I have a string in the following form:
'123string89continues...'
What is the most elegant way to replace or trim the leading digits? Note, there can be more or less than 3 digits, but at least one will always be present.
'1string89continues...' # This can happen
'0123456789string89continues...' # This can happen
'string89continues...' # This cannot happen
'123string89continues...'[/\D.*/]
#⇒ "string89continues..."
Try this one
"123asdads".sub(/A\d+/, "")
=> "asdads"
"1asdads".sub(/A\d+/, "")
=> "asdads"
"asdads".sub(/A\d+/, "")
=> "asdads"
You can use slice! to delete a specific portion from a string:
string = '123string89continues...'
string.slice!(/\A\d+/) #=> "123"
string #=> "string89continues..."
Related
I'm trying to display an array of words from a user's post. However the method I'm using treats an apostrophe like whitespace.
<%= var = Post.pluck(:body) %>
<%= var.join.downcase.split(/\W+/) %>
So if the input text was: The baby's foot
it would output the baby s foot,
but it should be the baby's foot.
How do I accomplish that?
Accepted answer is too naïve:
▶ "It’s naïve approach".split(/[^'\w]+/)
#⇒ [
# [0] "It",
# [1] "s",
# [2] "nai",
# [3] "ve",
# [4] "approach"
# ]
this is because nowadays there is almost 2016 and many users might want to use their normal names, like, you know, José Østergaard. Punctuation is not only the apostroph, as you might notice.
▶ "It’s naïve approach".split(/[^'’\p{L}\p{M}]+/)
#⇒ [
# [0] "It’s",
# [1] "naïve",
# [2] "approach"
# ]
Further reading: Character Properties.
Along the lines of mudasobwa's answer, here's what \w and \W bring to the party:
chars = [*' ' .. "\x7e"].join
# => " !\"\#$%&'()*+,-./0123456789:;<=>?#ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~"
That's the usual visible lower-ASCII characters we'd see in code. See the Regexp documentation for more information.
Grabbing the characters that match \w returns:
chars.scan(/\w+/)
# => ["0123456789",
# "ABCDEFGHIJKLMNOPQRSTUVWXYZ",
# "_",
# "abcdefghijklmnopqrstuvwxyz"]
Conversely, grabbing the characters that don't match \w, or that match \W:
chars.scan(/\W+/)
# => [" !\"\#$%&'()*+,-./", ":;<=>?#", "[\\]^", "`", "{|}~"]
\w is defined as [a-zA-Z0-9_] which is not what you want to normally call "word" characters. Instead they're typically the characters we use to define variable names.
If you're dealing with only lower-ASCII characters, use the character-class
[a-zA-Z]
For instance:
chars = [*' ' .. "\x7e"].join
lower_ascii_chars = '[a-zA-Z]'
not_lower_ascii_chars = '[^a-zA-Z]'
chars.scan(/#{lower_ascii_chars}+/)
# => ["ABCDEFGHIJKLMNOPQRSTUVWXYZ", "abcdefghijklmnopqrstuvwxyz"]
chars.scan(/#{not_lower_ascii_chars}+/)
# => [" !\"\#$%&'()*+,-./0123456789:;<=>?#", "[\\]^_`", "{|}~"]
Instead of defining your own, you can take advantage of the POSIX definitions and character properties:
chars.scan(/[[:alpha:]]+/)
# => ["ABCDEFGHIJKLMNOPQRSTUVWXYZ", "abcdefghijklmnopqrstuvwxyz"]
chars.scan(/\p{Alpha}+/)
# => ["ABCDEFGHIJKLMNOPQRSTUVWXYZ", "abcdefghijklmnopqrstuvwxyz"]
Regular expressions always seem like a wonderful new wand to wave when extracting information from a string, but, like the Sorcerer's Apprentice found out, they can create havoc when misused or not understood.
Knowing this should help you write a bit more intelligent patterns. Apply that to what the documentation shows and you should be able to easily figure out a pattern that does what you want.
You can use below RegEx instead of /\W+/
var.join.downcase.split(/[^'\w]+/)
/\W/ refers to all non-word characters, apostrophe is one such non-word character.
To keep the code as close to original intent, we can use /[^'\w]/ - this means that all characters that are not apostrophe and word character.
Running that string through irb with the same split call that you wrote in your comment gets this:
irb(main):008:0> "The baby's foot".split(/\W+/)
=> ["The", "baby", "s", "foot"]
However, if you use split without an explicit delimiter, you get the split you're looking for:
irb(main):009:0> "The baby's foot".split
=> ["The", "baby's", "foot"]
Does that get you what you're looking for?
How do I parse out a string in rails? I have my form for submitting a height. Example: 5'9 I want the comma parsed and the 59 saved within the database
If you want to ignore anything other than numbers, use this regex
"5'9".gsub(/\D/, '')
# => "59"
"5 9".gsub(/\D/, '')
# => "59"
"5 feet 9 inches".gsub(/\D/, '')
# => "59"
'5" 9'.gsub(/\D/, '')
# => "59"
Regex Explanation: \D stands for any character other than a digit.
