I'm doing a mobile app using ionic 1, and for printing in thermal printers I'm using the cordova-plugin-datecs-printer plugin.
Everything works ok except with the word wrapping. The printer cut words at the end of the line. Is there a way to adjust or enable word wrap to avoid incomplete words?
Here is my code:
tale += '{center}{b}Le Petit Prince{/b}{br}{br}';
tale += '{left}Once when I was six years old I saw a magnificent picture in a book, called True Stories from Nature, about the primeval forest. It was a picture of a boa constrictor in the act of swallowing an animal. Here is a copy of the drawing.&n the book it said: "Boa constrictors swallow their prey whole, without chewing it. After that they are not able to move, and they sleep through the six months that they need for digestion.';
//Replace & with line breaks
var text1 = tale.replace(/&/g, '{br}{br}');
cordovaPrinter.printText(text1);
And this is the result: :(
Take a look at this snippet, I think it's self explanatory.
var yourString = "The quick brown fox jumps over the lazy dog"; //replace with your string.
var finalString = "";
var maxLength = 16 // maximum number of characters to extract
var count = 0;
var maxTries = ((yourString.length / maxLength) * 2);
while (yourString.length > maxLength) {
count++;
if (count > maxTries) {
break;
}
//trim the string to the maximum length
var trimmedString = yourString.substr(0, maxLength);
//re-trim if we are in the middle of a word
trimmedString = trimmedString.substr(0, Math.min(trimmedString.length, trimmedString.lastIndexOf(" ")));
yourString = yourString.replace(trimmedString + " ", "");
finalString += trimmedString + '<br>';
}
finalString += yourString;
console.log(yourString);
console.log(finalString);
Related
Based on straight SQL searches in a previous app, I am adding CoreData searching to a new app. These searches are in a custom dictionary db that the app contains; this function does the work:
public func wordMatcher (pad: Int, word: Array<String>, substitutes : Set<String> ) {
let context = CoreDataManager.shared.persistentContainer.viewContext
var query: Array<String>
var foundPositions : Set<Int> = []
var searchTerms : Array<String> = []
if word.count >= 4 {
for i in 0..<word.count {
for letter in substitutes {
query = word
query[i] = letter
searchTerms.append(query.joined())
let rq: NSFetchRequest<Word> = Word.fetchRequest()
rq.predicate = NSPredicate(format: "name LIKE %#", query.joined())
rq.fetchLimit = 1
do {
if try context.fetch(rq).count != 0 {
foundPositions.insert(i)
break
}
} catch {
}
}
// do aggregated searchTerms search here instead of individual searches?
}
}
}
The NSFetchRequest focuses on one permutation at a time. But I'm accumulating the search string fragments in the array searchTerms because I don't know if it would be more efficient to construct a single query connected with ORs, and I also don't know how to do that in CoreData.
The focus is on the positions in the original term word: I need to indicate if any given location has at least one of the substitutes as a valid fit. So to implement the aggregate searchTerms approach, a FetchRequest would have to happen for each location in the base term.
A second complication is the one referred to in the title of the question. I am using LIKE because the search term in the FetchRequest could be a substring in a longer word. However, the maximum number of letters is 11, and pad is the starting point of the original term in that field of 11 spaces.
So if pad is 3, then I would need to allow for 0..<pad preceding characters. And because there may be trailing characters, I would also want results with 0..<(11 - (pad + word.count)) alphabetic characters after the last letter in the search term.
Regex seems like one way to do this, but I haven't found a clear example of how to do this in this case, and especially with the multiple search terms (if that's the way to go). The limits of SQLite in the previous version forced constructing multiple queries with increasing numbers of "_" underscores to indicate the padding characters; that tended to really explode the number of queries.
BTW, substitutes is limited to an absolute maximum of 9 values, and in practice is usually below 5, so things are a little more manageable.
I would like to get a grip on this, and so if anyone can provide direction or examples that can make this a reasonably efficient function, the help is appreciated greatly.
EDIT:
I've realized that I need a result for each position in the target string, with cases where the leading and trailing spaces also may need to contain a substitute as well.
