I have a string that can be '+', '-', '*' or '/', and two numbers. I need to apply the operation denoted by the string to the numbers. I tried:
op = '+'
(&op.to_sym).call 1, 2
but it won't parse it. please help.
Ruby is not a Polish notation language.
1.send(op, 2)
Related
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 $n return "%", "")
I need to extract a string 'MT/23232' I have written the below code, but
it's not working, Can any one help me here?
'Policy created with MT/1212'
'Policy created with MT/121212'
'Policy created with MT/21212121212'
I have written this code
msg="MT/33235"
id = msg.scan(/MT/\d+/\d+/)[0]
But it's not working for me, Can any one help me to extract this string?
You need to escape the forward slash which exists next to MT in your regex and you don't need to have a forward slash after \d+ . And also i suggest you to add a lookbehind, so that you get a clean result. (?<=\s) Positive lookbehind which asserts that the match must be preceded by a space character.
msg.scan(/(?<=\s)MT\/\d+/)[0]
If you don't care about the preceding character then the below regex would be fine.
msg.scan(/MT\/\d+/)[0]
Example:
> msg = 'Policy created with MT/21212121212'
=> "Policy created with MT/21212121212"
> msg.scan(/(?<=\s)MT\/\d+/)[0]
=> "MT/21212121212"
> msg.match(/(?<=\s)MT\/\d+/)[0]
=> "MT/21212121212"
your_string.scan(/\sMT.*$/).last.strip
If your required substring can be anywhere in the string, then:
your_string.scan(/\bMT\/\d+\b/).last.strip # "\b" is for word boundaries
Or you can specify the acceptable digits this way:
your_string.scan(/\bMT\/[0-9]+\b/).last.strip
Lastly, if the string format is going to remain as you specified, then:
your_string.split.last
I have to check if a 4 character string is all valid hex, I found another question which demonstrates exactly what I want to do but it's Java: Regex to check string contains only Hex characters
How can I accomplish this?
I read the ruby docs for Regular expressions, but I don't understand how to return a true or false based on this match?
In ruby regex \h matches a hex digit and \H matches a non-hex digit.
So
!str[/\H/] is what you're looking for.
if str =~ /^[0-9A-F]+$/
does the trick. If you want case insensitive then:
str =~ /^[0-9A-F]+$/i
pattern = /^[[:xdigit:]]+$/
And then just check:
if pattern === your_string
if /^[[:xdigit:]]+$/ === your_string
or with
if your_string.match?(/^[[:xdigit:]]+$/)
Apparently I still don't understand exactly how it works ...
Here is my problem: I'm trying to match numbers in strings such as:
910 -6.258000 6.290
That string should gives me an array like this:
[910, -6.2580000, 6.290]
while the string
blabla9999 some more text 1.1
should not be matched.
The regex I'm trying to use is
/([-]?\d+[.]?\d+)/
but it doesn't do exactly that. Could someone help me ?
It would be great if the answer could clarify the use of the parenthesis in the matching.
Here's a pattern that works:
/^[^\d]+?\d+[^\d]+?\d+[\.]?\d+$/
Note that [^\d]+ means at least one non digit character.
On second thought, here's a more generic solution that doesn't need to deal with regular expressions:
str.gsub(/[^\d.-]+/, " ").split.collect{|d| d.to_f}
Example:
str = "blabla9999 some more text -1.1"
Parsed:
[9999.0, -1.1]
The parenthesis have different meanings.
[] defines a character class, that means one character is matched that is part of this class
() is defining a capturing group, the string that is matched by this part in brackets is put into a variable.
You did not define any anchors so your pattern will match your second string
blabla9999 some more text 1.1
^^^^ here ^^^ and here
Maybe this is more what you wanted
^(\s*-?\d+(?:\.\d+)?\s*)+$
See it here on Regexr
^ anchors the pattern to the start of the string and $ to the end.
it allows Whitespace \s before and after the number and an optional fraction part (?:\.\d+)? This kind of pattern will be matched at least once.
maybe /(-?\d+(.\d+)?)+/
irb(main):010:0> "910 -6.258000 6.290".scan(/(\-?\d+(\.\d+)?)+/).map{|x| x[0]}
=> ["910", "-6.258000", "6.290"]
str = " 910 -6.258000 6.290"
str.scan(/-?\d+\.?\d+/).map(&:to_f)
# => [910.0, -6.258, 6.29]
If you don't want integers to be converted to floats, try this:
str = " 910 -6.258000 6.290"
str.scan(/-?\d+\.?\d+/).map do |ns|
ns[/\./] ? ns.to_f : ns.to_i
end
# => [910, -6.258, 6.29]
I'm trying to remove non-letters from a string. Would this do it:
c = o.replace(o.gsub!(/\W+/, ''))
Just gsub! is sufficient:
o.gsub!(/\W+/, '')
Note that gsub! modifies the original o object. Also, if the o does not contain any non-word characters, the result will be nil, so using the return value as the modified string is unreliable.
You probably want this instead:
c = o.gsub(/\W+/, '')
Remove anything that is not a letter:
> " sd 190i.2912390123.aaabbcd".gsub(/[^a-zA-Z]/, '')
"sdiaaabbcd"
EDIT: as ikegami points out, this doesn't take into account accented characters, umlauts, and other similar characters. The solution to this problem will depend on what exactly you are referring to as "not a letter". Also, what your input will be.
Keep in mind that ruby considers the underscore _ to be a word character. So if you want to keep underscores as well, this should do it
string.gsub!(/\W+/, '')
Otherwise, you need to do this:
string.gsub!(/[^a-zA-Z]/, '')
That will work most of the cases, except when o initially does not contain any non-letter, in which case gsub! will return nil.
If you just want a replaced string, it can be simpler:
c = o.gsub(/\W+/, '')
Using \W or \w to select or delete only characters won't work. \w means A-Z, a-z, 0-9, and "_":
irb(main):002:0> characters = (' ' .. "\x7e").to_a.join('')
=> " !\"\#$%&'()*+,-./0123456789:;<=>?#ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~"
irb(main):003:0> characters.gsub(/\W+/, '')
=> "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz"
So, stripping using \W preserves digits and underscores.
If you want to match characters use /[A-Za-z]+/, or the POSIX character class [:alpha:], i.e. /[[:alpha:]]+/, or /\p{ALPHA}/.
The final format is the Unicode property for 'A'..'Z' + 'a'..'z' in ASCII, and gets extended when dealing with Unicode, so if you have multibyte characters you should probably use that.
use Regexp#union to create a big matching object
allowed = Regexp.union(/[a-zA-Z0-9]/, " ", "-", ":", ")", "(", ".")
cleanstring = dirty_string.chars.select {|c| c =~ allowed}.join("")
I don't see what that o.replace is in there for if you have a string:
string = 't = 4 6 ^'
And you do:
string.gsub!(/\W+/, '')
You get:
t46
If you want to get rid of the number characters too, you can do:
string.gsub!(/\W+|\d+/, '')
And you get:
t