如何使用Applescript退出所有正在运行的用户应用程序?
答案 0 :(得分:10)
没关系......我想我找到了答案:
tell application "System Events" to set the visible of every process to true
set white_list to {"Finder"}
try
tell application "Finder"
set process_list to the name of every process whose visible is true
end tell
repeat with i from 1 to (number of items in process_list)
set this_process to item i of the process_list
if this_process is not in white_list then
tell application this_process
quit
end tell
end if
end repeat
on error
tell the current application to display dialog "An error has occurred!" & return & "This script will now quit" buttons {"Quit"} default button 1 with icon 0
end try
答案 1 :(得分:1)
tell application "System Events" to set the visible of every process to true
set white_list to {"Finder"}
try
tell application "Finder"
set process_list to the name of every process whose visible is true
end tell
repeat with i from 1 to (number of items in process_list)
set this_process to item i of the process_list
if this_process is not in white_list then
tell application this_process
quit
end tell
end if
end repeat
on error
tell the current application to display dialog "An error has occurred!" & return & "This script will now quit" buttons {"Quit"} default button 1 with icon 0
end try
答案 2 :(得分:1)
经过一些谷歌搜索,我发现了一个更好的方法:
background only
来构建初始应用列表,而不是
visible is true
。不同之处在于其他脚本将失败
退出使用⌘H隐藏的应用。-- get list of open apps
tell application "System Events"
set allApps to displayed name of (every process whose background only is false) as list
end tell
-- leave some apps open
set exclusions to {"AppleScript Editor", "Automator", "Finder", "LaunchBar"}
-- quit each app
repeat with thisApp in allApps
set thisApp to thisApp as text
if thisApp is not in exclusions then
tell application thisApp to quit
end if
end repeat
答案 3 :(得分:0)
tell application "System Events" to set quitapps to name of every application process whose visible is true and name is not "Finder"
repeat with closeall in quitapps
quit application closeall
end repeat