I'd like to use Nokogiri to extract all nodes in an element that contain a specific attribute name.
e.g., I'd like to find the 2 nodes that contain the attribute "blah" in the document below.
#doc = Nokogiri::HTML::DocumentFragment.parse <<-EOHTML
<body>
<h1 blah="afadf">Three's Company</h1>
<div>A love triangle.</div>
<b blah="adfadf">test test test</b>
</body>
EOHTML
I found this suggestion (below) at this website: http://snippets.dzone.com/posts/show/7994, but it doesn't return the 2 nodes in the example above. It returns an empty array.
# get elements with attribute:
elements = #doc.xpath("//*[#*[blah]]")
Thoughts on how to do this?
Thanks!
I found this here
elements = #doc.xpath("//*[#*[blah]]")
This is not a useful XPath expression. It says to give you all elements that have attributes that have child elements named 'blah'. And since attributes can't have child elements, this XPath will never return anything.
The DZone snippet is confusing in that when they say
elements = #doc.xpath("//*[#*[attribute_name]]")
the inner square brackets are not literal... they're there to indicate that you put in the attribute name. Whereas the outer square brackets are literal. :-p
They also have an extra * in there, after the #.
What you want is
elements = #doc.xpath("//*[#blah]")
This will give you all the elements that have an attribute named 'blah'.
You can use CSS selectors:
elements = #doc.css "[blah]"
Related
Given this XML, what XPath returns all elements whose prop attribute contains Foo (the first three nodes):
<bla>
<a prop="Foo1"/>
<a prop="Foo2"/>
<a prop="3Foo"/>
<a prop="Bar"/>
</bla>
//a[contains(#prop,'Foo')]
Works if I use this XML to get results back.
<bla>
<a prop="Foo1">a</a>
<a prop="Foo2">b</a>
<a prop="3Foo">c</a>
<a prop="Bar">a</a>
</bla>
Edit:
Another thing to note is that while the XPath above will return the correct answer for that particular xml, if you want to guarantee you only get the "a" elements in element "bla", you should as others have mentioned also use
/bla/a[contains(#prop,'Foo')]
This will search you all "a" elements in your entire xml document, regardless of being nested in a "blah" element
//a[contains(#prop,'Foo')]
I added this for the sake of thoroughness and in the spirit of stackoverflow. :)
This XPath will give you all nodes that have attributes containing 'Foo' regardless of node name or attribute name:
//attribute::*[contains(., 'Foo')]/..
Of course, if you're more interested in the contents of the attribute themselves, and not necessarily their parent node, just drop the /..
//attribute::*[contains(., 'Foo')]
descendant-or-self::*[contains(#prop,'Foo')]
Or:
/bla/a[contains(#prop,'Foo')]
Or:
/bla/a[position() <= 3]
Dissected:
descendant-or-self::
The Axis - search through every node underneath and the node itself. It is often better to say this than //. I have encountered some implementations where // means anywhere (decendant or self of the root node). The other use the default axis.
* or /bla/a
The Tag - a wildcard match, and /bla/a is an absolute path.
[contains(#prop,'Foo')] or [position() <= 3]
The condition within [ ]. #prop is shorthand for attribute::prop, as attribute is another search axis. Alternatively you can select the first 3 by using the position() function.
Have you tried something like:
//a[contains(#prop, "Foo")]
I've never used the contains function before but suspect that it should work as advertised...
John C is the closest, but XPath is case sensitive, so the correct XPath would be:
/bla/a[contains(#prop, 'Foo')]
If you also need to match the content of the link itself, use text():
//a[contains(#href,"/some_link")][text()="Click here"]
/bla/a[contains(#prop, "foo")]
try this:
//a[contains(#prop,'foo')]
that should work for any "a" tags in the document
For the code above...
//*[contains(#prop,'foo')]
I'm trying to perform html scrapping of a webpage. I like to fetch the three alternate text (alt - highlighted) from the three "img" elements.
I'm using the following code extract the whole "img" element of slide-1.
from lxml import html
import requests
page = requests.get('sample.html')
tree = html.fromstring(page.content)
text_val = tree.xpath('//a[class="cover-wrapper"][id = "slide-1"]/text()')
print text_val
I'm not getting the alternate text values displayed. But it is an empty list.
HTML Script used:
This is one possible XPath :
//div[#id='slide-1']/a[#class='cover-wrapper']/img/#alt
Explanation :
//div[#id='slide-1'] : This part find the target <div> element by comparing the id attribute value. Notice the use #attribute_name syntax to reference attribute in XPath. Missing the # symbol would change the XPath selector meaning to be referencing a -child- element with the same name, instead of an attribute.
