I am new to ruby. I am trying for a regex pattern matching for my input. My requirement is that my input should strictly adhere to the following format
CHECK ID#<number>
(Eg. my input should be CHECK ID#3213)
How do i frame the pattern for this?
If you want to extract the ID number use this
"CHECK ID#123".scan(/CHECK ID#(\d+)/).last.first.to_i # => 123
Because you just need one result there is not need to use .scan or .match
"CHECK ID#123"[/CHECK ID#(\d+)/, 1].to_i
How about this:
match = "CHECK ID#1221".match /^CHECK ID#(\d+)$/
puts match[1] if match
=> 1221
Related
I have a string like this:
var = "Renewal Quote RQ00041233 (Payment Pending) Policy R38A014294-1"
I have to extract "Payment Pending" from that string using only the information included in another single string.
The following:
var[/\((.*)\)/, 1]
will extract what I want. I can include the string representation of the regex in the string to be given, and construct the regular expression from it using Regexp.new, but I have no way to achieve the information 1 used as the second argument of [].
Without the second argument 1,
regex_string = '\((.*)\)'
var[Regexp.new(regex_string)]
fetches the string "(Payment Pending)"instead of the expected "Payment Pending".
Can someone help me?
Not sure what you are trying to do, but you can get rid of capturing groups using a different regex:
var[/(?<=\().*(?=\))/]
# => "Payment Pending"
or
var[Regexp.new('(?<=\().*(?=\))')]
# => "Payment Pending"
/\((.*)\)/ is just shorthand for Regexp.new('\((.*)\)').
String#[] takes a regex and a capture group as two separate arguments. var[/\((.*)\)/, 1] is var[Regex, 1].
The important thing to realize is 1 is passed to var[], not the regex.
re = Regexp.new('\((.*)\)')
match = var[re, 1]
Note: you might want to require a named capture group rather than a numbered one. It's very easy to accidentally include an extra capture group in a regex.
Assuming there are no nested parenthesis in the string, one way to do that without using a regular expression is as follows.
instance_eval "var[(i=var.index('(')+1)..var.index(')',i)-1]"
#=> "Payment Pending"
See String#index, particularly the reference to the optional second argument, "offset".
Is there any way using regular expressions to match and replace with a "variable string" like...
foo_1_a => bar_1_b
foo_2_a => bar_2_b
foo_3_a => bar_3_b
...
Using some expression with a variable "var" for example
"replace foo_var => [0-9]_a with bar_var_b "
Specifically I'm trying to take in one regex/replacement from command line using Ruby and executing all these replacements. Thanks.
If I understand you correctly, you are looking for back reference replace string. This is usually done by \1 or $1. The number 1 is the previously matched group's order.
So match foo_2_a by foo_(\d+)_a. Here parenthesis creates a group. And its the first group. So replace it with bar_\1_b. \1 will contain 2
More about Back Reference.
Here we go.
result = "foo_1_a".match(/_([0..1])_/){ "bar_#{$1}_b" }
puts result # "bar_1_b"
Apparently I still don't understand exactly how it works ...
Here is my problem: I'm trying to match numbers in strings such as:
910 -6.258000 6.290
That string should gives me an array like this:
[910, -6.2580000, 6.290]
while the string
blabla9999 some more text 1.1
should not be matched.
The regex I'm trying to use is
/([-]?\d+[.]?\d+)/
but it doesn't do exactly that. Could someone help me ?
It would be great if the answer could clarify the use of the parenthesis in the matching.
Here's a pattern that works:
/^[^\d]+?\d+[^\d]+?\d+[\.]?\d+$/
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]
In Ruby, what regex will strip out all but a desired string if present in the containing string? I know about /[^abc]/ for characters, but what about strings?
Say I have the string "group=4&type_ids[]=2&type_ids[]=7&saved=1" and want to retain the pattern group=\d, if it is present in the string using only a regex?
Currently, I am splitting on & and then doing a select with matching condition =~ /group=\d/ on the resulting enumerable collection. It works fine, but I'd like to know the regex to do this more directly.
Simply:
part = str[/group=\d+/]
If you want only the numbers, then:
group_str = str[/group=(\d+)/,1]
If you want only the numbers as an integer, then:
group_num = str[/group=(\d+)/,1].to_i
Warning: String#[] will return nil if no match occurs, and blindly calling nil.to_i always returns 0.
You can try:
$str =~ s/.*(group=\d+).*/\1/;
Typically I wouldn't really worry too much about a complex regex. Simply break the string down into smaller parts and it becomes easier:
asdf = "group=4&type_ids[]=2&type_ids[]=7&saved=1"
asdf.split('&').select{ |q| q['group'] } # => ["group=4"]
Otherwise, you can use regex a bunch of different ways. Here's two ways I tend to use:
asdf.scan(/group=\d+/) # => ["group=4"]
asdf[/(group=\d+)/, 1] # => "group=4"
Try:
str.match(/group=\d+/)[0]
I have following string:
"xxxxx GL=>G0 yyyyyy "
I want to extract GL and G0 using ruby regular expression.
Thanks.
Well, this is rather vague. Do you want to pull out key/value pairs when separated by => ?
The following regexp may suit your needs:
matches = /.*(\w{2})=>(\w{2}).*/.match("xxxxxx GL=>G0 yyyyy ")
puts matches[1] // GL
puts matches[2] // G0
This assumes that your key/values are 2 characters long separated by a => sign. It does not permit spaces between the characters and the => sign. Let me know if this is what you need. Otherwise, provide a more detailed description of what strings you may need to parse.