I am trying to convert the type of string to long in the following code:
PaymentReceived = String.Format(new CultureInfo("en-IN", true), "{0:n}", t.PaymentReceived),
Here t.PaymentReceived is of type long, and the PaymentReceived is of type string but I want it to be of type long.
I am using this to convert the PaymentReceived value into comma separated value.
I am trying to do as of my knowledge like
PaymentReceived = Convert.ToInt64( String.Format(new CultureInfo("en-IN", true), "{0:n}", t.PaymentReceived))
But the error is Additional information: Input string was not in a correct format.
So please help me with another solution, thank you.
The formatter n, adds additional non-numeric characters. For en-IN culture, that means a number like 1000 ends up as 1,000.00.
The Convert.ToInt64 method requires that the string be 100% numeric, including no period, which might be fine for Convert.ToDecimal, but a long is not a float. Therefore, emphatically, your string is not formatted correctly, and the error is both obvious and correct. I'm not sure what your ultimate goal here is, but it makes no sense to convert a long to a formatted string and then immediately convert it back to a long, anyways.
Assuming you have only the string and you need to format it as a long, then you need to ensure that it's formatted as a long should be. That requires:
Split on the decimal point and take just the left side:
str = str.Split(new[] { '.' })[0];
Replace any commas with empty strings:
str = str.Replace(",", "");
That assumes you know the format will something like 1,000.00. Otherwise, you may want to use a regex to replace all non-numeric characters with an empty string, instead. However, you still need to split on the decimal. Otherwise, if you just removed all non-numeric characters from something like 1,000.00, then you'd end up with 100000, a number 100 times larger than the actual string number. Also, this is all dependent on the culture. Some cultures use , as the decimal separator and . and delimiter in large numbers. If you need to handle various cultures, you'll need to adjust accordingly.
Related
I've been looking for a good way to see if a string of items are all numbers, and thought there might be a way of specifying a range from 0 to 9 and seeing if they're included in the string, but all that I've looked up online has really confused me.
def validate_pin(pin)
(pin.length == 4 || pin.length == 6) && pin.count("0-9") == pin.length
end
The code above is someone else's work and I've been trying to identify how it works. It's a pin checker - takes in a set of characters and ensures the string is either 4 or 6 digits and all numbers - but how does the range work?
When I did this problem I tried to use to_a? Integer and a bunch of other things including ranges such as (0..9) and ("0..9) and ("0".."9") to validate a character is an integer. When I saw ("0-9) it confused the heck out of me, and half an hour of googling and youtube has only left me with regex tutorials (which I'm interested in, but currently just trying to get the basics down)
So to sum this up, my goal is to understand a more semantic/concise way to identify if a character is an integer. Whatever is the simplest way. All and any feedback is welcome. I am a new rubyist and trying to get down my fundamentals. Thank You.
Regex really is the right way to do this. It's specifically for testing patterns in strings. This is how you'd test "do all characters in this string fall in the range of characters 0-9?":
pin.match(/\A[0-9]+\z/)
This regex says "Does this string start and end with at least one of the characters 0-9, with nothing else in between?" - the \A and \z are start-of-string and end-of-string matchers, and the [0-9]+ matches any one or more of any character in that range.
You could even do your entire check in one line of regex:
pin.match(/\A([0-9]{4}|[0-9]{6})\z/)
Which says "Does this string consist of the characters 0-9 repeated exactly 4 times, or the characters 0-9, repeated exactly 6 times?"
Ruby's String#count method does something similar to this, though it just counts the number of occurrences of the characters passed, and it uses something similar to regex ranges to allow you to specify character ranges.
The sequence c1-c2 means all characters between c1 and c2.
Thus, it expands the parameter "0-9" into the list of characters "0123456789", and then it tests how many of the characters in the string match that list of characters.
This will work to verify that a certain number of numbers exist in the string, and the length checks let you implicitly test that no other characters exist in the string. However, regexes let you assert that directly, by ensuring that the whole string matches a given pattern, including length constraints.
Count everything non-digit in pin and check if this count is zero:
pin.count("^0-9").zero?
