How do I count the number of tabs in cypress given that the class names of both the tabs are walkme-1-tab-name and walkme-2-tab-name?
As class names are slightly different, I want to count the number of classes which matches such a pattern - in the above example it is 2.
Partial matching requires some special characters in the selector
Starts with, using ^ char
cy.get('[class^="walkme"]')
.should('have.length', 2)
Ends with, using $ char
cy.get('[class$="tab-name"]')
.should('have.length', 2)
Contains, using * char
cy.get('[class*="tab"]')
.should('have.length', 2)
Combination
cy.get('[class^="walkme"][class$="tab-name"]')
.should('have.length', 2)
Related
I'm looking to convert Less mixin calls to their equivalents in Scss:
.mixin(); should become #mixin();
.mixin(0); should become #mixin(0);
.mixin(0; 1; 2); should become #mixin(0, 1, 2);
I'm having the most difficulty with the third example, as I essentially need to match n groups separated by semicolons, and replace those with the same groups separated by commas. I suppose this relies on some sort of repeating groups functionality in regexes that I'm not familiar with.
It's not simply enough to simply replace semicolons within paren - I need a regex that will only match the \.[\w\-]+\(.*\) format of mixins, but obviously with some magic in the second match group to handle the 3rd example above.
I'm doing this in Ruby, so if you're able to provide replacement syntax that's compatible with gsub, that would be awesome. I would like a single regex replacement, something that doesn't require multiple passes to clean up the semicolons.
I suggest adding two capturing groups round the subvalues you need and using an additional gsub in the first gsub block to replace the ; with , only in the 2nd group.
See
s = ".mixin(0; 1; 2);"
puts s.gsub(/\.([\w\-]+)(\(.*\))/) { "##{$1}#{$2.gsub(/;/, ',')}" }
# => #mixin(0, 1, 2);
The pattern details:
\. - a literal dot
([\w\-]+) - Group 1 capturing 1 or more word chars ([a-zA-Z0-9_]) or -
(\(.*\)) - Group 2 capturing a (, then any 0+ chars other than linebreak symbols as many as possible up to the last ) and the last ). NOTE: if there are multiple values, use lazy matching - (\(.*?\)) - here.
Here you go:
less_style = ".mixin(0; 1; 2);"
# convert the first period to #
less_style.gsub! /^\./, '#'
# convert the inner semicolons to commas
scss_style = less_style.gsub /(?<=[\(\d]);/, ','
scss_style
# => "#mixin(0, 1, 2);"
The second regex is using positive lookbehinds. You can read about those here: http://www.regular-expressions.info/lookaround.html
I also use this neat web app to play around with regexes: http://rubular.com/
This will get you a single pass through gsub:
".mixin(0; 1; 2);".gsub(/(?<!\));|\./, ";" => ",", "." => "#")
=> "#mixin(0, 1, 2);"
It's an OR regex with a hash for the replacement parameters.
Assuming from your example that you just want to replace semicolons not following close parens(negative lookbehind): (?<!\));
You can modify/build on this with other expressions. Even add more OR conditions to the regex.
Also, you can use the block version of gsub if you need more options.
I have enabled XPath 2.0 configuration synapse.xpath.dom.failover.enabled=true in synapse.properties but still unable to get string padding done. Is there any expression to achieve it?
Edit :
The length of a particular string needs to be 10 chars, if it is lesser than it, we have to pad it with the special character '%'.
Eg., Input = 'WSO2', after padding it should be 'WSO2%%%%%%'
Thanks in advance
This can be achieved using XPath 1.0 like so, assuming that "WSO2" will be replaced by dynamic input string in the actual implementation :
substring(concat('WSO2', '%%%%%%%%%%'), 1, 10)
The above XPath basically works by concatenating string of 10 specific for-padding characters to the original input string, and then substring the result to get only the first 10 characters. Found this trick in the following XSL question : XSL left-right justification with Padding
To put this in a more generic formula :
substring(concat('input_string', '%%%%....'), 1, n)
input_string : string to which padding operation will applied
% : character used for padding, repeated n times
n : fixed number of characters expected in the output string
The solution from #har07 is fine if you have a reasonable upper bound on the value of n, but if you don't, you can create a string containing '%' repeated $n times using
XPath 3.0: string-join((1 to $n)!"%")
XPath 2.0: string-join(for $x in 1 to $n return "%", "")
I have a string:
'my_array1: ["1445","374","1449","378"], my_array2: ["1445","374", "1449","378"]'
I need to match all sets of digits from my_array2: [...] and count how many of them there.
I need to do something like this with regex and ruby MatchData
string = 'my_array1: ["1445","374", "1449","378"], my_array2: ["1445","374", "1449","378"]'
matches = string.match(/my_array2\:\s[\[,]\"(\d+)\"/)
count_matches = matches.size
Expected result should be 4.
What is the correct way of doing it?
If you are guaranteed that the content of my_array2 is always numeric you could simply use split twice. First you splitby my_array2: [" and then split by ,. This should give you the amount of items you are after.
