通过忽略分隔符前后的空格来拆分逗号分隔的字符串，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

尝试以下操作：

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 删除字符串开头和结尾的空格，打印结果。然后将生成的输出重定向回 readarray 以创建主机名数组。

Answer 2

你会尝试以下方法吗：

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]

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

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