Question

这应该使用Curl在远程服务器上调用php脚本。如果您插入值而不是变量，那么curl命令将起作用，因此我知道语法在这里不可用。

但是，对我来说奇怪的是，当它抓住变量时，结果总是成功的。有任何想法吗？预先感谢！

#!/bin/bash
# this script should call the curl command to run the php script
# on the web server and log the user in openvpn client

echo Enter Account Number:

read accountNum

echo Enter Password:

read pwVar

# curl command to run php script on web server
authResult=$(curl -X POST IPADDRESSOFSERVERREMOVED -d '{"accountNumber":"$accountNum","password":"$pwVar"}' -H 'Content-Type: application/json')
authResult2=$(echo "$authResult")

if [ authResult2=$(echo "$authResult" | grep "successfully") ]
then 
    echo yep it works
else 
    echo no it dont 
fi

Answer 1

非常抱歉重复。这样重写就可以了：

#!/bin/bash
# this script should call the curl command to run the php script
# on the web server and log the user in openvpn client
echo Enter Account Number:
read accountNum
echo Enter Password:
read pwVar
# curl command to run php script on web server
authResult=$(curl -X POST ADDRESSREMOVEDHERE -d '{"accountNumber":"'$accountNum'","password":"'$pwVar'"}' -H 'Content-Type: application/json')

if [ [ $authResult = *"success"* ] ]; then
    echo yep it works
else
    echo no it dont
fi

Bash-传递给curl和grep命令的变量不起作用

1 个答案: