I have a String with different value every time:
String words = "My name is Rob Joe";
I want to get only last word
Joe
but every time I don't know how many words the string consists of
String words = "My name is Rob Joe";
var array = words.split(" "); // <-- [My, name, is, Rob, Joe]
print(array.last); // output 'Joe'
In Flutter(Dart), you can get the last word of a string by splitting the string into an array of substrings using the split method and then accessing the last element of the resulting array. Here's an example:
String words = "My name is Rob Joe";
List<String> wordsArray = myString.split(" ");
String lastWord = wordsArray[wordsArray.length - 1];
print(lastWord); // "Joe"
Instead of splitting, you can also use substring and lastIndexOf:
final words = "My name is Rob Joe";
final lastWord = words.substring(words.lastIndexOf(" ") + 1);
print(lastWord);
If you want to find the last word, you should first properly define what a "word" is.
It's clearly obvious here, which is why it's doubly important to write it down, because something else may be just as obvious to someone else.
(Read: Nothing is obvious. Document it all!)
But let's say that a word is a maximal contiguous sequence of ASCII letters.
Then that's what you should look for.
Splitting on space characters works for this string, but won't if you have punctuation, or trailing whitespace, or any number of other complications.
I'd probably use a RegExp:
// Matches a word. If used properly, only matches entire words.
var wordRE = RegExp(r"[a-zA-Z]+");
// Assume at least one word in `words`. Otherwise need more error handling.
var lastWord = wordRe.allMatches(words).last[0]!;
This can be a little inefficient, if the string is long.
Another approach that might be more efficient, depending on the RegExp implementation, is to search backwards:
/// Captures first sequence of [a-zA-Z]+ looking backwards from end.
var lastWordRE = RegExp(r"$(?<=([a-zA-Z]+)[^a-zA-Z]*)");
var lastWord = lastWordRE.firstMatch(words)?[1]!;
If you don't want to rely on RegExps (which are admittedly not that readable, and their performance is not always predictable), you can search for letters manually:
String? lastWord(String words) {
var cursor = words.length;
while (--cursor >= 0) {
if (_isLetter(words, cursor)) {
var start = 0;
var end = cursor + 1;
while (--cursor >= 0) {
if (!_isLetter(words, prev)) {
start = cursor + 1;
break;
}
}
return words.substring(start, end);
}
}
return null;
}
bool _isLetter(String string, int index) {
var char = string.codeUnitAt(index) | 0x20; // lower-case if letter.
return char >= 0x61 /*a*/ && char <= 0x7a /*z*/;
}
But first of all, decide what a word is.
Some very real words in common sentences might contain, e.g., ' or -, but whether they matter to you or not depends on your use-case.
More exotic cases may need you to decide whether"e.g." is one word or two? Is and/or? Is i18n?
Depends on what it'll be used for.
Related
I'm a beginner in dart.
void main() {
var abf = '+37.4054-122.0999/';
var abf2;
abf2 = abf.replaceAll("+"," ");
var abf1 = abf2.split(RegExp('(?=[+-])'));
print (abf1[0]);
print (abf1[1]);
}
The above code splits abf into two values for me
I want to remove the ending '/'. I tried many split methods using other variables but it's not removing the '/' even though its removing the '+'.
It's not really clear what you're trying to do with the split.
But if you're looking the remove the / this should work:
String number = '+37.4054-122.0999/';
number = number.replaceAll("/"," ");
You can create substring from this while you like to remove last element.
String abf = '+37.4054-122.0999/';
final result = abf.substring(0, abf.length - 1);
print(result);
Dart's List class has a built-in removeLast method. Maybe you can try to split the string and then removing the last element:
String str = "str";
String newStr = str.split(''). removeLast().join('');
How do I get the last n-characters in a string?
I've tried using:
var string = 'Dart is fun';
var newString = string.substring(-5);
But that does not seem to be correct
var newString = string.substring(string.length - 5);
Create an extension:
extension E on String {
String lastChars(int n) => substring(length - n);
}
Usage:
var source = 'Hello World';
var output = source.lastChars(5); // 'World'
While #Alexandre Ardhuin is correct, it is important to note that if the string has fewer than n characters, an exception will be raised:
Uncaught Error: RangeError: Value not in range: -5
It would behoove you to check the length before running it that way
String newString(String oldString, int n) {
if (oldString.length >= n) {
return oldString.substring(oldString.length - n)
} else {
// return whatever you want
}
}
While you're at it, you might also consider ensuring that the given string is not null.
oldString ??= '';
If you like one-liners, another options would be:
String newString = oldString.padLeft(n).substring(max(oldString.length - n, 0)).trim()
If you expect it to always return a string with length of n, you could pad it with whatever default value you want (.padLeft(n, '0')), or just leave off the trim().
