How can I get the int value from a string such as 423e - i.e. a string that contains a number but also maybe a letter?
Integer.parseInt() fails since the string must be entirely a number.
Replace all non-digit with blank: the remaining string contains only digits.
Integer.parseInt(s.replaceAll("[\\D]", ""))
This will also remove non-digits inbetween digits, so "x1x1x" becomes 11.
If you need to confirm that the string consists of a sequence of digits (at least one) possibly followed a letter, then use this:
s.matches("[\\d]+[A-Za-z]?")
The NumberFormat class will only parse the string until it reaches a non-parseable character:
((Number)NumberFormat.getInstance().parse("123e")).intValue()
will hence return 123.
Unless you're talking about base 16 numbers (for which there's a method to parse as Hex), you need to explicitly separate out the part that you are interested in, and then convert it. After all, what would be the semantics of something like 23e44e11d in base 10?
Regular expressions could do the trick if you know for sure that you only have one number. Java has a built in regular expression parser.
If, on the other hands, your goal is to concatenate all the digits and dump the alphas, then that is fairly straightforward to do by iterating character by character to build a string with StringBuilder, and then parsing that one.
You can also use Scanner :
Scanner s = new Scanner(MyString);
s.nextInt();
Just go through the string, building up an int as usual, but ignore non-number characters:
int res = 0;
for (int i=0; i < str.length(); i++) {
char c = s.charAt(i);
if (c < '0' || c > '9') continue;
res = res * 10 + (c - '0');
}
Perhaps get the size of the string and loop through each character and call isDigit() on each character. If it is a digit, then add it to a string that only collects the numbers before calling Integer.parseInt().
Something like:
String something = "423e";
int length = something.length();
String result = "";
for (int i = 0; i < length; i++) {
Character character = something.charAt(i);
if (Character.isDigit(character)) {
result += character;
}
}
System.out.println("result is: " + result);
Related
I intend to replace strings with normal strings into split strings, but there is a difference in length between the normal strings which amounts to 62 and the split length turns out to be 117, so when we write the 'a' button it doesn't change to 'π' is there another way of writing replace string easier?
public static String doublestruck(String input){
String normal = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
String split = "πππππππππ π‘ππππππππππππ π‘π’π£π€π₯π¦π§π¨π©πͺπ«πΈπΉβπ»πΌπ½πΎβπππππβπβββπππππππβ€";
String output = "";
char letter;
for(int i = 0; i < input.length(); i++){
letter = input.charAt(i);
int a = normal.indexOf(letter);
output += (a != -1) ? split.charAt(a):letter;
}
return new StringBuilder(output).toString();
}
The letters like π (U+1D7DC) are not in the Basic Multilingual Pane and thus take up two char values in Java.
Instead of charAt you need to use codePointAt and to find the correct offset you need to use offsetByCodePoint instead of directly using the same index. So split.charAt(a) needs to be replaced by split.codePointAt(spli.offsetByCodePoint(0, a)).
I need to replace first and middle char in string but without builder and etc, just with replace but idk how to make it.
String char = JOptionPane.showInputDialog(null, "Input string with more than 3 char");
if (char.length() < 3) {
JOptionPane.showMessageDialog(null, "Wrong input");
I just made this code and that is it, idk how to continue.
Example: input - pniut
I tried with smth like char.length / 2 but cant.
You can convert your string to a character array, and then swap the characters at 0 and middle position. Then convert the array back to String. e.g. I hard coded 2 here but like you mentioned in comments, you will need to figure out the character at the middle position.
String str = "input";
int mid = -1;
if(str.length() % 2 == 0) {
str.length() / 2 - 1
} else {
str.length() / 2;
}
char[] arr = str.toCharArray();
char temp = '0';
temp = arr[0];
arr[0] = arr[mid];
arr[mid] = temp;
String.valueOf(arr);
The value of the middle character, you will need to find out, like you said in the comments.
Since String objects are immutable, converting the original String to a char[] via toCharArray(), replace the characters, then making a new String from char[] via the String(char[]) constructor would work as shown below:
char[] c = character.toCharArray();
// Change characters at desired indicies
c[0] = 'p'; // first character
c[character.length()/2] = 'i'; // approximate middle character
String newString = new String(c);
System.out.println(newString); // "pniut"
Simple answer: not possible (for generic cases).
Meaning: all variants of String.replace() work by replacing one thing with another. There is no notion of using an index anywhere. So you can't say "replace index 1 with A" and "index 3 with B".
The simply solution is to push the string into a char[], to then swap/replace individual characters via index.
I'm betting the goal of the lesson is to learn how to use the API. So would start here Java API. Go to java.lang.String.
I would focus on the .toCharArray() method and the constructor that takes a char[] as an argument. You need to do this because a String is immutable, and cannot be changed. A char[], however can be altered, allowing you to modify the first and middle slots. You can then take your altered array and convert it back into a String.
I have inputs like
AS23456SDE
MFD324FR
I need to get First Character values like
AS, MFD
There should no first two or first 3 characters input can be changed. Need to get first characters before a number.
Thank you.
