I have a String as follows:
var _quotedText = "Text1[John;1234]Text2";
Which I want to split into a list as follows:
final List<String> _splitQuotedText = _quotedText.split(RegExp(r"\[([A-z].+);([0-9]+)\]"));
My list comes out as:
['Text1', 'Text2']
But I'd like it to actually comes out as:
['Text1', '[John;1234]Text2']
Meaning the delimiter should be included with the match, and ideally the delimiter would be considered the "start" for a given item.
Is there any kind of straightforward way to tackle this?
A simple way to do this is to use regex's lookahead function.
A lookahead will match whatever is before your lookahead, we can use it to match the space before your delimiter and use that as a delimiter:
A lookahead looks like this:
(?= ... )
So your new regex looks like this:
(?=\[([A-z].+);([0-9]+)\])
Here is your example:
void main() {
var myString = "Text1[John;1234]Text2";
List<String> myList = myString.split(RegExp(r"(?=\[([A-z].+);([0-9]+)\])"));
print(myList); // ['Text1', '[John;1234]Text2']
}
Related
I want to remove white space within a string like extra spaces in between the words. I have tried trim() method. but it only removes the leading and trailing whitespace I want to remove the spaces in between the string and I want to convert the first letter of the each word to capital .
example : var name = ' Aneesh devala ' to Aneesh Devala
I have tried this answers but it's not suitable for mine.
I hope this code will work for you.
String getCapitalizedName(String name) {
final names = name.split(' ');
String finalName = '';
for (var n in names) {
n.trim();
if (n.isNotEmpty) {
finalName += '${n[0].toUpperCase()}${n.substring(1)} ';
}
}
return finalName.trim();}
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 :)
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 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!
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