I want to replace part of the string with asterisk (* sign).
How can I achieve that? Been searching around but I can't find a solution for it.
For example, I getting 0123456789 from backend, but I want to display it as ******6789 only.
Please advise.
Many thanks.
Try this:
void main(List<String> arguments) {
String test = "0123456789";
int numSpace = 6;
String result = test.replaceRange(0, numSpace, '*' * numSpace);
print("original: ${test} replaced: ${result}");
}
Notice in dart the multiply operator can be used against string, which basically just creates N version of the string. So in the example, we are padding the string 6 times with'*'.
Output:
original: 0123456789 replaced: ******6789
try using replaceRange. It works like magic, no need for regex. its replaces your range of values with a string of your choice.
//for example
prefixMomoNum = prefs.getString("0267268224");
prefixMomoNum = prefixMomoNum.replaceRange(3, 6, "****");
//Output 026****8224
You can easily achieve it with a RegExp that matches all characters but the last n char.
Example:
void main() {
String number = "123456789";
String secure = number.replaceAll(RegExp(r'.(?=.{4})'),'*'); // here n=4
print(secure);
}
Output: *****6789
Hope that helps!
Related
For interaction with an API, I need to pass the course code in <string><space><number> format. For example, MCTE 2333, CCUB 3621, BTE 1021.
Yes, the text part can be 3 or 4 letters.
Most users enter the code without the space, eg: MCTE2333. But that causes error to the API. So how can I add a space between string and numbers so that it follows the correct format.
You can achieve the desired behaviour by using regular expressions:
void main() {
String a = "MCTE2333";
String aStr = a.replaceAll(RegExp(r'[^0-9]'), ''); //extract the number
String bStr = a.replaceAll(RegExp(r'[^A-Za-z]'), ''); //extract the character
print("$bStr $aStr"); //MCTE 2333
}
Note: This will produce the same result, regardless of how many whitespaces your user enters between the characters and numbers.
Try this.You have to give two texfields. One is for name i.e; MCTE and one is for numbers i.e; 1021. (for this textfield you have to change keyboard type only number).
After that you can join those string with space between them and send to your DB.
It's just like hack but it will work.
Scrolling down the course codes list, I noticed some unusual formatting.
Example: TQB 1001E, TQB 1001E etc. (With extra letter at the end)
So, this special format doesn't work with #Jahidul Islam's answer. However, inspired by his answer, I manage to come up with this logic:
var code = "TQB2001M";
var i = course.indexOf(RegExp(r'[^A-Za-z]')); // get the index
var j = course.substring(0, i); // extract the first half
var k = course.substring(i).trim(); // extract the others
var formatted = '$j $k'.toUpperCase(); // combine & capitalize
print(formatted); // TQB 1011M
Works with other formats too. Check out the DartPad here.
Here is the entire logic you need (also works for multiple whitespaces!):
void main() {
String courseCode= "MMM 111";
String parsedCourseCode = "";
if (courseCode.contains(" ")) {
final ensureSingleWhitespace = RegExp(r"(?! )\s+| \s+");
parsedCourseCode = courseCode.split(ensureSingleWhitespace).join(" ");
} else {
final r1 = RegExp(r'[0-9]', caseSensitive: false);
final r2 = RegExp(r'[a-z]', caseSensitive: false);
final letters = courseCode.split(r1);
final numbers = courseCode.split(r2);
parsedCourseCode = "${letters[0].trim()} ${numbers.last}";
}
print(parsedCourseCode);
}
Play around with the input value (courseCode) to test it - also use dart pad if you want. You just have to add this logic to your input value, before submitting / handling the input form of your user :)
I have read many questions on stackoverflow and didn't get what I want. I have a String with value: "1.000,00" and I want to make it a double and look exactly like this "1000.0" but I can't do it right. Using the code below, I just get "1.0" and this is incorrect. How to do this in the right way? I am using Flutter.
String t = "1.000,00";
double f = NumberFormat().parse(t);
userIncome = f;
You just need to set the defaultLocale property.
Intl.defaultLocale = 'pt_BR';
String t = "1.000,00";
double f = NumberFormat().parse(t);
print(f); // Prints 1000.0
In your case, is not just a "number", is a currency.
A simple way to deal with it is to convert into a number and parse:
String t = "1.000,00";
// the order of replaces here is important
String formated = t.replace(".", "").replace(",", ".");
userIncome = Double.valueOf(formated);
But, if your application is focus on currency you should take a look in this:
https://www.baeldung.com/java-money-and-currency
I have this Flutter bit of code here, which is a large String. It would be different every time, but the format would stay the same since it's a template:
"William\nWilliam description here...\n$^170^ usd" + Uuid().v4()
I want to extract the 170 part, and then convert it to interger, so I can remove it from list of ints. I have tried a lot of code, but it isn't working for a few reasons, one is I can't extract the actual number from the String between the ^ and ^, and then I can't convert it to interger. Here's the try function (incomplete).
deleteSumItem(item) {
final regEx = RegExp(r'\^\d+(?:\^\d+)?'); //not sure if this is right regex for the String template
final priceValueMatch = regEx.firstMatch(item); //this doesn't return the particular number extracted
_totalPrice.remove(priceValueMatch); //i get error here that it isn't a int
_counter = _counter - priceValueMatch; //then remove it from interger as int
}
The function would take that String ("William\nWilliam description here...\n$^170^ usd" + Uuid().v4()) template (the number would be different between the ^ ^, but the template is same), then convert it to interger and remove from list as int.
