I am still new at java. I have this basic split string function as below. I need to capture the substrings post split. My question is how to move individually split parts into separate variables instead of printing them? Do I need another array to move them separately? Is there another simpler way to achieve this? In part I, for simplicity, I am assuming the delimiters to be spaces. Appreciate your help!
public class SplitString {
public static void main(String[] args) {
String phrase = "First Second Third";
String delims = "[ ]+";
String[] tokens = phrase.split(delims);
String first;
String middle;
String last;
for (int i = 0; i < tokens.length; i++)
{
System.out.println(tokens[i]);
//I need to move first part to first and second part to second and so on
}
}
}
array[index] accesses the indexth element of array, so
first = tokens[0]; // Array indices start at zero, not 1.
second = tokens[1];
third = tokens[2];
You should really check the length first, and if the string you're splitting is user input, tell the user what went wrong.
if (tokens.length != 3) {
System.err.println(
"I expected a phrase with 3 words separated by spaces,"
+ " not `" + phrase + "`");
return;
}
If the number of Strings that you will end up with after the split has taken place is known, then you can simply assign the variables like so.
String first = tokens[0];
String middle = tokens[1];
String last = tokens[2];
If the number of tokens is not known, then there is no way (within my knowledge) to assign each one to a individual variable.
If you're assuming that your String is three words, then it's quite simple.
String first = tokens[0];
String middle = tokens[1];
String last = tokens[2];
if(tokens.length>3)
{
String first=tokens[0];
String middle=tokens[1];
String last=tokens[2];
}
Thanks everyone...all of you have answered me in some way or the other. Initially I was assuming only 3 entries in the input but turns out it could vary :) for now i'll stick with the simple straight assignments to 3 variables until I figure out another way! Thanks.
Related
I'm making a guess the String program game, and I need to convert the users inputted string from letters and whitespaces/spaces to question marks(?). How would I go about that?
I used a for loop and it gave me what I wanted but I can't use the output of the for loop outside the loop.
System.out.println("Please enter a string of words.");
String userString = scan.nextLine();
userString=userString.toLowerCase();
System.out.println(userString);
int s1 = userString.length();
System.out.println(s1);
for(int a=0;a<s1;a++) {
System.out.print("?");
}
I entered "this is a string" for userString.
When I run the program, it outputs:
????????????????
Is there a way to turn this output into a String?
Or is there a better way of converting String values into "?"?
There is no need to print ? for every character. You can just replace every char with a ? and generate a new string without printing.
String newUserString = userString.replaceAll(".","?");
If you want to replace every character of a String with ?, you can try
String newString = userString.replaceAll(".", "?");
The "." is a regular expression that will match any character except a newline, which your String won't contain anyway.
from what i understood it i think you want to store the "?" ,
to do that
create a string var outside the for loop
and append '?' in it.
String temp="";
for(int i=0;i<s1;i++)
{
temp=temp+"?";
}
System.out.println(temp);
this should store the string in temp.
Java 11 provides repeat() method.
You can use
“?”.repeat(s1) inside sysout
System.out.print(“?”.repeat(s1)) and can remove the for-loop
It will repeat the ? s1 times.
replace all is the correct answer, but if for some reason you HAD to do a for loop you could use StringBuilder and append a ? for each character in the string like so:
String userInput = "myString";
StringBuilder myBuilder = new StringBuilder();
for(int x = 0; x<userInput.length(); x++){
myBuilder.append("?");
}
String myNewString = myBuilder.toString();
But the replace all answers above are the right way to do it.
I need to replace parts of a string by looking up the System properties.
For example, consider the string It was {var1} beauty killed {var2}
I need to parse the string, and replace all the words contained within the parenthesis by looking up their value in System properties. If System.getProperty() returns null, then simply replace with empty character. This is pretty straightforward when I know the variables well ahead. But the string that I need to parse is not defined ahead. I wouldn't know how many number of variables are in the string and what the variable names are. Assuming a simple, well formatted string (no nested parenthesis, open - close matches), what is the simplest or the most elegant way to parse through the string and replace all the character sequences that are enclosed in the parenthesis?
