python如何请求本地网站链接?报错

遇到一个问题：

本地服务器上部署了一个PHP发送邮件的程序

在浏览器中通过访问:http://www.xxx.com/api/demo.php，正常执行，可以收到邮件

而在linux命令行下执行php /data/htdocs/www/api/demo.php，返回一个错误：找不到mysql_connect()函数，请教一下同事，得知貌似我程序中包含文件的路径为相对路径，而在命令行下运行不支持。

于是我想到了使用wget命令，wget http://www.xxx.com/api/demo.php，不断请求，也不见发出邮件

我又换了另外一种方式用curl，系统提示：curl: (7) couldn't connect to host

。。。。。。

我又尝试另外一种方法，用python写脚本来请求，结果又给哥报了一大堆错误：

python代码：

import urllib
urllib.urlopen('http://www.xxxx.com/api/demo.php')

rTraceback (most recent call last):
  File "<stdin>", line 1, in ?
  File "/usr/lib64/python2.4/urllib.py", line 82, in urlopen
    return opener.open(url)
  File "/usr/lib64/python2.4/urllib.py", line 190, in open
    return getattr(self, name)(url)
  File "/usr/lib64/python2.4/urllib.py", line 313, in open_http
    h.endheaders()
  File "/usr/lib64/python2.4/httplib.py", line 804, in endheaders
    self._send_output()
  File "/usr/lib64/python2.4/httplib.py", line 685, in _send_output
    self.send(msg)
  File "/usr/lib64/python2.4/httplib.py", line 652, in send
    self.connect()
  File "/usr/lib64/python2.4/httplib.py", line 636, in connect
    raise socket.error, msg
IOError: [Errno socket error] (110, 'Connection timed out')

唉呀呀，真是百思不得其姐啊~

求各位大婶指点