I am new to cucumber with capybara. I got an application to test whose flow is:'after submitting a form, an email will be sent to the user which contains the link to another app. In order to access the app we have to open the mail and click the link, which will redirect to the app.'. I don't have access to the mail Id. Is there any way to extract that link and continue with the flow?
Please, give some possible way to do it.
Regards,
Abhisek Das
In your test, use whatever means you need in order to trigger the sending of the email by your application. Once the email is sent, use a regular expression to find the URL from the link within the email body (note this will work only for an email that contains a single link), and then visit the path from that URL with Capybara to continue with your test:
path_regex = /(?:"https?\:\/\/.*?)(\/.*?)(?:")/
email = ActionMailer::Base.deliveries.last
path = email.body.match(path_regex)[1]
visit(path)
Regular expression explained
A regular expression (regex) itself is demarcated by forward slashes, and this regex in particular consists of three groups, each demarcated by pairs of parentheses. The first and third groups both begin with ?:, indicating that they are non-capturing groups, while the second is a capturing group (no ?:). I will explain the significance of this distinction below.
The first group, (?:"https?\:\/\/.*?), is a:
non-capturing group, ?:
that matches a single double quote, "
we match a quote since we anticipate the URL to be in the href="..." attribute of a link tag
followed by the string http
optionally followed by a lowercase s, s?
the question mark makes the preceding match, in this case s, optional
followed by a colon and two forward slashes, \:\/\/
note the backslashes, which are used to escape characters that otherwise have a special meaning in a regex
followed by a wildcard, .*?, which will match any character any number of times up until the next match in the regex is reached
the period, or wildcard, matches any character
the asterisk, *, repeats the preceding match up to an unlimited number of times, depending on the successive match that follows
the question mark makes this a lazy match, meaning the wildcard will match as few characters as possible while still allowing the next match in the regex to be satisfied
The second group, (\/.*?) is a capturing group that:
matches a single forward slash, \/
this will match the first forward slash after the host portion of the URL (e.g. the slash at the end of http://www.example.com/) since the slashes in http:// were already matched by the first group
followed by another lazy wildcard, .*?
The third group, (?:"), is:
another non-capturing group, ?:
that matches a single double quote, "
And thus, our second group will match the portion of the URL starting with the forward slash after the host and going up to, but not including, the double quote at the end of our href="...".
When we call the match method using our regex, it returns an instance of MatchData, which behaves much like an array. The element at index 0 is a string containing the entire matched string (from all of the groups in the regex), while elements at subsequent indices contain only the portions of the string matched by the regex's capturing groups (only our second group, in this case). Thus, to get the corresponding match of our second group—which is the path we want to visit using Capybara—we grab the element at index 1.
You can use Nokogiri to parse the email body and find the link you want to click.
Imagine you want to click a link Change my password:
email = ActionMailer::Base.deliveries.last
html = Nokogiri::HTML(email.html_part.body.to_s)
target_url = html.at("a:contains('Change my password')")['href']
visit target_url
I think this is more semantic and robust that using regular expressions. For example, this would work if the email has many links.
If you're using or willing to use the capybara-email gem, there's now a simpler way of doing this. Let's say you've generated an email to recipient#email.com, which contains the link 'fancy link'.
Then you can just do this in your test suite:
open_email('recipient#email.com') # Allows the current_email method
current_email.click_link 'fancy link'
Related
I am trying to write a cypress test to confirm that the user is brought to the correct page after submitting a form. But I am having trouble using the regex when the digit is nested in the URL... see below for what I am looking to test for.
cy.url().should('match', /some-cool-page/with-stuff/ALLOW ANY DIGIT HERE/final/)
It should be just a regex, nothing special in Cypress.
cy
.url()
.should('match', /\/some-cool-page\/with-stuff\/\d+\/final\//)
Regex is surrounded by //, so if you want to match /, it needs to be escaped.
A digit could be matched with \d, if you want to match one or more, you add + after it.