So I'm moving to this:
public func wordMatcher (pad: Int, word: Array<String>, substitutes : Set<String> ) {
let context = CoreDataManager.shared.persistentContainer.viewContext
var pad_ = pad
var query: Array<String>
var foundPositions : Set<Int> = []
let rq: NSFetchRequest<Word> = Word.fetchRequest()
rq.fetchLimit = 1
let subs = "[\(substitutes.joined())]"
// if word.count >= 4 { // because those locations will be blocked off anyway otherwise
let start = pad > 0 ? -1 : 0
let finish = 11 - (pad + word.count) > 0 ? word.count + 1 : word.count
for i in start..<finish {
query = word
var _pad = 11 - (pad + word.count)
if i == -1 {
query = Array(arrayLiteral: subs) + query
pad_ -= 1
} else if i > word.count {
query.append(subs)
_pad -= 1
} else {
pad_ = pad
query[i] = subs
}
let endPad = _pad > 0 ? "{0,\(_pad)}" : ""
let predMatch = ".\(query.joined())\(endPad)"
print(predMatch)
rq.predicate = NSPredicate(format:"position <= %# AND word MATCHES %#", pad_, predMatch)
do {
if try context.fetch(rq).count != 0 {
foundPositions.insert(i)
}
} catch {
}
// }
}
lFreq = foundPositions
}
This relies on a regex substitution, inserted into the original target string. What I'll have to find out is if this is fast enough at the edge cases, but it may not be critical even in the worst case.
predMatch will end up looking something like "ab[xyx]d{0,3}", and I think I can get rid of the position section by changing it to be "{0,2}ab[xyx]d{0,3}". But I guess I'm going to have to try to find out.
I've been studying for a coding exam by doing the HackerRank test cases, for the most part I've been doing well, but I get hung up on some easy cases and you all help me when I can't see the solution. I'm working on this problem:
https://www.hackerrank.com/challenges/ctci-ransom-note
A kidnapper wrote a ransom note but is worried it will be traced back to him. He found a magazine and wants to know if he can cut out whole words from it and use them to create an untraceable replica of his ransom note. The words in his note are case-sensitive and he must use whole words available in the magazine, meaning he cannot use substrings or concatenation to create the words he needs.
Given the words in the magazine and the words in the ransom note, print Yes if he can replicate his ransom note exactly using whole words from the magazine; otherwise, print No.
Input Format
The first line contains two space-separated integers describing the respective values of (the number of words in the magazine) and (the number of words in the ransom note).
The second line contains space-separated strings denoting the words present in the magazine.
The third line contains space-separated strings denoting the words present in the ransom note.
Each word consists of English alphabetic letters (i.e., to and to ).
The words in the note and magazine are case-sensitive.
Output Format
Print Yes if he can use the magazine to create an untraceable replica of his ransom note; otherwise, print No.
Sample Input
6 4
give me one grand today night
give one grand today
Sample Output
Yes
Explanation
All four words needed to write an untraceable replica of the ransom note are present in the magazine, so we print Yes as our answer.
And here is my solution:
import Foundation
func main() -> String {
let v = readLine()!.components(separatedBy: " ").map{Int($0)!}
var a = [String](); var b = [String]()
if v[0] < v[1] { return "No"}
for i in 0 ..< 2 {
if i == 0 {
a = (readLine()!).components(separatedBy: " ")
} else { b = (readLine()!).components(separatedBy: " ") }
}
// Get list of elements that intersect in each array
let filtered = Set(a).intersection(Set(b))
// Map set to set of Boolean where true means set a has enough words to satisfy set b's needs
let checkB = filtered.map{ word in reduceSet(b, word: word) <= reduceSet(a, word: word) }
// If mapped set does not contain false, answer is Yes, else No
return !checkB.contains(false) ? "Yes" : "No"
}
func reduceSet(_ a: [String], word: String) -> Int {
return (a.reduce(0){ $0 + ($1 == word ? 1 : 0)})
}
print(main())
I always time out on three of the 20 test-cases with this solution. So the solution seems to solve all the test cases, but not within their required time constraints. These are great practice, but it's so extremely frustrating when you get stuck like this.
I should note that I use Sets and the Set(a).intersection(Set(b)) because when I tried mapping an array of Strings, half the test-cases timed out.
