能否用AppleScript检测屏幕是否有窗口？Bash脚本需求咨询

当然可以！AppleScript完全能搞定这个需求，而且能轻松和你的bash脚本结合起来用。咱们分两部分来实现：先检测屏幕上是否有任何可见窗口，再搞定你提到的额外需求——检测是否存在活动窗口（当前聚焦的窗口）。

一、检测屏幕是否存在任何可见窗口 #

这段AppleScript会遍历系统中所有运行的进程，检查每个进程是否可见且拥有窗口，只要找到符合条件的就会返回true：

tell application "System Events"
    set hasVisibleWindows to false
    repeat with proc in every process
        if visible of proc is true and (count of windows of proc) > 0 then
            set hasVisibleWindows to true
            exit repeat
        end if
    end repeat
    return hasVisibleWindows
end tell

在bash脚本里调用非常简单，用osascript命令执行即可：

has_windows=$(osascript -e 'tell application "System Events" set hasVisibleWindows to false repeat with proc in every process if visible of proc is true and (count of windows of proc) > 0 then set hasVisibleWindows to true exit repeat end if end repeat return hasVisibleWindows end tell')

if [ "$has_windows" = "true" ]; then
    echo "屏幕上存在可见窗口"
else
    echo "屏幕上没有任何可见窗口"
fi

二、额外需求：检测是否存在活动窗口 #

如果需要进一步判断当前是否有处于聚焦状态的活动窗口，可以用这段脚本：

tell application "System Events"
    set activeWindowExists to exists (first window of (first process whose frontmost is true) where visible is true)
    return activeWindowExists
end tell

对应的bash调用示例：

has_active_window=$(osascript -e 'tell application "System Events" set activeWindowExists to exists (first window of (first process whose frontmost is true) where visible is true) return activeWindowExists end tell')

if [ "$has_active_window" = "true" ]; then
    echo "存在活动窗口"
else
    echo "没有活动窗口（或当前前台应用无可见窗口）"
fi

三、整合两个检测的bash脚本示例 #

要是你想一次完成两种检测，直接把逻辑整合到一个bash脚本里就行：

#!/bin/bash

# 检测是否有可见窗口
has_windows=$(osascript -e 'tell application "System Events" set hasVisibleWindows to false repeat with proc in every process if visible of proc is true and (count of windows of proc) > 0 then set hasVisibleWindows to true exit repeat end if end repeat return hasVisibleWindows end tell')

# 检测是否有活动窗口
has_active_window=$(osascript -e 'tell application "System Events" set activeWindowExists to exists (first window of (first process whose frontmost is true) where visible is true) return activeWindowExists end tell')

echo "屏幕窗口状态检测结果："
if [ "$has_windows" = "true" ]; then
    echo "- 存在可见窗口"
    if [ "$has_active_window" = "true" ]; then
        echo "- 存在活动窗口"
    else
        echo "- 无活动窗口"
    fi
else
    echo "- 无任何可见窗口"
fi

这个脚本会清晰输出当前屏幕的窗口状态，完全满足你的需求~

内容的提问来源于stack exchange，提问作者Utku Birkan