getting a KeyError:'href' any ideas?

[homepricemaps@gmail.com]

i have an
href which looks like this:


<td class="all">
<a class="btn" name="D1" href="http://www.cnn.com">
</a>


here is my code

for incident in row('td', {'class':'all'} ):
n = incident.findNe xtSibling('a', {'class': 'btn'})
link = incident.findNe xtSibling['href'] + "','"



and the full error is this:


File "/home/felafela/BeautifulSoup.p y", line 301, in __getitem__
return self._getAttrMa p()[key]
KeyError: 'href'


any idea what i'm doing wrong here with the syntax? thanks in advance

[Mike Meyer]

Re: getting a KeyError:'href' any ideas?

"homepricemaps@ gmail.com" <homepricemaps@ gmail.com> writes:

Please use less whitespace in your posts in the future. There's really
no need to put two blank lines between sections.
[color=blue]
> i have an
> href which looks like this:
> <td class="all">
> <a class="btn" name="D1" href="http://www.cnn.com">
> </a>
> here is my code
> for incident in row('td', {'class':'all'} ):
> n = incident.findNe xtSibling('a', {'class': 'btn'})
> link = incident.findNe xtSibling['href'] + "','"
> any idea what i'm doing wrong here with the syntax? thanks in advance[/color]

It's not the syntax, it's the logic. the a element is not a sibling of
the td element, it's a child. findNextSibling is going to return the
next td, assuming there is one. Trying to index
incident.findNe xtSibling in the next line is also broken. That's a
method, not an indexable object. So that line will also break.

<mike
--
Mike Meyer <mwm@mired.or g> http://www.mired.org/home/mwm/
Independent WWW/Perforce/FreeBSD/Unix consultant, email for more information.

[LordLaraby]

Re: getting a KeyError:'href' any ideas?

You wrote:[color=blue]
> i have an
> href which looks like this:
> <td class="all">
> <a class="btn" name="D1" href="http://www.cnn.com">
> </a>
> here is my code
> for incident in row('td', {'class':'all'} ):
> n = incident.findNe xtSibling('a', {'class': 'btn'})
> link = incident.findNe xtSibling['href'] + "','"
> any idea what i'm doing wrong here with the syntax? thanks in advance[/color]
Since you already found the anchor tag with the 'btn' class attribute,
just grab it's href attribute like so:
for incident in row('td', {'class':'all'} ):
n = incident.findNe xtSibling('a', {'class': 'btn'})
link = n['href'] + "','"

Works for me...

LL

[homepricemaps@gmail.com]

Re: getting a KeyError:'href' any ideas?

what is the syntax used to find a child of td?

Mike Meyer wrote:[color=blue]
> "homepricemaps@ gmail.com" <homepricemaps@ gmail.com> writes:
>
> Please use less whitespace in your posts in the future. There's really
> no need to put two blank lines between sections.
>[color=green]
> > i have an
> > href which looks like this:
> > <td class="all">
> > <a class="btn" name="D1" href="http://www.cnn.com">
> > </a>
> > here is my code
> > for incident in row('td', {'class':'all'} ):
> > n = incident.findNe xtSibling('a', {'class': 'btn'})
> > link = incident.findNe xtSibling['href'] + "','"
> > any idea what i'm doing wrong here with the syntax? thanks in advance[/color]
>
> It's not the syntax, it's the logic. the a element is not a sibling of
> the td element, it's a child. findNextSibling is going to return the
> next td, assuming there is one. Trying to index
> incident.findNe xtSibling in the next line is also broken. That's a
> method, not an indexable object. So that line will also break.
>
> <mike
> --
> Mike Meyer <mwm@mired.or g> http://www.mired.org/home/mwm/
> Independent WWW/Perforce/FreeBSD/Unix consultant, email for more information.[/color]

[homepricemaps@gmail.com]

Re: getting a KeyError:'href' any ideas?

ok the syntax for next is this:

b = n.findNext



LordLaraby wrote:[color=blue]
> You wrote:[color=green]
> > i have an
> > href which looks like this:
> > <td class="all">
> > <a class="btn" name="D1" href="http://www.cnn.com">
> > </a>
> > here is my code
> > for incident in row('td', {'class':'all'} ):
> > n = incident.findNe xtSibling('a', {'class': 'btn'})
> > link = incident.findNe xtSibling['href'] + "','"
> > any idea what i'm doing wrong here with the syntax? thanks in advance[/color]
> Since you already found the anchor tag with the 'btn' class attribute,
> just grab it's href attribute like so:
> for incident in row('td', {'class':'all'} ):
> n = incident.findNe xtSibling('a', {'class': 'btn'})
> link = n['href'] + "','"
>
> Works for me...
>
> LL[/color]