At least, as of Dart SDK 2.8.1, that is the case. I know they are working on improving null safety and this might change in the future.
var newString = string.substring((string.length - 5).clamp(0, string.length));
note: I am using clamp in order to avoid Value Range Error. By that you are also immune to negative n-characters if that is somehow calculated.
In fact I wonder that dart does not have such clamp implemented within the substring method.
If you want to be null aware, just use:
var newString = string?.substring((string.length - 5).clamp(0, string.length));
I wrote my own solution to get any no of last n digits from a string of unknown length, for example the 5th to the last digit from an n digit string,
String bin='408 408 408 408 408 1888';// this is the your string
// this function is to remove space from the string and then reverse the
string, then convert it to a list
List reversed=bin.replaceAll(" ","").split('').reversed.toList();
//and then get the 0 to 4th digit meaning if you want to get say 6th to last digit, just pass 0,6 here and so on. This second reverse function, return the string to its initial arrangement
var list = reversed.sublist(0,4).reversed.toList();
var concatenate = StringBuffer();
// this function is to convert the list back to string
list.forEach((item){
concatenate.write(item);
});
print(concatenate);// concatenate is the string you need
I'm using Regex to search for a word in a textView. I implemented a textField and two switch as options (Whole words and Match case). All work fine when you enter a plain word in the search filed but I get an error when I enter a special character like \ or *.
The error I get is like this one:
Error Domain=NSCocoaErrorDomain Code=2048 "The value “*” is invalid." UserInfo={NSInvalidValue=*}
Is there a way to avoid this problem and have the code handle all the text like plain text?
Because I would like to search also for special characters, I would like to prefer to not interdict to enter them. At the beginning I though to programmatically add an escape backslash to all special character before to perform a search, but maybe there are some more smart approaches?
Here is the code I'm using (based on this tutorial: NSRegularExpression Tutorial: Getting Started)
struct SearchOptions {
let searchString: String
var replacementString: String
let matchCase: Bool
let wholeWords: Bool
}
extension NSRegularExpression {
convenience init?(options: SearchOptions) {
let searchString = options.searchString
let isCaseSensitive = options.matchCase
let isWholeWords = options.wholeWords
// handle case sensitive option
var regexOption: NSRegularExpressionOptions = .CaseInsensitive
if isCaseSensitive { // if it is match case remove case sensitive option
regexOption = []
}
// put the search string in the pattern
var pattern = searchString
// if it's whole word put the string between word boundary \b
if isWholeWords {
pattern = "\\b\(searchString)\\b" // the second \ is used as escape
}
do {
try self.init(pattern: pattern, options: regexOption)
} catch {
print(error)
}
}
}
You may use NSRegularExpression.escapedPatternForString:
Returns a string by adding backslash escapes as necessary to protect any characters that would match as pattern metacharacters.
Thus, you need
var pattern = NSRegularExpression.escapedPatternForString(searchString)
Also, note that this piece:
if isWholeWords {
pattern = "\\b\(searchString)\\b"
might fail if a user inputs (text) and wishes to search for it as a whole word. The best way to match whole words is by means of lookarounds disallowing word chars on both ends of the search word:
if isWholeWords {
pattern = "(?<!\\w)" + NSRegularExpression.escapedPatternForString(searchString) + "(?!\\w)"
Is there a function to capitalize each word in a string or is this a manual process?
For e.g. "bob is tall"
And I would like "Bob Is Tall"
Surely there is something and none of the Swift IOS answers I have found seemed to cover this.
Are you looking for capitalizedString
Discussion
A string with the first character in each word changed to its corresponding uppercase value, and all remaining characters set to their corresponding lowercase values.
and/or capitalizedStringWithLocale(_:)
Returns a capitalized representation of the receiver using the specified locale.
For strings presented to users, pass the current locale ([NSLocale currentLocale]). To use the system locale, pass nil.
