我想使用命令行来找出哪个应用程序打开了 http 和 https 链接。我需要在脚本中执行此操作。我可以通过执行“defaults read com.apple.LaunchServices”来查看答案,但我不想自己解析该字典。
答案1
Apple Script 具有内置解析属性列表文件的功能。macscripter.net 上的 StefanK 提供了一个代码片段这在前段时间已经帮助了我。您可以轻松地将其保存并作为脚本执行,我将其存储在我的用户 bin 目录中(我将其添加到我的$PATH
):
#!/usr/bin/osascript
tell (system attribute "sysv") to set MacOS_version to it mod 4096 div 16
if MacOS_version is 5 then
set {a1, a2} to {1, 2}
else
set {a1, a2} to {2, 1}
end if
set pListpath to (path to preferences as Unicode text) & "com.apple.LaunchServices.plist"
tell application "System Events"
repeat with i in property list items of property list item 1 of contents of property list file pListpath
if value of property list item a2 of i is "http" then
return value of property list item a1 of i
end if
end repeat
return "com.apple.Safari"
end tell
答案2
其他选择:
VERSIONER_PERL_PREFER_32_BIT=1 perl -MMac::InternetConfig -le 'print +(GetICHelper "http")[1]'
tell application "System Events"
try
value of property list item "LSHandlerRoleAll" of (property list item 1 of property list item "LSHandlers" of property list file ((path to preferences as text) & "com.apple.LaunchServices.plist") where value of property list items contains "http")
on error
"com.apple.safari"
end try
end tell
http://www.hamsoftengineering.com/codeSharing/defaultApplication/defaultApplication.html:
$ DefaultApplication -url http:
/Applications/Safari.app