/a[#class='cover-wrapper'] : from each <div> element found by the previous bit of the XPath, find child element <a> that has class attribute value equals 'cover-wrapper'
/img/#alt : then from each of such <a> elements, find child element <img> and return its alt attribute
You might want to change the id filter to be starts-with(#id,'slide-') if you meant to return the all 3 alt attributes in the screenshot.
Try this:
//a[#class="cover-wrapper"]/img/#alt
So, I am first selecting the node having a tag and class as cover-wrapper and then I select the node img and then the attribute alt of img.
To find the whole image element :
//a[#class="cover-wrapper"]
I think you want:
//div[#class="showcase-wrapper"][#id="slide-1"]/a/img/#alt
What XPATH I need to extract the text inside SPAN that is preceded by a specific label inside a STRONG, both inside a P?
For example to extract website and email addresses from a page that looks like this:
<p>
<strong>Website:</strong>
<span>www.example.com</span>
</p>
<p>
<strong>Contact email:</strong>
<span>email#example.com</span>
</p>
This shall do:
//p/span[preceding::*[1][self::strong and . = 'Contact email:']]
Here, you are selecting all p/span elements with first preceding element strong, where label is Contact email:
Website:
//p/span[preceding::strong[1]/text()='Website:']
Email:
//p/span[preceding::strong[1]/text()='Contact email:']
It is also important to note that, by using preceding axes as shown in the other two answers, the XPath will mistakenly return span element that is formed like the following :
<strong>Website:</strong>
<p>
<span>www.example.com</span>
</p>
You can use preceding-sibling axes instead to avoid the mistake mentioned above :
//p/span[preceding-sibling::*[1][self::strong and . = 'Website:']]
preceding-sibling axes only consider elements that is located before context element (the span in this case), and is sibling (share the same parent) of the context element.
Hello I want to ask a question
I scrape a website with xpath ,and the result is like this:
[u'<tr>\r\n
<td>address1</td>\r\n
<td>phone1</td>\r\n
<td>map1</td>\r\n
</tr>',
u'<tr>\r\n
<td>address1</td>\r\n
<td>telephone1</td>\r\n
<td>map1</td>\r\n
</tr>'...
u'<tr>\r\n
<td>address100</td>\r\n
<td>telephone100</td>\r\n
<td>map100</td>\r\n
</tr>']
now I need to use xpath to analyze this results again.
I want to save the first to address,the second to telephone,and the last one to map
But I can't get it.
Please guide me.Thank you!
Here is code,it's wrong. it will catch another thing.
store = sel.xpath("")
for s in store:
address = s.xpath("//tr/td[1]/text()").extract()
tel = s.xpath("//tr/td[2]/text()").extract()
map = s.xpath("//tr/td[3]/text()").extract()
As you can see in scrappy documentation to work with relative XPaths you have to use .// notation to extract the elements relative to the previous XPath, if not you're getting again all elements from the whole document. You can see this sample in the scrappy documentation that I referenced above:
For example, suppose you want to extract all <p> elements inside <div> elements. First, you would get all <div> elements:
divs = response.xpath('//div')
At first, you may be tempted to use the following approach, which is wrong, as it actually extracts all <p> elements from the document, not only those inside <div> elements:
for p in divs.xpath('//p'): # this is wrong - gets all <p> from the whole document
This is the proper way to do it (note the dot prefixing the .//p XPath):
for p in divs.xpath('.//p'): # extracts all <p> inside
So I think in your case you code must be something like:
for s in store:
address = s.xpath(".//tr/td[1]/text()").extract()
tel = s.xpath(".//tr/td[2]/text()").extract()
map = s.xpath(".//tr/td[3]/text()").extract()
Hope this helps,
I have the plenty of links like this:
<b>Edit issue >></b>
Trying to extract the href' content I use Xpath expression:
//a[contains(#href,'/edit_flat')]
but it returns me null. What am I doing wrong ?
//a[contains(#href,'/edit_flat')] selects a elements anywhere in the document tree that have an href attribute containing the '/edit_flat' string.
These matching elements do have this very "href" attribute, but the XPath expression you are using returns "only" the a elements, if there are any.
To actually return the matching elements' attribute's values, you need an extra step, with / and #href. So what you want is:
//a[contains(#href,'/edit_flat')]/#href
Suggestion:
What you really want is probably to select links which href begin with the substring "/edit_flat", so it's safer to use:
.//a[starts-with(#href,'/edit_flat')]/#href