我写了下面的脚本:
#!/bin/bash
if [ $# -ne 1 ];then
echo "Usage: ./script <input-file>"
exit 1
fi
while read user pass; do
curl -iL --data-urlencode user="$user" --data-urlencode password="$pass" http://foo.com/signin 1>/dev/null 2>&1
if [ $? -eq 0 ];then
echo "ok"
elif [ $? -ne 0 ]; then
echo "failed"
fi
done < $1
问题:
每当我运行错误的用户并传递结果对我来说都是ok...
如何确定我的参数是否正确？
谢谢
最佳答案
这是因为您正在从 curl 命令获取输出。使用随机用户/密码输入该命令得到这个:
$ curl -iL --data-urlencode user=BLAHBLAH --data-urlencode password=BLAH http://foo.com/signin
HTTP/1.1 301 Moved Permanently
Server: nginx/1.0.5
Date: Wed, 19 Feb 2014 05:53:36 GMT
Content-Type: text/html
Content-Length: 184
Connection: keep-alive
Location: http://www.foo.com/signin
. . .
. . .
<body>
<!-- This file lives in public/500.html -->
<div class="dialog">
<h1>We're sorry, but something went wrong.</h1>
</div>
</body>
</html>
因此，
$ echo $?
0
但是把url修改成垃圾:
$ curl -iL --data-urlencode user=BLAHBLAH --data-urlencode password=BLAH http://foof.com/signin
curl: (6) Could not resolve host: foof.com
$ echo $?
6
关于linux - 如何使用 bash 确保 user & pass 在 curl 中是正确的，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/21871985/