Since you seem to be looking for answers outside regex and since Chris already spelled out how the count method was being implemented in the example above, I'll try to add one more idea for testing whether a string is an Integer or not:
pin.to_i.to_s == pin
What we're doing is converting the string to an integer, converting that result back to a string, and then testing to see if anything changed during the process. If the result is =>true, then you know nothing changed during the conversion to an integer and therefore the string is only an Integer.
EDIT:
The example above only works if the entire string is an Integer and won’t properly deal with leading zeros. If you want to check to make sure each and every character is an Integer then do something like this instead:
pin.prepend(“1”).to_i.to_s(1..-1) == pin
Part of the question seems to be exactly HOW the following portion of code is doing its job:
pin.count("0-9")
This piece of the code is simply returning a count of how many instances of the numbers 0 through 9 exist in the string. That's only one piece of the relevant section of code though. You need to look at the rest of the line to make sense of it:
pin.count("0-9") == pin.length
The first part counts how many instances then the second part compares that to the length of the string. If they are equal (==) then that means every character in the string is an Integer.
Sometimes negation can be used to advantage:
!pin.match?(/\D/) && [4,6].include?(pin.length)
pin.match?(/\D/) returns true if the string contains a character other than a digit (matching /\D/), in which case it it would be negated to false.
One advantage of using negation here is that if the string contains a character other than a digit pin.match?(/\D/) would return true as soon as a non-digit is found, as opposed to methods that examine all the characters in the string.
How can I test if a certain character of a string variable is a digit in SPSS (and then apply some operations, depending on the result)?
So let's for example say, I have a variable that reflects the street number. Some street numbers have additional character at the end e.g. "12b". Now let's further assume that I extracted the last character (that could be a digit, or the additional letter) into a string variable. After that I'd like to check if this character is a digit or a letter. How can this be done?
I managed to do this with the MAX function, where "mychar" is the character variable to be checked:
COMPUTE digitcheck = (MAX(mychar,"9")="9").
If the content of "mychar" is a digit [0-9] the result of the MAX function will be "9" otherwise the MAX function will return the letter and the equality test fails.
In this way you can also check if a whole string variable contains a letter or not. It looks pretty ugly though, because you have to compare every single character of your string variable.
compute justdigits = (MAX((CHAR.SUBSTR(mystr,1,1), CHAR.SUBSTR(mystr,2,1), CHAR.SUBSTR(mystr,3,1), ..., CHAR.SUBSTR(mystr,n,1),"9")="9").
If you try to turn a letter into a number then it becomes a missing value. Therefore, to test whether a character is a digit, you can do this:
if not missing(number(YourCharacter,f1)) .....
The same test can determine whether a string has only a number in it or not:
compute OnlyNumber=(not missing(number(YourString,f10))).
Note: using the number command on strings will produce warning messages which you can of course ignore.
I have a Ruby string variable with the value 1.14.2.ab3-4.dl0.rhel
However, I want to discard everything after the second decimal so that I get the value as 1.14
I am using the following command:
str.split(".")[0] but it doesn't seem to work
When you split by . on your string you get:
['1', '14', '2', 'ab3-4', 'dl0', 'rhel']
From this you can get the first two items joined by period:
str.split(".")[0..1].join(".")
# or
str.split(".").first(2).join(".")
With a regexp, you could just look for the first number with 2 decimals :
"1.14.2.ab3-4.dl0.rhel"[/\d+\.\d{2}/]
#=> "1.14"
#maxple's answer only works when the substring of interest is at the beginning of the string. As that was not part of the specification (only in the example), I don't think that's a reasonable assumption. (#Eric did not make that assumption.)
There is also ambiguity about your statement, "discard everything after the second decimal". #maxple interpreted that as after the second decimal point (but also discarded the second decimal point), whereas #Eric assumed it meant after the second decimal digit. This is what happens when questions are imprecise.
