CODEWITHSUNDEEP
rubyregexcucumberwebrat
ac_
ac_

Reputation: 1167

Is there a simpler way to write the following cucumber test?

I'm usisng cucmber to test a php app and it's working quite well actually.

I have a cucmber feature that uses the following step to check for the presence of a link in a page:

Then /^I should see a link that contains "(.+)"$/ do |link|
    assert !!(response_body =~ 
      /<a ([\w\.]*="[\w\.]*" )*href="(http:\/\/)?([\w\.\/]*)?(#{link})/m), response_body
end

Now, this works but it's butt ugly and complicated.

Originally I tried using the xpath thing:

response_body.should have_xpath("//a[@href=\"#{link}\"]")

But then if I check for a link to 'blah.com' then it won't match 'http://blah.com" - which kind of defeats the whole purpose of the test. Hence the reason I switched to regex.

So is there a simpler way to write the test which doesn't rely on complicated regular expressions?

Cheers.

EDIT: After lots of hair-pulling... I did find a less messy way to find images on my page:

response_body.should include(image)

Where the image string is set to something like 'myimage.png' - of course, this will break if the actual text 'myimage.png' is on the page and not the image. There must be a better way. I was considering Hpricot to see if I can parse the html and pull out the attribute I want to test, then test that with a regex but that all seems so... bloated.

Upvotes: 2

Views: 509

Answers (2)

ac_
ac_

Reputation: 1167

Then /^I should see the image "(.+)"$/ do |image|    
  response_body.should have_selector("img[src*='#{image}']")
end

Then /^I should see a link that contains "(.+)"$/ do |link|    
  response_body.should have_selector("a[href*='#{link}']")
end

Thanks AlistairH - your advice worked! :)

I still don't undestand why it's searching html with a css selector syntax but maybe that was just a design choice they guy who wrote it took because it looked easier than regex...? I don't know.

Upvotes: 0

AlistairH
AlistairH

Reputation: 3229

Something like this should work:

response_body.should have_css("a[href*='#{link}']")

See this for details: http://www.w3.org/TR/css3-selectors/#attribute-substrings

EDIT: Looks like the equivalent method for webrat is have_selector, so:

response_body.should have_selector("a[href*='#{link}']")

Upvotes: 1

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