I have ruby app that uses first matched string by regex. my_url.match(/my_regex/).first
As a strings I have a list of urls that contain adress or postcode and from each of them I need to extract postocode or adreess by using regex
Samples of urls:
http://www.adresses.co.uk/avon/bath-city
http://www.adresses.co.uk/postcode/rm107jj
My regex:
\.co\.uk\/postcode\/([^\/]*)|\.co\.uk\/(?!postcode)([^\/]*\/[^\/]*)
My problem is that for non postcode urls a first matched data by this regex is nil see_on_rubular
How to rewrite or change this reflex so it will skip nil matches or to make first matches non nils. I need to solve it with regex not in ruby coding please.
Here's a regex that captures in group #1 everything after postcode/ if it's present, or else everything after .co.uk/:
\.co\.uk\/(?:postcode\/)?([^\/\n]+(?:\/[^\/\n]+)?)
(DEMO)
Note that this will give unexpected results if there are unwanted path elements at the end of a postcode link, such as:
http://www.adresses.co.uk/postcode/rm107jj/oops
UPDATE: Based on the comments, it looks like you want to match just the last path element. But we can't simply capture the second element, because there might be only one:
http://www.adresses.co.uk/west-midlands
We can, however, make the first element optional:
\.co\.uk\/(?:[^\/\n]+\/)?([^\/\n]+)
Notice how I used a non-capturing group for the optional portion, so the part you want is still captured in group #1.
...
I need create a regular expression for validate the first and second name of a person. The second name is optional because there are people without second name. The space character can be between the two names, but it can not be the end of string
example
"Juan Perez" is valid
"Juan Perez " is invalid because there is a space character the end of the string
You could use the below regex which uses an optional group.
^[A-Za-z]+(?:\s[A-Za-z]+)?$
(?:\s[A-Za-z]+)? optional group will do a match if there is a space followed by one or more alphabets. It won't match the string if there is only a single space exists. And also it asserts that there must be an alphabet present at the last.
DEMO
Update:
If the user name contains not only the firstname but also middlename,lastname ... then you could use the below regex. * repeats the previous token zero or more times where ? after a token (not of * or +) will turn the previous token as an optional one.
^[A-Za-z]+(?:\s[A-Za-z]+)*$
How about a way that doesn't require repeating the char class:
^\b(\s?[[:alpha:]]+)+$
I'm trying to match a pattern in a url that does not include a number.
For example:
/painters/1-joe-bob/dashboard
I would only want to match urls that are the following:
/painters
/painters/string
If the url includes /painters/1-something then there should be no match.
I've been trying the following with no luck:
\/{1}(painters|contractors)\/?[^0-9][a-z]*
This still matches on /painters/ or /contractors/
Please advise.
You can try this regex. It uses a negative lookahead to disallow a match if a number comes after your second forward slash.
^\/(painters|contractors)\/(?![0-9])
Note that if you don't want number anywhere in the string you can use a negative lookahead right at the beginning.
^(?!.*[0-9])\/(painters|contractors)\/
This construct will disallow any string containing numbers.
I am trying to replace a specific pattern in a text string.
That pattern is a href containing the word "sak".
My script currently looks like this:
ccontent=ccontent.sub(/<a .+?href=\"([^\"]+)\"[^\>]*>Sak<\/a>/, '')
The problem is that this replaces the entire string. (the string contains two links).
The problem is somewhere around the `a .+?" symbols, it runs through the link i want to Replace entirely and goes into the next link and replaces that whole link as well.
But I want it to STOP when the first pattern match is reached so that it only erases "sak" link.
How do i make the pattern match stop at the first time it reaches the 'href'?
Your expression is greedy, because .+? will actually keep matching any character as long as the pattern still matches.
Just use the [^>]* character set you're already using at the end of the regex:
ccontent.sub(/<a [^>]*href=\"([^\"]+)\"[^>]*>Sak<\/a>/, '')