I am trying to find if two phone numbers are same or not (Two same phone number may not be in same format , as +11234567890 is same as 1234567890 and 0011234567890)
I tried PhoneNumberUtils.Compare like this:
if(PhoneNumberUtils.compare("+11234567890", "34567890"))
{
Toast.makeText(getApplicationContext(), "Are same", Toast.LENGTH_LONG).show();
}
But it returns true for "+11234567890", "34567890" while they are not same.
Is there any better method to do this?
The best way to solve this problem is using Google's libphonenumber library
PhoneNumberUtil pnu = PhoneNumberUtil.getInstance();
MatchType mt = pnu.isNumberMatch("+11234567890", "34567890");
if( mt == MatchType.NSN_MATCH || mt == MatchType.EXACT_MATCH )
{
Toast.makeText(getApplicationContext(), "are Same" , Toast.LENGTH_LONG).show();
}
if we use MatchType.SHORT_NSN_MATCH it will return same result as PhoneNumberUtils.compare
According to the documentation:
Compare phone numbers a and b, and return true if they're identical
enough for caller ID purposes.
The number is the same because the only difference is the prefix, which is not necessary for compare purposes.
And if you really want to distinguish between a phone number and a phone number plus its a prefix you should string compare method.
String number1 = "+11234567890";
String number2 = "34567890";
number1.compareTo(number2);
Related
This question already has answers here:
Calculate string value in javascript, not using eval
(12 answers)
Closed 4 months ago.
When the text was '2+3+5+1', the logic was easy
Split('+') so the string is converted to an array.
loop over the array and calculate the sum.
check the code below
void main() {
const text = '2+3+5+1';
final array = text.split('+');
int res =0;
for (var i=0; i<= array.length -1; i++){
res+=int.parse(array[i]);;
}
print(array);
print(res);
}
Now this String "2+3-5+1" contains minus.
how to get the right response using split method?
I am using dart.
note: I don't want to use any library (math expression) to solve this exercice.
Use the .replace() method.
text = text.replace("-", "+-");
When you run through the loop, it will calculate (-).
You can split your string using regex text.split(/\+|\-/).
This of course will fail if any space is added to the string (not to mention *, / or even decimal values).
const text = '20+3-5+10';
const arr = text.split(/\+|\-/)
let tot = 0
for (const num of arr) {
const pos = text.indexOf(num)
if (pos === 0) {
tot = parseInt(num)
} else {
switch (text.substr(text.indexOf(num) - 1, 1)) {
case '+':
tot += parseInt(num)
break
case '-':
tot -= parseInt(num)
break
}
}
}
console.log(tot)
I see 2 maybe 3 options, definitely there are hundreds
You don't use split and you just iterate through the string and just add or subtract on the way. As an example
You have '2+3-5+1'. You iterate until the second operator (+ or -) on your case. When you find it you just do the operation that you have iterated through and then you just keep going. You can do it recursive or not, doesn't matter
"2+3-5+1" -> "5-5+1" -> "0+1" -> 1
You use split on + for instance and you get [ '2', '3-5', '1' ] then you go through them with a loop with 2 conditions like
if(isNaN(x)) res+= x since you know it's been divided with a +
if(!isNaN(x)) res+= x.split('-')[0] - x.split('-')[1]
isNaN -> is not a number
Ofc you can make it look nicer. If you have parenthesis though, none of this will work
You can also use regex like split(/[-+]/) or more complex, but you'll have to find a way to know what operation follows each digit. One easy approach would be to iterate through both arrays. One of numbers and one of operators
"2+3-5+1".split(/[-+]/) -> [ '2', '3', '5', '1' ]
"2+3-5+1".split(/[0-9]*/).filter(x => x) -> [ '+', '-', '+' ]
You could probably find better regex, but you get the idea
You can ofc use a map or a switch for multiple operators
im trying to make locals with Chinese letters
local 屁 = p
or
屁 = p
none of those work
any ways to do it?
