[zorp-hu] Zorp http filterURL

[SZALAY Attila]

Hi!

On 2002 Jul 18, Kosa Attila wrote:
>   def filterURL(self, method, url, version):
>     if (url == "http://webmail.xxx.hu")
>       self.request_url = "http://129.184.93.3/"
>       return HTTP_REQ_ACCEPT
>     if (url == "http://www.xxx.hu")
>       return HTTP_REQ_ACCEPT
>     if (url == "https://www.xxx.hu")
>       return HTTP_REQ_ACCEPT
>     self.error_info = "Content denied by policy."
>     return HTTP_REQ_REJECT
> 
> Jul 17 19:56:06 thewall Http[9042]: zorp version 1.4.3 starting up
> Jul 17 19:56:06 thewall Http[9044]: (zorp/nosession):   File "/etc/zorp/test.py", line 37
> Jul 17 19:56:06 thewall Http[9044]: (zorp/nosession):     if (url == "http://webmail.xxx.hu")
> Jul 17 19:56:06 thewall Http[9044]: (zorp/nosession):                                       ^
> Jul 17 19:56:06 thewall Http[9044]: (zorp/nosession): SyntaxError: invalid syntax
> Jul 17 19:56:06 thewall Http[9042]: (zorp/nosession): Error opening policy file /etc/zorp/test.py
> Jul 17 19:56:06 thewall Http[9042]: zorp version 1.4.3 going down.

Mint az latszik, a sor vegen van a gondja.

A gond egyebkent az, hogy python-ban a if utan ':'-t kell rakni, ugyanugy
mint def utan, azaz:
	def filterURL(self, method, url, version):
		if (url == "http://webmail.xxx.hu"):
                                                   ^
			self.request_url = "http://129.184.93.3/"
			return HTTP_REQ_ACCEPT

-- 
Szalay Attila                     BalaBit IT Biztonságtechnikai Kft.
tel:(36-1)-371-05-40              1116 Bp. Csurgoi ut 20/b
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