If the substring is at the beginning of the string, and you mean to discard the second decimal point and everything after, here are two ways to do that.
str = "1.14.2.ab3-4.dl0.rhel"
1. Modify #Eric's regex:
str[/\A\d+\.\d+/]
#=> "1.14"
2. Convert the string to a float and then back to a string:
str.to_f.to_s
#=> "1.14"
#1 returns nil if the desired substring does not exist, whereas #2 returns "0.0". As long as "0.0" is not a valid substring, either can be used to determine if the substring exists, and if it does, return the substring.
You could also use the partition method in String: https://ruby-doc.org/core-2.2.0/String.html#method-i-partition
"1.14.2.ab3-4.dl0.rhel".partition(/\d+\.\d{2}/)[1]
=> "1.14"
I have read a multiline file and converted it to a list with the following code:
Lines = string:tokens(erlang:binary_to_list(Binary), "\n"),
I converted it to a string to do some work on it:
Flat = string:join(Lines, "\r\n"),
I finished working on the string and now I need to convert it back to a multiline list, I tried to repeat the first snippet shown above but that never worked, I tried string:join and that didnt work.. how do i convert it back to a list just like it used to be (although now modified)?
Well that depends on the modifications you made on the flattened string.
string:tokens/2 will always explode a string using the separator you provide. So as long as your transformation preserves a specific string as separator between the individual substrings there should be no problem.
However, if you do something more elaborate and destructive in your transformation then the only way is to iterate on the string manually and construct the individual substrings.
Your first snippet above contains a call to erlang:binary_to_list/1 which first converts a binary to a string (list) which you then split with the call to string:tokens/2 which then join together with string:join/2. The result of doing the tokens then join as you have written it seems to be to convert it from a string containing lines separated by \n into one containing lines separated by \r\n. N.B. that this is a flat list of characters.
Is this what you intended?
What you should do now depends on what you mean by "I need to convert it back to a multiline list". Do you mean everything in a single list of characters (string), or in a nested list of lines where each line is a list of characters (string). I.e. if you ended up with
"here is line 1\r\nhere is line 2\r\nhere is line 3\r\n"
this already is a multiline line list, or do you mean
["here is line 1","here is line 2","here is line 3"]
Note that each "string" is itself a list of characters. What do you intend to do with it afterwards?
You have your terms confused. A string in any language is a sequence of integer values corresponding to a human-readable characters. Whether the representation of the value is a binary or a list does not matter, both are technically strings because of the data they contain.
That being said, you converted a binary string to a list string in your first set of instructions. To convert a list into a binary, you can call erlang:list_to_binary/1, or erlang:iolist_to_binary/1 if your list is not flat. For instance:
BinString = <<"this\nis\na\nstring">>.
ListString = "this\nis\na\nstring" = binary_to_list(BinString).
Words = ["this", "is", "a", "string"] = string:tokens(ListString, "\n").
<<"thisisastring">> = iolist_to_binary(Words).
Rejoined = "this\r\nis\r\na\r\nstring" = string:join(Words, "\r\n").
BinAgain = <<"this\r\nis\r\na\r\nstring">> = list_to_binary(Rejoined).
For your reference, the string module always expects a flat list (e.g., "this is a string", but not ["this", "is", "a", "string"]), except for string:join, which takes a list of flat strings.
I'm using Delphi XE2 and use the following code to enter the letter Y into a bookmark in a Word (2010) template.
Doc.Bookmarks.Item('NS').Range.InsertAfter('Y');
Except in the document, instead of the letter Y, the number 89 appears.
Is the fault likely to be from my code or in the Word document? Any direction gratefully received.
Your literal 'Y' is a character literal rather than a string string literal. The ASCII code for Y is 89.
So, you are passing a Char rather than a string. When Word needs to get a string representation of that integer it simply converts the integer 89 to its textual representation, the string '89'.
To get around the problem you can do this:
var
Text: string;
....
Text := 'Y';
Doc.Bookmarks.Item('NS').Range.InsertAfter(Text);
The idea is that we ensure that we pass a string to InsertAfter() rather than a character. Remember that InsertAfter() receives a variant parameter and so you do need to be careful about the type of the payload stored in the variant.