You can't do this as "屁" is not a valid Lua identifier.
Lua identifiers can only have letters, numbers and underscores and must not start with a number.
However, you can create a table with a key 屁:
local chinese_letters = {
["屁"] = p
}
And access it as chinese_letters["屁"], for example
local chinese_letters = {
["屁"] = 10
}
print(chinese_letters["屁"])
By the way, the correct name for these chinese characters is Hanzi
a = (any random number)
Is there a way could I get the third number (12345) without converting it into a string?
If not, what would be a good way to get it by converting it into a string?
You can use the sub function
function getDigit(value,digitPlace)
return tonumber(tostring(value):sub(digitPlace,digitPlace))
end
This will get the third digit of a as a number:
a = 12345
print(getDigit(a,3))
You can get that using simple maths.
function getDigitInt(value, digit)
-- get rid of the sign
value = math.abs(value)
-- how many digits does the number have?
local numDigits = math.floor(math.log(value, 10)) + 1
-- does the requested digit exist?
if digit > numDigits or digit < 1 then
print("digit does not exist")
return
end
-- return the requested digit
return math.floor(value / 10^(numDigits - digit)) % 10
end
-- test
for i = 0, 8 do print(getDigitInt(1234567, i)) end
Add more error handling as needed. Also this can only handle integers of course. But I'm sure you will find out how to apply this idea to decimals as well.
You can convert the number into array and find the any place easily like blow
public int GetDigitsPlace(int number, int digitPlace) {
string t = number.ToString();
int[] nArr = new int[t.Length];
for(int i = 0; i < nArr.Length; i++) {
nArr[i] = int.Parse(t[i]);
}
return nArr[digitPlace];
}
I want to make sure players can't put letters or symbols in the value, how do I check if it is a number only?
function seed1()
ESX.UI.Menu.CloseAll()
ESX.UI.Menu.Open('dialog', GetCurrentResourceName(), 'amountseed1', {
title = 'Shop'
}, function(data, menu)
local amount = tostring(data.value)
if amount == nil then
...
else
[[What should i put here to check its only contain number ?]]
end
end, function(data, menu)
menu.close()
end)
end
I can put something like this, but maybe this isn't a good way to do that:
else
if amount > 0 and amount < 9999 then
...
else
print('Invalid amount or amount higher than 9999')
end
end
Since you only care about the number, there is no need to convert the value to string:
local amount = tostring(data.value)
-- ^^^^^^^^ partially useless
Instead, go for the number right away:
local amount = tonumber(data.value)
if amount == nil then
-- Not a number
else
-- A number
end
In the end, remember that tonumber attempts to convert the value to a number and returns nil in case of a failure.
Simply replace tostring with tonumber. This will turn strings into numbers if possible, and return nil if it can't.
Keep in mind: tonumber won't just take the largest valid prefix of a string, so tonumber("20 foo") will return nil and not 20. It also supports all ways to write number literals in Lua, so tonumber("2.3e2") will return 230 and tonumber("0xff") will return 255.
I tried this code but the result is just uppercase first letter (not each letter)
String ucwords(String input) {
if (input == null) {
throw new ArgumentError("string: $input");
}
if (input.length == 0) {
return input;
}
return input[0].toUpperCase() + input.substring(1);
}
I suggest you read the documentation for toUpperCase(). It gives an example of exactly what you want to do.
Your code should read:
return input.toUpperCase();
Well, i didn't understand you question completely, but if you want to put each letter in Uppercase, this means all in Uppercase, well you can just use .toUpperCase() like this:
//This will print 'LIKE THIS'
print('like this'.toUpperCase());
But if you want to put an specific letter in UpperCase, you can just use the Text_Tools package:
https://pub.dev/packages/text_tools
Example
//This will put the letter in position 2 in UpperCase, will print 'liKe this'
print(TextTools.toUppercaseAnyLetter(text: 'like this', position: 2));
If this be of any help