bash 从 shell 脚本启动 Web 浏览器的干净方法？
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Clean way to launch the web browser from shell script?
提问by nicoulaj
In a bash script, I need to launch the user web browser. There seems to be many ways of doing this:
在 bash 脚本中，我需要启动用户 Web 浏览器。似乎有很多方法可以做到这一点：
$BROWSERxdg-opengnome-openon GNOMEwww-browserx-www-browser- ...
$BROWSERxdg-opengnome-open在 GNOME 上www-browserx-www-browser- ...
Is there a more-standard-than-the-others way to do this that would work on most platforms, or should I just go with something like this:
是否有一种比其他方法更标准的方法可以在大多数平台上运行，或者我应该使用这样的方法：
#/usr/bin/env bash
if [ -n $BROWSER ]; then
$BROWSER 'http://wwww.google.com'
elif which xdg-open > /dev/null; then
xdg-open 'http://wwww.google.com'
elif which gnome-open > /dev/null; then
gnome-open 'http://wwww.google.com'
# elif bla bla bla...
else
echo "Could not detect the web browser to use."
fi
采纳答案by Philipp
xdg-openis standardized and should be available in most distributions.
xdg-open是标准化的，应该在大多数发行版中可用。
Otherwise:
除此以外：
evalis evil, don't use it.- Quote your variables.
- Use the correct test operators in the correct way.
eval是邪恶的，不要使用它。- 引用你的变量。
- 以正确的方式使用正确的测试运算符。
Here is an example:
下面是一个例子：
#!/bin/bash
if which xdg-open > /dev/null
then
xdg-open URL
elif which gnome-open > /dev/null
then
gnome-open URL
fi
Maybe this version is slightly better (still untested):
也许这个版本稍微好一点（仍未测试）：
#!/bin/bash
URL=
[[ -x $BROWSER ]] && exec "$BROWSER" "$URL"
path=$(which xdg-open || which gnome-open) && exec "$path" "$URL"
echo "Can't find browser"
回答by jfs
python -mwebbrowser http://example.com
works on many platforms
适用于许多平台
回答by mbs400
OSX:
操作系统：
$ open -a /Applications/Safari.app http://www.google.com
or
或者
$ open -a /Applications/Firefox.app http://www.google.com
or simply...
或者干脆……
$ open some_url
回答by Joan Rieu
You could use the following:
您可以使用以下内容：
x-www-browser
It won't run the user's but rather the system's default X browser.
它不会运行用户的而是系统的默认 X 浏览器。
See: this thread.
请参阅：此线程。
回答by jamescampbell
Taking the other answers and making a version that works for all major OS's as well as checking to ensure that a URL is passed in as a run-time variable:
获取其他答案并制作适用于所有主要操作系统的版本，并检查以确保 URL 作为运行时变量传入：
#!/bin/bash
if [ -z ]; then
echo "Must run command with the url you want to visit."
exit 1
else
URL=
fi
[[ -x $BROWSER ]] && exec "$BROWSER" "$URL"
path=$(which xdg-open || which gnome-open) && exec "$path" "$URL"
if open -Ra "safari" ; then
echo "VERIFIED: 'Safari' is installed, opening browser..."
open -a safari "$URL"
else
echo "Can't find any browser"
fi
回答by Raphi
This may not apply exactly to what you want to do, but there is a really easy way to create and launch a server using the http-servernpm package.
这可能并不完全适用于您想要做的事情，但是有一种使用http-servernpm 包创建和启动服务器的非常简单的方法。
Once installed (just npm install http-server -g) you can put
一旦安装（只是npm install http-server -g）你可以把
http-server -o
http-server -o
in your bash script and it will launch a server from the current directory and open a browser to that page.
在您的 bash 脚本中，它将从当前目录启动服务器并打开浏览器到该页面。
