Xpath expression returns null - xpath

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

Related

Xpath expression (nokogiri) to get tag's child element?

From my xml, I can get this :
<home>
<creditors>
<count>2</count>
</creditors>
</home>
OR even this :
<home>
<creditors>
<moreThan>2</moreThan>
</creditors>
</home>
Which xpath expression can I use to get "<count>2</count>" instead of getting only "2" OR to get "<moreThan>2</moreThan>" instead of getting "2" ?
This XPath,
//creditors/count
will select all count child elements of all creditors elements in the XML document.
Update per OP's request in comments for a single XPath that selects both count and moreThan elements:
This XPath,
//creditors/*[self::count or self::moreThan]
will select all count or moreThan child elements of all creditors elements in the XML document.
Assuming that your xpath expression is OK, you just need to convert the element to string:
doc.xpath("home/creditors/*").to_s
=> "<count>2</count>"
Please check with queries returning more than one element, to make sure that it's desired behaviour.

XPath - Nested path scraping

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

Selenium, PHPUnit: when I try get style attribute I get empty result instead of value

I met a problem when I wanted to get style attribute of element.
$styleValue = $this->getAttribute("//ul#style");
But result of var_dump($styleValue) is
string(1) ";"
But I expected "margin-left: -2432px;"
So, where am I wrong? How can I get style attribute of element?
Your XPath expression is wrong. If you want to read the #style attribute of the <ul/> tag, you have to use two step expressions: stepping into the list, then into the attribute. Each must be seperated using a /, the # only denominates an attribute.
//ul/#style

use YQL with substring-before in xpath

I am trying to get a string before '--' within a paragraph in an html page using the xpath and send it to yql
for example i want to get the date from the following article:
<div>
<p>Date --- the body of the article</p>
</div>
I tried this query in yql:
select * from html where url="article url" and xpath="//div/p/text()/[substring-before(.,'--')]"
but it does not work.
how can I get the date of the article which is before the '--'
You can simply use:
substring-before(//div/p,'--')
Use:
substring-before(/div/p/text(), '--')
This XPath expression evaluates to the string immediately preceding '--' in the first text node in the XML document, that is a child of a p that is a child of the div top element.
In case you want to get this value for every such text node, you have to use an expression like:
substring-before((//div/p/text())[$k], '--')
and evaluate this expression $N times, for $k = 1,2, ..., $N
where $N is count(//div/p/text())
Do note: Try to avoid using the // XPath pseudo-operator always when the structure of the XML document is statically known. Using // usually results in big inefficiency (O(N^2)) that are felt especially painful on big XML documents.

Use Nokogiri to get all nodes in an element that contain a specific attribute name

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]"

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