如何编写用于检查网站实时的shell脚本

Question

我正在编写一个shell脚本来监控网站是否存在，并在下面发送电子邮件提醒是我的代码

#!/bin/bash
if [[ curl -s --head  --request GET http://opx.com/opx/version | grep "200 OK" > /dev/null] && [ curl -s --head --request GET http://oss.com/version | grep "200 OK" > /dev/null ]]
then echo "The HTTP server on opx.com and oss.com is up!" #> /dev/null
else
msg="The HTTP server  opx.com Or oss.com is down "
email="opx-noc@opx.com"

curl --data "body=$msg &to=$email &subject=$msg" https://opx.com/email/send
fi;

如果我运行此代码我

./Monitoring_Opx_Oss: line 2: conditional binary operator expected
./Monitoring_Opx_Oss: line 2: syntax error near `-s'
./Monitoring_Opx_Oss: line 2: `if [[ curl -s --head  --request GET http://opx.com/opx/version | grep "200 OK" > /dev/null] && [ curl -s --head --request GET http://oss.com/version | grep "200 OK" > /dev/null ]] '

请纠正我......

Answer 1

更改它：

if [ $(curl -s --head  --request GET http://opx.opera.com/opx/version | grep "200 OK" > /dev/null) ] && [ $(curl -s --head --request GET http://oss.opera.com/version | grep "200 OK" > /dev/null) ]

要检查if内部命令的状态，您必须执行

if [ $(command) ]

当你使用

时

if [ command]

还要注意[ ]周围的空格：if [_space_ command _space_ ]

更新

基于Ansgar Wiechers's comment，您还可以使用以下内容：

if curl -s --head  --request GET http://opx.com/opx/version | grep "200 OK" > /dev/null && curl -s --head --request GET http://oss.com/version | grep "200 OK" > /dev/null;

即，

if command && command