Remove nth character from a string variable in Linux bash shellscript - bash

I have seen many methods for removing some characters based on regular string or special match using tr etc. But I didn't find any way to delete a character based on accurate index in a string.
For example: In var="hello world",I want to delete the 5th char 'o' and then var should be like "hell world". The code should apply to all other normal strings. Thanks in advance.

One method is:
n=5
var="hello world"
var=${var:0:n-1}${var:n}

Related

Replace non-word characters, unless given sequence matches

I have a string like this:
"Jim-Bob's email ###hl###address###endhl### is: jb#example.com"
I want to replace all non-word characters (symbols and whitespace), except the ### delimiters.
I'm currently using:
str.gsub(/[^\w#]+/, 'X')
which yields:
"JimXBobXsXemailX###hl###address###endhl###XisXjb#exampleXcom"
In practice, this is good enough, but it offends me for two reasons:
The # in the email address is not replaced.
The use of [^\w] instead of \W feels sloppy.
How do I replace all non-word characters, unless those characters make up the ###hl### or ###endhl### delimiter strings?
str.gsub(/(###.*?###|\w+)|./) { $1 || "X" }
# => "JimXBobXsXemailX###hl###address###endhl###XisXXjbXexampleXcom"
This approach uses the fact that alternations work like case structure: the first matching one consumes the corresponding string, then no further matching is done on it. Thus, ###.*?### will consume a marker (like ###hl###; nothing else will be matched inside it. We also match any sequence of word characters. If any of those are captured, we can just return them as-is ($1). If not, then we match any other character (i.e. not inside a marker, and not a word character) and replace it with "X".
Regarding your second point, I think you are asking too much; there is no simple way to avoid that.
Regarding the first point, a simple way is to temporarily replace "###" with a character that you will never use (let's say you are using a system without "\r", so that that character is not used; we can use that as a temporal replacement).
"Jim-Bob's email ###hl###address###endhl### is: jb#example.com"
.gsub("###", "\r").gsub(/[^\w\r]/, "X").gsub("\r", "###")
# => "JimXBobXsXemailX###hl###address###endhl###XisXXjbXexampleXcom"

String split by certain condition

I'd like to split this string by ' ' only if it has ':':
"A:Hey B:Are C:You there"
C:You there should be not split. The result should be:
["A:Hey", "B:Are", "C:You there"]
How can I do this?
\s+(?=\S*:)
You can split by this.
See demo.
https://regex101.com/r/hF7zZ1/4
This basically use a lookahead to make sure that the space which is being split upon is followed by non space characters and : .So it will work as you want.

ruby remove variable length string from regular expression leaving hyphen

I have a string such as this: "im# -33.870816,151.203654"
I want to extract the two numbers including the hyphen.
I tried this:
mystring = "im# -33.870816,151.203654"
/\D*(\-*\d+\.\d+),(\-*\d+\.\d+)/.match(mystring)
This gives me:
33.870816,151.203654
How do I get the hyphen?
I need to do this in ruby
Edit: I should clarify, the "im# " was just an example, there can be any set of characters before the numbers. the numbers are mostly well formed with the comma. I was having trouble with the hyphen (-)
Edit2: Note that the two nos are lattidue, longitude. That pattern is mostly fixed. However, in theory, the preceding string can be arbitrary. I don't expect it to have nos. or hyphen, but you never know.
How about this?
arr = "im# -33.2222,151.200".split(/[, ]/)[1..-1]
and arr is ["-33.2222", "151.200"], (using the split method).
now
arr[0].to_f is -33.2222 and arr[1].to_f is 151.2
EDIT: stripped "im#" part with [1..-1] as suggested in comments.
EDIT2: also, this work regardless of what the first characters are.
If you want to capture the two numbers with the hyphen you can use this regex:
> str = "im# -33.870816,151.203654"
> str.match(/([\d.,-]+)/).captures
=> ["33.870816,151.203654"]
Edit: now it captures hyphen.
This one captures each number separetely: http://rubular.com/r/NNP2OTEdiL
Note: Using String#scan will match all ocurrences of given pattern, in this case
> str.scan /\b\s?([-\d.]+)/
=> [["-33.870816"], ["151.203654"]] # Good, but flattened version is better
> str.scan(/\b\s?([-\d.]+)/).flatten
=> ["-33.870816", "151.203654"]
I recommend you playing around a little with Rubular. There's also some docs about regegular expressions with Ruby:
http://www.ruby-doc.org/docs/ProgrammingRuby/html/language.html#UJ
http://www.regular-expressions.info/ruby.html
http://www.ruby-doc.org/core-1.9.3/Regexp.html
Your regex doesn't work because the hyphen is caught by \D, so you have to modify it to catch only the right set of characters.
[^0-9-]* would be a good option.

Using Ruby on a string, how can I slice between two parts of the string using RegEx?

I just want to save the text between two specific points in a string into a variable. The text would look like this:
..."content"=>"The text I want to save to a variable"}]...
I suppose I would have to use scan or slice, but not exactly sure how to pull out just the text without grabbing the RegEx identifiers before and after the text. I tried this, but it didn't work:
var = mystring.slice(/\"content\"\=\>\".\"/)
This should do the job
var = mystring[/"content"=>"(.*)"/, 1]
Note that:
.slice aliases []
none of the characters you escaped are special regexp characters where you're using them
you can "group" the bit you want to keep with ()
.slice / [] take a second parameter to pick a matched group
your_text = '"content"=>"The text I want to save to a variable"'
/"content"=>"(?<hooray>.*)"/ =~ your_text
Afterwards, hooray local variable will be magically set to contain your text. Can be used to set multiple variables.
This regex will match your string:
/\"content\"=>\"(.*)\"/
you can try rubular.com for testing
It looks like you're trying to truncate a sentence. You can split the sentence either on punctuation, or even on words.
mystring.split(".")
mystring.split("word")

Regular expression Unix shell script

I need to filter all lines with words starting with a letter followed by zero or more letters or numbers, but no special characters (basically names which could be used for c++ variable).
egrep '^[a-zA-Z][a-zA-Z0-9]*'
This works fine for words such as "a", "ab10", but it also includes words like "b.b". I understand that * at the end of expression is problem. If I replace * with + (one or more) it skips the words which contain one letter only, so it doesn't help.
EDIT:
I should be more precise. I want to find lines with any number of possible words as described above. Here is an example:
int = 5;
cout << "hello";
//some comments
In that case it should print all of the lines above as they all include at least one word which fits the described conditions, and line does not have to began with letter.
Your solution will look roughly like this example. In this case, the regex requires that the "word" be preceded by space or start-of-line and then followed by space or end-of-line. You will need to modify the boundary requirements (the parenthesized stuff) as needed.
'(^| )[a-zA-Z][a-zA-Z0-9]*( |$)'
Assuming the line ends after the word:
'^[a-zA-Z][a-zA-Z0-9]+|^[a-zA-Z]$'
You have to add something to it. It might be that the rest of it can be white spaces or you can just append the end of line.(AFAIR it was $ )
Your problem lies in the ^ and $ anchors that match the start and end of the line respectively. You want the line to match if it does contain a word, getting rid of the anchors does what you want:
egrep '[a-zA-Z][a-zA-Z0-9]+'
Note the + matches words of length 2 and higher, a * in that place would signel chars too.

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