I have a variable which has strings stored in the format "domain\alias" and I want to split this in two different strings domain and alias.
I have two solutions for the above case, but none of them are working in my case.
solution 1: separating alias from the string.
for this I am using the code below:
int index = name.IndexOf("\") + 1;
string piece = name.Substring(index);
where name is the variable which stores the string in the format "domain\alias"
This solution doesn't work for '\' however it works in case of '.'
solution 2:
separating domain from the string.
Here I got a solution below:
var domainFormattedString = #"fareast\v-sidmis";
var parts = domainFormattedString.Split('\\');
var domainString = parts[0];
return domainString;
this works, but it needs a string prefixed with #symbol and i have my string stored in the variable name for which this solution doesn't work.
Someone please help me to extract the two substrings from my variable name.
EDIT 1: Thanks all for your help! I figured out the issue...when i explicitly declare a string as: var x = "domian\alias" it creates and issue as \ is treated as a escape character by c# so i had to append # at the beginning. But I got to know that when a string is read from a user, the solution works!
\ has a special meaning so you need to override the escape sequence to be treated as normal character with another escape character.
string input = #"domain\alias";
int inputindex= input.IndexOf("\\");
string domain = input.Substring(0, inputindex);
string alias = input.Substring(inputindex+1);
Hope It helps eventhough better late than never :)
Related
I'm trying to create a raw string that contains three quotes in itself.
The resulting string x should contain something like """abc""".
I've been able to create the string with the following code, but was wondering if there's a simpler solution for this.
val x = """${'"'.toString().repeat(3)}abc${'"'.toString().repeat(3)}"""
There's no easy way to use a triple quote directly in a string literal.
One workaround I've sometimes used is to make an interim variable to hold the triple-quote string.
val quotes = "\"\"\""
val result = "${quotes}abc${quotes}"
I think a simpler way would be to escape them manually, so like:
val x = "\"\"\"abc\"\"\""
In Apex, I want to remove all the special characters in a string except for "+". This string is actually a phone number. I have done the following.
String sampleText = '+44 597/58-31-30';
sampleText = sampleText.replaceAll('\\D','');
System.debug(sampleText);
So, what it prints is 44597583130.
But I want to keep the sign + as it is represents 00.
Can someone help me with this ?
Possible solutions
String sampleText = '+44 597/58-31-30';
// exclude all characters which you want to keep
System.debug(sampleText.replaceAll('[^\\+|\\d]',''));
// list explicitly each char which must be replaced
System.debug(sampleText.replaceAll('/|-| ',''));
Output in both case will be the same
|DEBUG| +44597583130
|DEBUG| +44597583130
Edit
String sampleText = '+0032 +497/+59-31-40';
System.debug(sampleText.replaceAll('(?!^\\+)[^\\d]',''));
|DEBUG|+0032497593140
I have string variable which is
temp = '1\2\3\4'
I would like to add a prefix 'r' to the string variable and get
r'1\2\3\4'
so that I can split the string based on '\'. I tried the following:
r'temp'
'r' + temp
r + temp
But none of the above works. Is there a simple to do it? I'm using python 3. I also tried to encode the string, using
temp.encode('string-escape')
But it returns the following error
LookupError: unknown encoding: string-escape
r is a prefix for string literals. This means, r"1\2\3\4" will not interpret \ as an escape when creating the string value, but keep \ as an actual character in the string. Thus, r"1\2\3\4" will have seven characters.
You already have the string value: there is nothing to interpret. You cannot have the r prefix affect a variable, only a literal.
Your temp = "1\2\3\4" will interpret backslashes as escapes, create the string '1\x02\x03\x04' (a four-character string), then assign this string to the variable temp. There is no way to retroactively reinterpret the original literal.
EDIT: In view of the more recent comments, you do not seem to, in fact, have a string "1\2\3\4". If you have a valid path, you can split it using
path.split(r'\')
or
path.split('\\')
but you probably also don't need that; rather, you may want to split a path into directory and file name, which is best done by os.path functions.
Wouldn't it just be re.escape(temp)?
Take for example the use case of trying to generate a pattern on the fly involving word boundaries. Then you can do this
r'\b' + re.escape(temp) + r'\b'
just to prefix r in variable in search, Please do this r+""+temp.
e.g.-
import re
email_address = 'Please contact us at: support#datacamp.com'
searchString = "([\w\.-]+)#([\w\.-]+)"
re.serach(r""+searchString, email_address)
I have the input as
var = primarynode.domain.local
and now I Need only primarynode from it.
I was looking both split and tokenize but not able to do it in one line code. does anyone know how to do it in one line code?
Well assuming that you want to just get the first word(before . )
from the input string.
You can use the tokenize operator of the String
If you have
def var = "primarynode.domain.local"
then you can do
def firstValue = var.tokenize(".")[0]
println firstValue
output
primarynode
The split method works, you just have to be aware that the argument is a regular expression and not a plain String. And since "." means "any character" in a regular expression, you'll need to escape it...
var = 'primarynode.domain.local'.split(/\./)[0]
...or use a character class (the "." is not special inside a character class)
var = 'primarynode.domain.local'.split(/[.]/)[0]
In Swift I can create a String variable such as this:
let s = "Hello\nMy name is Jack!"
And if I use s, the output will be:
Hello
My name is Jack!
(because the \n is a linefeed)
But what if I want to programmatically obtain the raw characters in the s variable? As in if I want to actually do something like:
let sRaw = s.raw
I made the .raw up, but something like this. So that the literal value of sRaw would be:
Hello\nMy name is Jack!
and it would literally print the string, complete with literal "\n"
Thank you!
The newline is the "raw character" contained in the string.
How exactly you formed the string (in this case from a string literal with an escape sequence in source code) is not retained (it is only available in the source code, but not preserved in the resulting program). It would look exactly the same if you read it from a file, a database, the concatenation of multiple literals, a multi-line literal, a numeric escape sequence, etc.
If you want to print newline as \n you have to convert it back (by doing text replacement) -- but again, you don't know if the string was really created from such a literal.
You can do this with escaped characters such as \n:
let secondaryString = "really"
let s = "Hello\nMy name is \(secondaryString) Jack!"
let find = Character("\n")
let r = String(s.characters.split(find).joinWithSeparator(["\\","n"]))
print(r) // -> "Hello\nMy name is really Jack!"
However, once the string s is generated the \(secondaryString) has already been interpolated to "really" and there is no trace of it other than the replaced word. I suppose if you already know the interpolated string you could search for it and replace it with "\\(secondaryString)" to get the result you want. Otherwise it's gone.