If you are not guaranteed that, you could still split by my_array2 and instead of splitting again, you use a pattern such as "\d+" (or "\d+(\.\d+)? if you have floating point values) and count.
An example of the expression is available here.
I have a homework-related question.
I have an array of objects. One of the properties of each of these objects is a string. I need to be able to sort the array alphabetically with respect to this string. I have the sort function written; I've tested it with integers in place of the string. The sorting works fine. However, I don't know how to compare two strings alphabetically. How would I do this?
If you don't want to use existing string compare functions, try to use the ASCII values of each letter for comparison. For example, 'A' = 41 and 'B' = 42, so 'B' > 'A'
Thus, if you have 2 strings like
char str1[] = "abc";
char str2[] = "def";
you can compare them positionwise, so you may start first with something like
if(str1[0] < str2[0])
...
if(str1[1] < str2[1])
...
and so on. Of course, you can improve this through using a for or while loop
can you help me with this:
I want a regular expression for my Ruby program to match a word with the below pattern
Pattern has
List of letters ( For example. ABCC => 1 A, 1 B, 2 C )
N Wild Card Charaters ( N can be 0 or 1 or 2)
A fixed word (for example “XY”).
Rules:
Regarding the List of letters, it should match words with
a. 0 or 1 A
b. 0 or 1 B
c. 0 or 1 or 2 C
Based on the value of N, there can be 0 or 1 or 2 wild chars
Fixed word is always in the order it is given.
The combination of all these can be in any order and should match words like below
ABWXY ( if wild char = 1)
BAXY
CXYCB
But not words with 2 A’s or 2 B’s
I am using the pattern like ^[ABCC]*.XY$
But it looks for words with more than 1 A, or 1 B or 2 C's and also looks for words which end with XY, I want all words which have XY in any place and letters and wild chars in any postion.
If it HAS to be a regex, the following could be used:
if subject =~
/^ # start of string
(?!(?:[^A]*A){2}) # assert that there are less than two As
(?!(?:[^B]*B){2}) # and less than two Bs
(?!(?:[^C]*C){3}) # and less than three Cs
(?!(?:[ABCXY]*[^ABCXY]){3}) # and less than three non-ABCXY characters
(?=.*XY) # and that XY is contained in the string.
/x
# Successful match
else
# Match attempt failed
end
This assumes that none of the characters A, B, C, X, or Y are allowed as wildcards.
I consider myself to be fairly good with regular expressions and I can't think of a way to do what you're asking. Regular expressions look for patterns and what you seem to want is quite a few different patterns. It might be more appropriate to in your case to write a function which splits the string into characters and count what you have so you can satisfy your criteria.
Just to give an example of your problem, a regex like /[abc]/ will match every single occurrence of a, b and c regardless of how many times those letters appear in the string. You can try /c{1,2}/ and it will match "c", "cc", and "ccc". It matches the last case because you have a pattern of 1 c and 2 c's in "ccc".
One thing I have found invaluable when developing and debugging regular expressions is rubular.com. Try some examples and I think you'll see what you're up against.
I don't know if this is really any help but it might help you choose a direction.
You need to break out your pattern properly. In regexp terms, [ABCC] means "any one of A, B or C" where the duplicate C is ignored. It's a set operator, not a grouping operator like () is.
What you seem to be describing is creating a regexp based on parameters. You can do this by passing a string to Regexp.new and using the result.
An example is roughly:
def match_for_options(options)
pattern = '^'
pattern << 'A' * options[:a] if (options[:a])
pattern << 'B' * options[:b] if (options[:b])
pattern << 'C' * options[:c] if (options[:c])
Regexp.new(pattern)
end
You'd use it something like this:
if (match_for_options(:a => 1, :c => 2).match('ACC'))
# ...
end
Since you want to allow these "elements" to appear in any order, you might be better off writing a bit of Ruby code that goes through the string from beginning to end and counts the number of As, Bs, and Cs, finds whether it contains your desired substring. If the number of As, Bs, and Cs, is in your desired limits, and it contains the desired substring, and its length (i.e. the number of characters) is equal to the length of the desired substring, plus # of As, plus # of Bs, plus # of Cs, plus at most N characters more than that, then the string is good, otherwise it is bad. Actually, to be careful, you should first search for your desired substring and then remove it from the original string, then count # of As, Bs, and Cs, because otherwise you may unintentionally count the As, Bs, and Cs that appear in your desired string, if there are any there.
You can do what you want with a regular expression, but it would be a long ugly regular expression. Why? Because you would need a separate "case" in the regular expression for each of the possible orders of the elements. For example, the regular expression "^ABC..XY$" will match any string beginning with "ABC" and ending with "XY" and having two wild card characters in the middle. But only in that order. If you want a regular expression for all possible orders, you'd need to list all of those orders in the regular expression, e.g. it would begin something like "^(ABC..XY|ACB..XY|BAC..XY|BCA..XY|" and go on from there, with about 5! = 120 different orders for that list of 5 elements, then you'd need more for the cases where there was no A, then more for cases where there was no B, etc. I think a regular expression is the wrong tool for the job here.