bash - IF ELSE声明

Question

我想问一个问题:“您想使用哪种格式？”，然后根据答案是xml还是json，打印适当的输出。但是，如果不是xml或json，则代码完成。

码:

#!/bin/bash
read -r -e -p "What format do you want to use? json/xml?: " format

if [[ "${format,,}" == "json" ]]; then
    curl=$curl"\"\Content-Type: application/json\"\
else [[ "${format,,}" == "xml" ]]; then
    curl=$curl"\"\Content-Type: application/xml\"\
elif [[ "${format,,}" == "no" ]]; then
  echo "Goodbye, you must specify a format to use Rest"
else 
  echo "Unknown"
  exit 0
fi
echo $curl

当我尝试运行它时，我收到此错误:

[root@osprey groups]# ./test3
What format do you want to use? json/xml?: xml
./test3: line 8: syntax error near unexpected token `then'
./test3: line 8: `elif [[ "${format,,}" == "no" ]]; then'

Answer 1

我会使用更长的时间，但是使用起来更加友好:

err() { echo "$@" >&2; return 1; }

show_help() {
cat - >&2 <<EOF
You must specify a format to use Rest because bla bla...
EOF
}

select_format() {
    PS3="Select the format do you want to use?> "
    select ans in json xml "quit program"; do
        case "$REPLY" in
            [0-9]*) answer=$ans;;
            *) answer=$REPLY;;
        esac
        case "${answer,,}" in
            '') ;;
            j|js|json) echo '-H "Content-Type: application/json"' ; return 0 ;;
            x|xml)     echo '-H "Content-Type: application/xml"'  ; return 0 ;;
            q|quit*)   err "Bye..." ; return 1;;
            h|help)  show_help ;;
            *) err "Unknown format $answer" ;;
        esac
    done
    [[ "$answer" ]] || return 1 #handles EOF (ctrl-D)
}

curl_args=()
curl_args+=( $(select_format) ) || exit 1
echo curl "${curl_args[@]}"

它显示一个菜单:

1) json
2) xml
3) quit program
Select the format do you want to use?> 

用户可以输入:

对应编号

jsont_li或 j， js， json和xml的 x， xml，xml的

h获取帮助

退出的

和q ...

错误(类型错误)答案已处理并再次询问...