how to ask google, altavista or another search engine...
Shagshag13
i'm looking for a python script able to ask to a web search engine (google, altavista or another search engine) and/or able to parse
a result page.
any other helps / advices / urls are welcome...
thanks in advance,
s13.
John Hunter
Shagshag13> hello, i'm looking for a python script able to ask to
Shagshag13> a web search engine (google, altavista or another
Shagshag13> search engine) and/or able to parse a result page.
Shagshag13> any other helps / advices / urls are welcome...
Shagshag13> thanks in advance,
Shagshag13> s13.
If you register (free) with google at
you can use pygoogle
http://diveintomark.org/projects/#pygoogle
Here's a simple example script (add your own license key)
import google
# Get your own from http://www.google.com/apis/
google.LICENSE_KEY = 'youRliCenseKEYHere'
data = google.doGoogleSearch('python')
# Search time
#print data.meta.searchTime
# Meta attributes
#print dir(data.meta)
print 'Found %d results' % len(data.results)
for result in data.results:
print 'Title: ', result.title
print 'URL: ', result.URL
print 'Summary: ', result.snippet
print
# Result attributes
#print dir(result)
# Spelling suggestion:
#print google.doSpellingSuggestion('pithon')
Paul Brian
It can all be done with the google web API
This is an xml-rpc based service that allows you to send through queries and
get back xml formatted results.
Its very impressive, although I am lost for "real world" automated
applications....
You need to register however, to prevent abuse i suppose:
https://www.google.com/accounts/
As for the other search engines, perhaps use urllib to download the pages
and just a simple string.find would result in most answers, with little
effort (or, for masochists, use htmltidy to convert to xhtml, and XPath to
get the bits, or even regular expressions ! )
------------------------------
Paul Brian
(07899) 877 295
paul1...@yahoo.com
"Shagshag13" <shags...@yahoo.fr> wrote in message
news:ah431f$q33vu$1...@ID-146704.news.dfncis.de...
Shagshag13
"John Hunter" <jdhu...@nitace.bsd.uchicago.edu>
thanks for that ! i just have registered !
but i'm also looking for others search engines wrappers like altavista or metacrawler, etc.
s13.
John Hunter
Shagshag13> but i'm also looking for others search engines
Shagshag13> wrappers like altavista or metacrawler, etc.
If you are willing to go to the dark side, you may find the WWW:Search
perl mods useful.
John Hunter
Shagshag13
but thanks for the advice.
s13.
"John Hunter" <jdhu...@nitace.bsd.uchicago.edu> a écrit dans le message de news: m2y9cag...@mother.paradise.lost...
Marcus Vinicius Laranjeira
Folks,
I use the mx.DateTime package, and I need to know if in one particular year
has a february with 28 or 29 days ? I don't know how to do that !
Cheers,
Marcus
John Hunter
Marcus> I use the mx.DateTime package, and I need to know if in
Marcus> one particular year has a february with 28 or 29 days ? I
Marcus> don't know how to do that !
Here's one way....
import mx.DateTime
def is_leapyear(year):
try:
d = mx.DateTime.DateTime(year,2,29)
return 1
except mx.DateTime.mxDateTime.RangeError:
return 0
print is_leapyear(2000)
print is_leapyear(2002)
Mark McEahern
> particular year
> has a february with 28 or 29 days ? I don't know how to do that !
import mx.DateTime
d = mx.DateTime.DateTimeFrom("2/1/2002")
print d.days_in_month == 28
d2 = d + mx.DateTime.RelativeDateTime(years=2)
print d2.days_in_month == 29
// m
-
Eric Brunel
> Folks,
>
> I use the mx.DateTime package, and I need to know if in one particular
> year has a february with 28 or 29 days ? I don't know how to do that !
A year is a leap year if it can be divided by 4, except if it can be
divided by 100, where it's a leap year only if it can be divided by 400. It
sounds awful to compute, but in Python, it's actually quite simple:
>>> isLeap = lambda x: x % 4 == 0 and (x % 100 != 0 or x % 400 == 0)
>>> isLeap(1980)
1
>>> isLeap(1991)
0
>>> isLeap(2000)
1
>>> isLeap(1900)
0
HTH
--
- Eric Brunel <eric....@pragmadev.com> -
PragmaDev : Real Time Software Development Tools - http://www.pragmadev.com
Andrew McNamara
>divided by 100, where it's a leap year only if it can be divided by 400. It
>sounds awful to compute, but in Python, it's actually quite simple:
>
>>>> isLeap = lambda x: x % 4 == 0 and (x % 100 != 0 or x % 400 == 0)
Or, if you have a pathological aversion to lambda:
>>> def isLeap(x):
... return x % 4 == 0 and (x % 100 != 0 or x % 400 == 0)
...
>>> isLeap(1980)
1
>>> isLeap(1991)
0
>>> isLeap(2000)
1
>>> isLeap(1900)
0
8-)
--
Andrew McNamara, Senior Developer, Object Craft
http://www.object-craft.com.au/
Marcus Vinicius Laranjeira
All,
I am using the boa constructor, and I need to put the logo of my company in
the wxPanel. This bitmap will be static and it will be there just for
decoration purposes.
Anyone can help me ?
Cheers,
Marcus
Roman Yakovenko
You can put this control on panel. Use "Bitmap" property to set desired bitmap.
Also you may insert you bitmap to code, see img2py.py file in wxPython\tools folder.
Roman.
All,
Cheers,
Marcus
Michael Gilfix
look at the numerous examples that come with wxPython. There should be
a static bitmap example there that would be most educational.
-- Mike
On Wed, Jul 31 @ 11:05, Marcus Vinicius Laranjeira wrote:
>
>
> All,
>
> I am using the boa constructor, and I need to put the logo of my company in
> the wxPanel. This bitmap will be static and it will be there just for
> decoration purposes.
>
> Anyone can help me ?
>
> Cheers,
>
> Marcus
>
>
> --
> http://mail.python.org/mailman/listinfo/python-list
`-> (m.laranjeira)
--
Michael Gilfix
mgi...@eecs.tufts.edu
For my gpg public key:
http://www.eecs.tufts.edu/~mgilfix/contact.html
John Baxter
Eric Brunel <eric....@pragmadev.com> wrote:
> A year is a leap year if it can be divided by 4, except if it can be
> divided by 100, where it's a leap year only if it can be divided by 400. It
> sounds awful to compute, but in Python, it's actually quite simple:
With ongoing discussion of a potential y4k exception exception exception
(not leap year) to adjust just a bit more finely. For most purposes, we
have time to deal with that one.
--John
Steve Holden
news:mailman.1028042640...@python.org...
Another way is to RTFM :-) From the mx.DateTime documentation on DateTime
instance variables:
is_leapyear
Returns 1 iff the instances value points to a leap year in the Gregorian
calendar.
regards
-----------------------------------------------------------------------
Steve Holden http://www.holdenweb.com/
Python Web Programming http://pydish.holdenweb.com/pwp/
-----------------------------------------------------------------------