ruby remove backslash from string

Question

How would I make this a regular string with no slash?

I have:

 "3e265629c7ff56a3a88505062dd526bd\""

I want:

"3e265629c7ff56a3a88505062dd526bd"

Answer 1

What you have:

"3e265629c7ff56a3a88505062dd526bd\""

Is equivalent to:

'3e265629c7ff56a3a88505062dd526bd"'

So to remove that:

string.tr!('"', '')

Remember all special characters are prefixed with backslash. This includes not only things like newline \n or linefeed \r, but also quotation mark " or backslash itself \\.

Answer 2

How is this?

s = "3e265629c7ff56a3a88505062dd526bd\""
s[/\w+/]
# => "3e265629c7ff56a3a88505062dd526bd"

Answer 3

If you want to delete everything except letters and numbers you can use 'tr' function.

For example:

"3e265629c7ff56a3a88505062dd526bd\"".tr('^A-Za-z0-9','')

This function replace everything but letters and numbers with a string with no characters. Here is a reference of the function.

Hope it works for you.

Answer 4

Just another way "3e265629c­7ff56a3a88­505062dd52­6bd\"".del­ete ?"

Answer 5

str = "3e265629c7ff56a3a88505062dd526bd\""
str.gsub(%r{\"}, '')
 => "3e265629c7ff56a3a88505062dd526bd"

Answer 6

Ruby's String#[] is your friend. Starting with:

foo = "3e...bd\""

These are alternate ways to get the value without the trailing embedded quote:

# delete it
foo[-1] = ''
foo['"'] = ''
foo[/"$/] = ''

Or:

# skip it
foo[0..-2]