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Linux 实现bashshell getops代码_Linux_Bash_Shell_Debian - Fatal编程技术网

Linux 实现bashshell getops代码

Linux 实现bashshell getops代码,linux,bash,shell,debian,Linux,Bash,Shell,Debian,我获得了创建两段独立代码的帮助: FIRST-检查是否存在T参数 TYPE="" while getopts "t:" opt; do case $opt in t) TYPE="$OPTARG" ;; esac done if [ -z "$TYPE" ]; then echo "No -t. Bye." exit 1 # error else echo "Valid -t!" exit 0 # error fi

我获得了创建两段独立代码的帮助:

FIRST-检查是否存在T参数

TYPE=""

while getopts "t:" opt; do
    case $opt in
        t)  TYPE="$OPTARG"
            ;;
    esac
done

if [ -z "$TYPE" ]; then
  echo "No -t. Bye."
  exit 1 # error
else
  echo "Valid -t!"
  exit 0 # error
fi
第二个-检查T标志是否有效(img、图像或文档)

就我个人而言,我无法将两者结合起来,让它们发挥作用。如何让这些脚本运行良好,这样我不仅可以检查是否有t参数,还可以检查它是否有效,并相应地打印一条消息

谢谢你能给我的帮助

#!/bin/bash
TYPE=""

while getopts "t:" opt; do
    case $opt in
        t)
            case $OPTARG in
                img|image|doc)
                    TYPE="$OPTARG"
                ;;
                *)
                    echo >&2 "Unsupported type..."
                    exit 1
                ;;
            esac
        ;;
    esac
done

if [ -z "$TYPE" ]; then
    echo "No -t. Bye."
    exit 1 # error
else
    echo "Valid -t!"
    exit 0 # error
fi
#!/bin/bash
TYPE=""

while getopts "t:" opt; do
    case $opt in
        t)
            case $OPTARG in
                img|image|doc)
                    TYPE="$OPTARG"
                ;;
                *)
                    echo >&2 "Unsupported type..."
                    exit 1
                ;;
            esac
        ;;
    esac
done

if [ -z "$TYPE" ]; then
    echo "No -t. Bye."
    exit 1 # error
else
    echo "Valid -t!"
    exit 0 # error
fi