如何在shell脚本中使用curl ?

I'm trying to run this shell script in order to install RVM in an Ubuntu box

我试图运行这个shell脚本，以便在Ubuntu框中安装RVM

#!/bin/bash
RVMHTTP="https://raw.github.com/wayneeseguin/rvm/master/binscripts/rvm-installer"
CURLARGS="-f -s -S -k"

bash < <(curl $CURLARGS $RVMHTTP)

but I get the following error

但是我得到了下面的错误

Syntax error: Redirection unexpected

语法错误:重定向意想不到的

Also tested not using the variables, but same result, could you tell what I'm missing?

也测试了不使用变量，但结果是一样的，你能告诉我遗漏了什么吗?

3 个解决方案

#1

#!/bin/bash                                                                                                                                                                                     
CURL='/usr/bin/curl'
RVMHTTP="https://raw.github.com/wayneeseguin/rvm/master/binscripts/rvm-installer"
CURLARGS="-f -s -S -k"

# you can store the result in a variable
raw="$($CURL $CURLARGS $RVMHTTP)"

# or you can redirect it into a file:
$CURL $CURLARGS $RVMHTTP > /tmp/rvm-installer

or:

或者:

Execute bash script from URL

从URL执行bash脚本

#2

Firstly, your example is looking quite correct and works well on my machine. You may go another way.

首先，您的示例看起来非常正确，并且在我的机器上运行良好。你可以走另一条路。

curl $CURLARGS $RVMHTTP > ./install.sh

All output now storing in ./install.sh file, which you can edit and execute.

所有输出现在存储在。/安装中。sh文件，您可以编辑和执行。

#3

url=”http://shahkrunalm.wordpress.com“
content=”$(curl -sLI “$url” | grep HTTP/1.1 | tail -1 | awk {‘print $2′})”
if [ ! -z $content ] && [ $content -eq 200 ]
then
echo “valid url”
else
echo “invalid url”
fi