Edit : This is what I have tried.
public static String getPrefix(String serial) {
StringBuilder prefix = new StringBuilder();
for(char c : serial.toCharArray()){
if(Character.isDigit(c)){
break;
}
else{
prefix.append(c);
}
}
return prefix.toString();
}
Here is a nice one line solution. It uses a regex to match the first non numeric characters in the string, and then replaces the input string with this match.
public String getFirstLetters(String input) {
return new String("A" + input).replaceAll("^([^\\d]+)(.*)$", "$1")
.substring(1);
}
System.out.println(getFirstLetters("AS23456SDE"));
System.out.println(getFirstLetters("1AS123"));
Output:
AS
(empty)
A simple solution could be like this:
public static void main (String[]args) {
String str = "MFD324FR";
char[] characters = str.toCharArray();
for(char c : characters){
if(Character.isDigit(c))
break;
else
System.out.print(c);
}
}
Use the following function to get required output
public String getFirstChars(String str){
int zeroAscii = '0'; int nineAscii = '9';
String result = "";
for (int i=0; i< str.lenght(); i++){
int ascii = str.toCharArray()[i];
if(ascii >= zeroAscii && ascii <= nineAscii){
result = result + str.toCharArray()[i];
}else{
return result;
}
}
return str;
}
pass your string as argument
I think this can be done by a simple regex which matches digits and java's string split function. This Regex based approach will be more efficient than the methods using more complicated regexs.
Something as below will work
String inp = "ABC345.";
String beginningChars = inp.split("[\\d]+",2)[0];
System.out.println(beginningChars); // only if you want to print.
The regex I used "[\\d]+" is escaped for java already.
What it does?
It matches one or more digits (d). d matches digits of any language in unicode, (so it matches japanese and arabian numbers as well)
What does String beginningChars = inp.split("[\\d]+",2)[0] do?
It applies this regex and separates the string into string arrays where ever a match is found. The [0] at the end selects the first result from that array, since you wanted the starting chars.
What is the second parameter to .split(regex,int) which I supplied as 2?
This is the Limit parameter. This means that the regex will be applied on the string till 1 match is found. Once 1 match is found the string is not processed anymore.
From the Strings javadoc page:
The limit parameter controls the number of times the pattern is applied and therefore affects the length of the resulting array. If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array's length will be no greater than n, and the array's last entry will contain all input beyond the last matched delimiter. If n is non-positive then the pattern will be applied as many times as possible and the array can have any length. If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded.
This will be efficient if your string is huge.
Possible other regex if you want to split only on english numerals
"[0-9]+"
public static void main(String[] args) {
String testString = "MFD324FR";
int index = 0;
for (Character i : testString.toCharArray()) {
if (Character.isDigit(i))
break;
index++;
}
System.out.println(testString.substring(0, index));
}
this prints the first 'n' characters before it encounters a digit (i.e. integer).
I have one String generated of random characters that will encrypt another String given by the user by adding the first character from the String with the first character of the given String. It's working fine, but if the user were to enter multiple words with spaces in between, I want to choose the next character of the first String rather than code the space itself. Is that possible? This is what I have:
(random is the coded string and sentenceUpper is string given by user)
public static void encrypt(String sentenceUpper){
String newSentence = "";
for(int i = 0; i < sentenceUpper.length(); i++){
char one = random.charAt(i);
char two = sentenceUpper.charAt(i);
if(one < 'A' || one > 'Z'){
two = sentenceUpper.charAt(1 + i);}
char result = (char)((one + two)%26 + 'A');
newSentence += "" + result;
}
EDIT FOR BETTER EXPLANATION:
I have:
String random = "WFAZYZAZOHS";
I would like to code user input:
String upperCase: "YOU GO";
So, I'm going to take Y + L = U, etc...
to get :
"UTUSEN
"
But I see that there's a space in "YOU GO" , So I'd like to change it to:
WFA ZY + YOU GO = UTU SE.
I hope that's better explained.
The simplest way to do this would probably be to use an if statement to run the code in the loop only if the character is not a space. If you don't want to skip the character in the random string, you would need a separate variable to track the current character index in that string.
Example: Put this after defining one and two and put the rest of the loop inside it:
if(two==' '){
...
}
Then, add the space in the output:
else{
newSentence+=" ";
}
for a college project, I am doing a spelling test for children and i need to give 1 mark for a minor spelling error. For this I am going to do if the spelling has 2 characters wrong. How can I compare the saved word to the inputed word?
char wLetter1 = word1.charAt(0);
char iLetter1 = input1.charAt(0);
char wLetter2 = word1.charAt(1);
char iLetter2 = input1.charAt(1);
I have started out with this where word1 is the saved word and input1 is the user input word.
However, if I add lots of these, if the word is 3 characters long but I am trying to compare the 4th character, I will get an error? Is there a way of knowing how many characters are in the string and only finding the characters of those letters?
Just use a for loop. Since I'm assuming this is about JavaScript, calling charAt() with an index out-of-bounds will just return the empty string "".
To avoid a out-of-bounds exception you'll have to iterate up until the lower of the lengths:
int errs = Math.abs(word1.length - input1.length);
int len = Math.min(word1.length, input1.length);
for (int i = 0; i < len; i++) {
if (word1.charAt(i) != input1.charAt(i)) errs++;
}
// errs now holds the number of character mismatches