+$)/,"",$i);print $i } }' <<< "$HOSTNAME")"

使用awk，将字段分隔符设置为"，"然后循环遍历每个字段，使用gsub从字符串的开头和结尾删除空格，打印结果。然后将结果输出重定向回readarray以创建hostname数组。

Answer 2

您可以试试以下操作吗:

str=" hostname1,  hostname 2 , hostname value 3    "
IFS=, read -r -a ary < <(sed 's/^ *//; s/ *$//; s/ *, */,/g' <<< "$str")
for i in "${ary[@]}"; do echo "hostname[$i]"; done

输出:

hostname[hostname1]
hostname[hostname 2]
hostname[hostname value 3]