There are a number of ways to do this. If you want to just remove the quote, you could use:
"5'9".gsub "'", ""
#=> "59"
or
"5'9".split("'").join("")
#=> "59"
If you want to save the 5 and the 9 in different attributes, you could try:
a = "5'9".split("'")
object.feet = a[0]
object.inches = a[1]
If you want to remove everything but the numbers you could use a regex:
"5'9".gsub /[^\d]/, ""
#=> "59"
If you have a different requirement, please update the question to add more detail.
You want to look at the sub or gsub methods
height.gsub! "'", ''
Where sub replaces the first instance, and gsub replaces all instances and you could even do this on the model:
before_validation :remove_apostrophes # or before_save
protected
def remove_apostrophes
self.property.gsub! "'", ''
end
How do I add a space after a certain amount of numbers?
If i have the number 02920555555 and wanted to add a space after the fifth number only, how would I go about that? Could I do this with a gsub and a regular expression? I would like to put it in a helper:
def area_code(tel)
tel.gsub()
end
Thanks
Just use the insert method:
'02920555555'.insert(5, ' ')
# => "02920 555555"
Use '\& ' as replacement string. (\& represents the matched string)
'02920555555'.gsub(/\d{4}/, '\& ')
# => "0292 0555 555"
UPDATE
to add a space after the 5th number only, use sub instead of gsub:
'02920555555'.sub(/\d{5}/, '\& ')
# => "02920 555555"
or using gsub with the pattern ^... (matches only at the beginning of the input string):
'02920555555'.gsub(/^\d{5}/, '\& ')
# => "02920 555555"
If you don't feel comfortable with regex then you can use the insert method:
'123456789'.insert(5, ' ')
The first parameter for the insert method is the position, (the index at which you want to insert) and second parameter is for the character that you want to insert.
So, our code is saying take number 123456789 and insert space at index 5.
output:
"12345 6789"
'0292055555'.gsub(/(\d{5})(\d*)/, "\\1 \\2")
# => 02920 55555
I am saving a price string to my database in a decimal-type column.
The price comes in like this "$ 123.99" which is fine because I wrote a bit of code to remove the "$ ".
However, I forgot that the price may include a comma, so "$ 1,234.99" breaks my code. How can I also remove the comma?
This is my code to remove dollar sign and space:
def price=(price_str)
write_attribute(:price, price_str.sub("$ ", ""))
# possible code to remove comma also?
end
You can get there two ways easily.
String's delete method is good for removing all occurrences of the target strings:
'$ 1.23'.delete('$ ,') # => "1.23"
'$ 123,456.00'.delete('$ ,') # => "123456.00"
Or, use String's tr method:
'$ 1.23'.tr('$, ', '') # => "1.23"
'$ 123,456.00'.tr('$ ,', '') # => "123456.00"
tr takes a string of characters to search for, and a string of characters used to replace them. Consider it a chain of gsub methods, one for each character.
BUT WAIT! THERE'S MORE! If the replacement string is empty, all characters in the search string will be removed.
I have a string and I need to check whether the last character of that string is *, and if it is, I need to remove it.
if stringvariable.include? "*"
newstring = stringvariable.gsub(/[*]/, '')
end
The above does not search if the '*' symbol is the LAST character of the string.
How do i check if the last character is '*'?
Thanks for any suggestion
Use the $ anchor to only match the end of line:
"sample*".gsub(/\*$/, '')
If there's the possibility of there being more than one * on the end of the string (and you want to replace them all) use:
"sample**".gsub(/\*+$/, '')
You can also use chomp (see it on API Dock), which removes the trailing record separator character(s) by default, but can also take an argument, and then it will remove the end of the string only if it matches the specified character(s).
"hello".chomp #=> "hello"
"hello\n".chomp #=> "hello"
"hello\r\n".chomp #=> "hello"
"hello\n\r".chomp #=> "hello\n"
"hello\r".chomp #=> "hello"
"hello \n there".chomp #=> "hello \n there"
"hello".chomp("llo") #=> "he"
"hello*".chomp("*") #=> "hello"
String has an end_with? method
stringvariable.chop! if stringvariable.end_with? '*'
You can do the following which will remove the offending character, if present. Otherwise it will do nothing:
your_string.sub(/\*$/, '')
If you want to remove more than one occurrence of the character, you can do:
your_string.sub(/\*+$/, '')
Of course, if you want to modify the string in-place, use sub! instead of sub
Cheers,
Aaron
You can either use a regex or just splice the string:
if string_variable[-1] == '*'
new_string = string_variable.gsub(/[\*]/, '') # note the escaped *
end
That only works in Ruby 1.9.x...
Otherwise you'll need to use a regex:
if string_variable =~ /\*$/
new_string = string_variable.gsub(/[\*]/, '') # note the escaped *
end
But you don't even need the if:
new_string = string_variable.gsub(/\*$/, '')