Any cleaner, or more efficient solutions will be greatly appreciated! Thank you!
Thanks to #Alexander - I was able to solve this issue using NSCountedSet instead of my custom reduce method. It's much cleaner and more efficient. Here is the solution:
import Foundation
func main() -> String {
let v = readLine()!.components(separatedBy: " ").map{Int($0)!}
var a = [String](); var b = [String]()
if v[0] < v[1] { return "No"}
for i in 0 ..< 2 {
if i == 0 {
a = (readLine()!).components(separatedBy: " ")
} else { b = (readLine()!).components(separatedBy: " ") }
}
let countA = NSCountedSet(array: a)
let countB = NSCountedSet(array: b)
let intersect = Set(a).intersection(Set(b))
let check = intersect.map{ countB.count(for: $0) <= countA.count(for: $0) }
return !check.contains(false) ? "Yes" : "No"
}
print(main())
Many thanks!
I took the leisure of making some improvements on your code. I put comments to explain the changes:
import Foundation
func main() -> String {
// Give more meaningful variable names
let firstLine = readLine()!.components(separatedBy: " ").map{Int($0)!}
let (magazineWordCount, ransomNoteWordCount) = (firstLine[0], firstLine[1])
// a guard reads more like an assertion, stating the affirmative, as opposed to denying the negation.
// it also
guard magazineWordCount > ransomNoteWordCount else { return "No" }
// Don't use a for loop if it only does 2 iterations, which are themselves hardcoded in.
// Just write the statements in order.
let magazineWords = readLine()!.components(separatedBy: " ")
let ransomNoteWords = readLine()!.components(separatedBy: " ") //You don't need ( ) around readLine()!
let magazineWordCounts = NSCountedSet(array: magazineWords)
let ransomNoteWordCounts = NSCountedSet(array: ransomNoteWords)
// intersect is a verb. you're looking for the noun, "intersection"
// let intersection = Set(a).intersection(Set(b))
// let check = intersect.map{ countB.count(for: $0) <= countA.count(for: $0) }
// You don't actually care for the intersection of the two sets.
// You only need to worry about exactly the set of words that
// exists in the ransom note. Just check them directly.
let hasWordWithShortage = ransomNoteWordCounts.contains(where: { word in
magazineWordCounts.count(for: word) < ransomNoteWordCounts.count(for: word)
})
// Don't negate the condition of a conditional expression. Just flip the order of the last 2 operands.
return hasWordWithShortage ? "No" : "Yes"
}
print(main())
with the comments removed:
import Foundation
func main() -> String {
let firstLine = readLine()!.components(separatedBy: " ").map{Int($0)!}
let (magazineWordCount, ransomNoteWordCount) = (firstLine[0], firstLine[1])
guard magazineWordCount > ransomNoteWordCount else { return "No" }
let magazineWords = readLine()!.components(separatedBy: " ")
let ransomNoteWords = readLine()!.components(separatedBy: " ")
let magazineWordCounts = NSCountedSet(array: magazineWords)
let ransomNoteWordCounts = NSCountedSet(array: ransomNoteWords)
let hasWordWithShortage = ransomNoteWordCounts.contains{ word in
magazineWordCounts.count(for: word) < ransomNoteWordCounts.count(for: word)
}
return hasWordWithShortage ? "No" : "Yes"
}
print(main())
It's simpler, and much easier to follow. :)
I've created a function that generates a RegularExpression in Swift 3.0. I'm close to what I want, but the backslash is causing me a lot of trouble.
I've looked at Swift Documentation and I thought changing the "\" to \u{005C} or u{005C} would resolve the issue, but it doesn't.
