告诉应用程序打开，然后退出时不是AppleScript的

Question

正常工作，这个脚本是打开的微软应用程序，然后3秒

tell application "Finder" 
set myFolder to ((startup disk as text) & "Applications:Microsoft Office 2011") as alias 
set myFiles to (every item of myFolder) as alias list 
open myFiles 
end tell 

delay 3 

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 
end tell 

但是当我运行该脚本，它在这之后停止后退出。 （编辑），它似乎关闭每一个应用程序除了Microsoft应用程序

tell application "Finder" 
set myFolder to ((startup disk as text) & "Applications:Microsoft Office 2011") as alias 
set myFiles to (every item of myFolder) as alias list 
open myFiles 

都打开和关闭应用程序的脚本工作的伟大分别，但我似乎不知道如何加入他们的行列。如果有人知道这是为什么发生，那会很好。由于

Answer 1

您遇到的问题是，告诉Finder打开一个应用程序是在脚本asyncrynous。意思是，直到应用程序加载前，它才会完成。使用延迟命令永远不会得到正确的结果。我建议使用“告诉应用程序xxx激活”，然后再继续。而且，它使您的脚本方式更清洁。

property myApps : {"Microsoft Word", "Microsoft Excel"} 

repeat with thisApp in myApps 
    try 
     tell application thisApp to activate 
    end try 
end repeat 

-- do whatever you need after all are open 

repeat with thisApp in myApps 
    try 
     tell application thisApp to quit 
    end try 
end repeat 

如果你想戒烟，除了查找每一个可见的应用程序，您还可以添加

tell application "Finder" to set myApps to name of (every process whose ((visible is true) and (name is not "Finder")))