使用Bash识别OSA对话框按钮点击状态异常排查

问题原因 #

你的Bash条件判断存在语法错误：在[[ ... ]]结构中，$selection="button returned:Restart Now"是赋值操作，而非字符串比较。赋值语句在Bash中会始终返回true，导致第一个if分支无论用户点击哪个按钮都会触发，所以重启相关代码会一直执行。

解决步骤 #

修正条件判断的语法，将赋值操作改为比较操作：

• 比较符号使用==（[[ ... ]]推荐的字符串比较符）
• 比较符号前后必须加空格，避免被解析为字符串拼接或赋值

修正后的代码：

#!/bin/sh

selection=$(osascript -e 'display dialog "A restart is required for this software to install properly..." buttons {"Test", "Restart Later", "Restart Now"} default button "Restart Now"')
echo $selection
if [[ $selection == "button returned:Restart Now" ]]; then
    echo "Would Restart..."
    #sudo shutdown -r now
elif [[ $selection == "button returned:Restart Later" ]]; then
    osascript -e 'display dialog "Software was not installed." buttons {"Ok"} default button "Ok" with icon caution'
else
    echo "Nothing selected!"
fi
exit 0;

优化方案 #

让osascript直接返回按钮的文本内容，而非完整的button returned:xxx字符串，能让判断逻辑更简洁：

#!/bin/sh

# 直接获取按钮返回的文本
selection=$(osascript -e 'display dialog "A restart is required for this software to install properly..." buttons {"Test", "Restart Later", "Restart Now"} default button "Restart Now"' -e 'return button returned of result')
echo $selection
if [[ $selection == "Restart Now" ]]; then
    echo "Would Restart..."
    #sudo shutdown -r now
elif [[ $selection == "Restart Later" ]]; then
    osascript -e 'display dialog "Software was not installed." buttons {"Ok"} default button "Ok" with icon caution'
else
    echo "Nothing selected!"
fi
exit 0;

另外注意：原代码中$(osascript ...)执行弹窗的写法可以简化为直接写osascript ...，不需要用子shell包裹，因为我们不需要捕获它的输出。

内容的提问来源于stack exchange，提问作者ezogg