Swift 3:
var lowercased = "hello there"
var stringCapitalized = lowercased.capitalized
//prints: "Hello There"
Since iOS 9 a localised capitalization function is available as capitalised letters may differ in languages.
if #available(iOS 9.0, *) {
"istanbul".localizedCapitalizedString
// In Turkish: "İstanbul"
}
An example of the answer provided above.
var sentenceToCap = "this is a sentence."
println(sentenceToCap.capitalizedStringWithLocale(NSLocale.currentLocale()) )
End result is a string "This Is A Sentence"
For Swift 3 it has been changed to capitalized .
Discussion
This property performs the canonical (non-localized) mapping. It is suitable for programming operations that require stable results not depending on the current locale.
A capitalized string is a string with the first character in each word changed to its corresponding uppercase value, and all remaining characters set to their corresponding lowercase values. A “word” is any sequence of characters delimited by spaces, tabs, or line terminators (listed under getLineStart(_:end:contentsEnd:for:)). Some common word delimiting punctuation isn’t considered, so this property may not generally produce the desired results for multiword strings.
Case transformations aren’t guaranteed to be symmetrical or to produce strings of the same lengths as the originals. See lowercased for an example.
There is a built in function for that
nameOfString.capitalizedString
This will capitalize every word of string. To capitalize only the first letter you can use:
nameOfString.replaceRange(nameOfString.startIndex...nameOfString.startIndex, with: String(nameOfString[nameOfString.startIndex]).capitalizedString)
Older Thread
Here is what I came up with that seems to work but I am open to anything that is better.
func firstCharacterUpperCase(sentenceToCap:String) -> String {
//break it into an array by delimiting the sentence using a space
var breakupSentence = sentenceToCap.componentsSeparatedByString(" ")
var newSentence = ""
//Loop the array and concatinate the capitalized word into a variable.
for wordInSentence in breakupSentence {
newSentence = "\(newSentence) \(wordInSentence.capitalizedString)"
}
// send it back up.
return newSentence
}
or if I want to use this as an extension of the string class.
extension String {
var capitalizeEachWord:String {
//break it into an array by delimiting the sentence using a space
var breakupSentence = self.componentsSeparatedByString(" ")
var newSentence = ""
//Loop the array and concatinate the capitalized word into a variable.
for wordInSentence in breakupSentence {
newSentence = "\(newSentence) \(wordInSentence.capitalizedString)"
}
// send it back up.
return newSentence
}
}
Again, anything better is welcome.
Swift 5 version of Christopher Wade's answer
let str = "my string"
let result = str.capitalized(with: NSLocale.current)
print(result) // prints My String
Is there a number format that would produce a localized number without the thousands separator?
Globalize.format("1000.12", "n?" )
I realize I could do:
Globalize.culture().numberFormat[","]="";
But I have some fields where I want it off and some where it should be on. For example... If the value is:
1000.123 -> Want it to show formatted to 1000,12 or 1000.12 depending on locale..But without the thousands separator.
You can use the "d" format instead of the "n" format to exclude the thousands separator.
Globalize.format(1000.12, "d");
Edit
Note that this will only work if you don't care about the decimal part.
If you care about the decimal part, as far as I know, you can't exclude the thousands separator except through one of the following methods:
Set the thousands character in the culture object to an empty string:
Globalize.culture().numberFormat[","] = "";
Globalize.format(1000.12, "n");
You could turn this into a utility function fairly easily:
function formatNumberNoThousands(num, format, culture) {
var numberFormat = Globalize.cultures[culture || Globalize.culture().name].numberFormat,
thousands = numberFormat[","];
numberFormat[","] = "";
try { return Globalize.format(num, format, culture); }
finally { numberFormat[","] = thousands; }
}
Perform a replace on the string result of the format:
Globalize.format(1000.12, "d").replace(new RegExp("\\" + Globalize.culture().numberFormat[","], "g"), "");
Which can also be easily turned into a utility function:
function formatNumberNoThousands(num, format, culture) {
return Globalize.format(num, format).replace(new RegExp("\\" + Globalize.culture(culture).numberFormat[","], "g"), "");
}
With this approach, if you know there will never be more than one thousands character in the formatted result you can remove the regexp. Otherwise if you plan on using this a lot or inside of a loop, you will want to cache the regexp and re-use it.