Here's the array I'm feeding my regex generation function:
var letterArray = ["a","","a","","","","","","",""]
Here's the relevant portion of my method:
var outputString = String()
// getMinimumWordLength returns 3
let minimumWordLength = getMinimumWordLength(letterArray: letterArray)
// for the array above, maximumWordLength returns 10
let maximumWordLength = letterArray.count
var index = 0
for letter in letterArray {
if index < minimumWordLength {
if letter as! String != "" {
outputString = outputString + letter.lowercased
} else {
// this puts an extra \ in my regex
outputString = outputString + "\\w" // first \ is an escape character, 2nd one gets read
// this puts an extra backslash in, too
// outputString = outputString + "\u{005C}w"
}
}
index += 1
}
outputString = outputString + ("{\(minimumWordLength),\(maximumWordLength)}$/")
return outputString
My desired output is:
a\wa{3,10}$/
My actual output is:
a\\wa{3,10}$/
If anyone has suggestions what I'm fouling up, I welcome them. Thank you for reading.
When string is printed in debugger, escape character will be displayed. When it is displayed for user, it will not.
Basic question
I have 2 strings. I want to add one string to another? Here's an example:
var secondString= "is your name."
var firstString = "Mike, "
Here I have 2 strings. I want to add firstString to secondString, NOT vice versa. (Which would be: firstString += secondString.)
More detail
I have 5 string
let first = "7898"
let second = "00"
let third = "5481"
let fourth = "4782"
var fullString = "\(third):\(fourth)"
I know for sure that third and fourth will be in fullString, but I don't know about first and second.
So I will make an if statement checking if second has 00. If it does, first and second won't go in fullString. If it doesn't, second will go intofullString`.
Then I will check if first has 00. If it does, then first won't go inside of fullString, and if not, it will go.
The thing is, I need them in the same order: first, second, third fourth. So in the if statement, I need a way to potentially add first and second at the beginning of fullString.
Re. your basic question:
secondString = "\(firstString)\(secondString)"
or
secondString = firstString + secondString
Here is a way to insert string at the beginning "without resetting" per your comment (first at front of second):
let range = second.startIndex..<second.startIndex
second.replaceRange(range, with: first)
Re. your "more detail" question:
var fullString: String
if second == "00" {
fullString = third + fourth
} else if first == "00" {
fullString = second + third + fourth
} else {
fullString = first + second + third + fourth
}
From the Apple documentation:
String values can be added together (or concatenated) with the addition operator (+) to create a new String value:
let string1 = "hello"
let string2 = " there"
var welcome = string1 + string2
// welcome now equals "hello there"
You can also append a String value to an existing String variable with the addition assignment operator (+=):
var instruction = "look over"
instruction += string2
// instruction now equals "look over there"
You can append a Character value to a String variable with the String type’s append() method:
let exclamationMark: Character = "!"
welcome.append(exclamationMark)
// welcome now equals "hello there!"
So you are pretty much free to add these in any way shape or form.
Which includes
secondstring += firststring
Edit to accommodate the new information:
Strings in Swift are mutable which means you can always add to a string in-place without recreating any objects.
Something like (pseudo-code)
if(second != "00")
{
fullstring = second + fullstring
//only do something with first if second != 00
if(first != "00")
{
fullstring = first + fullstring
}
}
So I have an array of elements of type float. And I have a UILabel which is supposed to display each of these elements of the array on DIFFERENT LINES.
The Array is
var history = [Float]()
I used a for loop to go through each element and append it to the UILabel
for(i=0; i<size; i++){
currentString = String(history[i])
result.text = result.text! + "\n" + currentString
}
result is the UILabel.
I tried using \n but it doesn't seem to recognise it. Any solutions for this in Swift. Thanks in advance!
You can try below to solve your issue.
let history = [1.0, 2.0, 3.0, 4.0]
var result = ""
self.lbl.text = ""
for var i=0; i < history.count; i++
{
let currentString = NSString(format: "%.2f", history[i])
self.lbl.text = self.lbl.text! + "\n" + (currentString as String)
}
And Line Number should be 0. I set that from XIB.
Thanks
The first thing I would check is if your label has (it is stupid mistake, but can happen to all of us)
label.numberOfLines = 0
Assuming that is true, instead of building the string by hand, you could use in-built function join(). Sadly, since it is not String array, you can't use it directly, but you have to create string array first
// Get both storages
var history = [Float]() // Filled with your data
var stringHistory = [String]()
// Convert each number to string
for value in history {
stringHistory.append(String(value))
}
// Finally, join array to one string using \n as separator
let finalString = "\n".join(stringHistory)
Hope some of this helps!