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Why is my django snippet not working?

开发者 https://www.devze.com 2023-03-23 19:14 出处:网络
I put together a function 开发者_JAVA技巧to link to news items with django as a filter. It works on dev_appserver but on production server it returns None Can you tell me why it\'s not working? Should

I put together a function 开发者_JAVA技巧to link to news items with django as a filter. It works on dev_appserver but on production server it returns None Can you tell me why it's not working? Should I investigate the except clause of the code where it currently just passes?

def news(n):
    url = os.environ.get('HTTP_HOST') if os.environ.get('HTTP_HOST') else os.environ['SERVER_NAME']
    tld = url[url.rfind('.'):]    
    try:        
        if url == 'localhost:8080':
            result = urlfetch.fetch('http://news.google.se/?output=rss')    
        elif tld != '.com' and tld != '.se' and tld != '.cl' :
            result = urlfetch.fetch('http://news.google.com'+tld+'/?output=rss') 
        else:      
            result = urlfetch.fetch('http://news.google.com/?output=rss')        
        if result.status_code == 200:
            dom = minidom.parseString(result.content)
            item_node = dom.getElementsByTagName("item")
            try:
                random_1=random.choice(item_node)
                rss1_link = random_1.childNodes[1].firstChild.data
                rss1_text = random_1.childNodes[0].firstChild.data
                return mark_safe('<a href="%s">%s</a>' % (rss1_link, rss1_text))
            except IndexError,e:
                return ''
    except urlfetch.Error, e:
        pass

register.filter(news)

Update: Now it returns an empty string on production but locally it works. It's something else that status 200 on production:

def news(n):
    url = os.environ.get('HTTP_HOST') if os.environ.get('HTTP_HOST') else os.environ['SERVER_NAME']
    tld = url[url.rfind('.'):]    
    try:        
        if url == 'localhost:8080':
            result = urlfetch.fetch('http://news.google.se/?output=rss')    
        elif tld != '.com' and tld != '.se' and tld != '.cl' :
            result = urlfetch.fetch('http://news.google.com'+tld+'/?output=rss') 
        else:        
            result = urlfetch.fetch('http://news.google.com/?output=rss')        
        if result.status_code == 200:
            dom = minidom.parseString(result.content)
            item_node = dom.getElementsByTagName("item")
            try:
                random_1=random.choice(item_node)
                rss1_link = random_1.childNodes[1].firstChild.data
                rss1_text = random_1.childNodes[0].firstChild.data
                return mark_safe('<a href="%s">%s</a>' % (rss1_link, rss1_text))        
            except IndexError,e:
                return ''
        else:
            return ''
    except urlfetch.Error, e:
        logging.error(str(e))
        return ''

EDIT: Here's the simplest reproduction that return a status 200 locally and a status 503 on production

def status(n):
    try:             
        result = urlfetch.fetch('http://news.google.com/?output=rss')       
        return str(result.status_code)
    except urlfetch.Error, e:
        return 'error'

Update: Here's the solution I currently use. It still needs improfement since there is a possibility for choosing 2 news items that are the same:

import random
def updateFeed(url):#to do, get srv from url and find number of entries
    srv = os.environ.get('HTTP_HOST') if os.environ.get('HTTP_HOST') else os.environ['SERVER_NAME']
    tld = srv[srv.rfind('.'):] 
    url = 'http://news.google.com/?output=rss'
    if srv.endswith('.com.br'):
        url = 'http://news.google.com.br/?output=rss'
    elif srv == 'localhost:8080' or srv.endswith('alltfunkar.com'):
        url = 'http://news.google.se/?output=rss'
    elif tld != '.com' and tld != '.se' and tld != '.cl' :
        url = 'http://news.google.com'+tld+'/?output=rss'
    query_args = { 'q': url, 'v':'1.0', 'num': '15', 'output': 'json' }
    qs = urllib.urlencode(query_args)
    loader = 'http://ajax.googleapis.com/ajax/services/feed/load'
    loadurl = '%s?%s' % (loader, qs)
    logging.info(loadurl)
    result = urlfetch.fetch(url=loadurl,headers={'Referer': '...'})
    if result.status_code == 200:
        news = simplejson.loads(result.content) 

        """ not working, using random.randrange instead
        some_key = random.choice(news.keys())
        something = news[some_key]
        """
        i = random.randrange(0,10)#to do: instead of 10, it should be number of entries
        title = news[u'responseData'][u'feed'][u'entries'][i][u'title']
        link = news[u'responseData'][u'feed'][u'entries'][i][u'link']
    return mark_safe('<a href="%s">%s</a>' % (link, title))


Python functions that don't explicitly return anything will return None. If this code is returning None on your production server, it's probably because it's hitting that last except: pass block, as you mentioned.

Without reading the actual code (which I haven't), I'd say replace that pass with return '' to safely swallow the urlfetch.Error or decide what you want to happen in that case and implement some new code for that block.

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