Try the following:
void main() {
RegExp regExp = RegExp(r'\^(\d+)\^');
String input = r"William\nWilliam description here...\n$^170^ usd";
String match = regExp.firstMatch(input).group(1);
print(match); // 170
int number = int.parse(match);
print(number); // 170
}
I have changed the RegExp so it does correctly capture the number in its own capture group. It looked like you got a little confused in the process of creating the RegExp but it could also be I am missing some details about the problem.
So I have a quite interesting exercise I've been trying to solve for a while now, but haven't come up with a reliable solution, so I thought you guys could help me out. I have this String that is composed of few random custom parts, for example:
"William\nWilliam description here...\n$170.00 usd") + Uuid().v4();
I need to extract the part after '$' and '.', in this case 170, but it can be any number between.
UPDATE
as I said in the last comment, if I wanted to do it in a function (find the price only), it could go something like this:
deleteSumItem(item) {
final regEx = RegExp(r'\$\d+(?:\.\d+)?');
const textToSearch = r'item';
final priceValueMatch = regEx.firstMatch(textToSearch);
print(priceValueMatch.group(0));
_totalPrice.remove(priceValueMatch);
_counter = _counter - priceValueMatch; //getting error here to convert to num
//but int.parse won't work either, then I get the String error
//RegExp can't be assigned to paremeter String
}
Also, this function returns null for regex, so there is some mistake I'm making, any thoughts?
deleteSumItem(item) {
final regEx = RegExp(r'\1\d+(?:\.\d+)?');
final priceValueMatch = regEx.firstMatch(r'item');
print('THIS IS REGEX: $priceValueMatch');} //priceValueMatch returns null
fix
deleteSumItem(item) {
RegExp regExp = RegExp(r'\^(\d+)\^');
String input = item;
String match = regExp.firstMatch("r" + '"' + input + '"').group(1);
print('Match: $match');
int number = int.parse(match);
print('Number: $number');
_totalPrice.remove(number);
_counter = _counter - number;}
Assuming you can answer 'yes' to the questions in my above comment, you can simply use regular expressions to find the price value in your string:
final regEx = RegExp(r'\$\d+(?:\.\d+)?');
const textToSearch = r'William\nWilliam description here...\n$170.00 cm';
final priceValueMatch = regEx.firstMatch(textToSearch);
print(priceValueMatch.group(0)); // this will print $170.00
The regular expression is looking for a dollar sign \$ followed by 1 or more digits d+ followed by optional decimal point and optional digits behind that decimal (?:\.\d+)?.
This actually ignores a lot of the questions in my above comment. This simply looks for a price value preceded by a dollar sign within the string you give it.
Here is another approach based on your comments. This is assuming the new line characters will always exist
const textToSearch = 'William\nWilliam description here...\n170.00 cm';
final lines = textToSearch.split('\n'); // Split on new line character
// If your template is always the same,
// then your number will be at the start of line 3:
print(lines[2]); // Will print: 170.00 cm
// If you want just your 170 value then:
final regEx = RegExp(r'\d+');
final priceValueMatch = regEx.firstMatch(lines[2]);
final priceInt = int.parse(priceValueMatch.group(0));
print(priceInt); // Will print: 170
I have this string: "C:\Procesos\rrhh\CorteDocumentos\Cortados\10001662-1_20060301_29_1_20190301.pdf" and im trying to get this part : "20190301". The problem is the lenght is not always the same. It would be:
"9001662-1_20060301_4_1_20190301".
I've tried this: item.ToString.Substring(66,8), but it doesn't work sometimes.
What can I do?.
This is a code example of what I said in my comment.
Sub Main()
Dim strFileName As String = ""
Dim di As New DirectoryInfo("C:\Users\Maniac\Desktop\test")
Dim aryFi As FileInfo() = di.GetFiles("*.pdf")
Dim fi As FileInfo
For Each fi In aryFi
Dim arrname() As String
arrname = Split(Path.GetFileNameWithoutExtension(fi.Name), "_")
strFileName = arrname(arrname.Count - 1)
Console.WriteLine(strFileName)
Next
End Sub
You could achieve this using a simple regular expressions, which has the added benefit of including pattern validation.
If you need to get exactly eight numbers from the end of file name (and after an underscore), you can use this pattern:
_(\d{8})\.pdf
And then this VB.NET line:
Regex.Match(fileName, "_(\d{8})\.pdf").Groups(1).Value
It's important to mention that Regex is by default case sensitive, so to prevent from being in a situations where "pdf" is matched and "PDF" is not, the patter can be adjusted like this:
(?i)_(\d{8})\.pdf
You can than use it directly in any expression window:
PS: You should also ensure that System.Text.RegularExpressions reference is in the Imports:
You can achieve it by this way as well :)
Path.GetFileNameWithoutExtension(Str1).Split("_"c).Last
Path.GetFileNameWithoutExtension
Returns the file name of the specified path string without the extension.
so with your String it will return to you - 10001662-1_20060301_29_1_20190301
then Split above String i.e. 10001662-1_20060301_29_1_20190301 based on _ and will return an array of string.
Last
It will return you the last element of an array returned by Split..
Regards..!!
AKsh