Only solution I could come up with is to traverse the string from the first character, note down the positions of the start and end positions of the parenthesis, replace the string between them, and then continue until reaching the end of the string. Is there simpler way to do this?
You can use the parentheses to break the initial string into substrings, and then replace every other substring.
String[] substituteValues = {"the", "str", "other", "another"};
int substituteValuesIndex = 0;
String test = "Here is {var1} string called {var2}";
// split the string up into substrings
test = test.replaceAll("\\}", "\\{");
String[] splitString = test.split("\\{");
// now sub in your values
for (int k=1; k < splitString.length; k = k+2) {
splitString[k] = substituteValues[substituteValuesIndex];
substituteValuesIndex++;
}
String result = "";
for (String s : splitString) {
result = result + s;
}
I have an ArrayList of Strings and it contains records such as:
this is a first sentence
hello my name is Chris
what's up man what's up man
today is tuesday
I need to clear this list, so that the output does not contain repeated content. In the case above, the output should be:
this is a first sentence
hello my name is Chris
what's up man
today is tuesday
as you can see, the 3rd String has been modified and now contains only one statement what's up man instead of two of them.
In my list there is a situation that sometimes the String is correct, and sometimes it is doubled as shown above.
I want to get rid of it, so I thought about iterating through this list:
for (String s: myList) {
but I cannot find a way of eliminating duplicates, especially since the length of each string is not determined, and by that I mean there might be record:
this is a very long sentence this is a very long sentence
or sometimes short ones:
single word singe word
is there some native java function for that maybe?
Assuming the String is repeated just twice, and with an space in between as in your examples, the following code would remove repetitions:
for (int i=0; i<myList.size(); i++) {
String s = myList.get(i);
String fs = s.substring(0, s.length()/2);
String ls = s.substring(s.length()/2+1, s.length());
if (fs.equals(ls)) {
myList.set(i, fs);
}
}
The code just split each entry of the list into two substrings (dividing by the half point). If both are equal, substitute the original element with only one half, thus removing the repetition.
I was testing the code and did not see #Brendan Robert answer. This code follows the same logic as his answer.
I would suggest using regular expressions. I was able to remove duplicates using this pattern: \b([\w\s']+) \1\b
public class Main {
static String [] phrases = {
"this is a first sentence",
"hello my name is Chris",
"what's up man what's up man",
"today is tuesday",
"this is a very long sentence this is a very long sentence",
"single word single word",
"hey hey"
};
public static void main(String[] args) throws Exception {
String duplicatePattern = "\\b([\\w\\s']+) \\1\\b";
Pattern p = Pattern.compile(duplicatePattern);
for (String phrase : phrases) {
Matcher m = p.matcher(phrase);
if (m.matches()) {
System.out.println(m.group(1));
} else {
System.out.println(phrase);
}
}
}
}
Results:
this is a first sentence
hello my name is Chris
what's up man
today is tuesday
this is a very long sentence
single word
hey
Assumptions:
Uppercase words are equal to lowercase counterparts.
String fullString = "lol lol";
String[] words = fullString.split("\\W+");
StringBuilder stringBuilder = new StringBuilder();
Set<String> wordsHashSet = new HashSet<>();
for (String word : words) {
// Check for duplicates
if (wordsHashSet.contains(word.toLowerCase())) continue;
wordsHashSet.add(word.toLowerCase());
stringBuilder.append(word).append(" ");
}
String nonDuplicateString = stringBuilder.toString().trim();
simple logic : split every word by token space i.e " " and now add it in LinkedHashSet , Retrieve back, Replace "[","]",","
String s = "I want to walk my dog I want to walk my dog";
Set<String> temp = new LinkedHashSet<>();
String[] arr = s.split(" ");
for ( String ss : arr)
temp.add(ss);
String newl = temp.toString()
.replace("[","")
.replace("]","")
.replace(",","");
System.out.println(newl);
o/p : I want to walk my dog
It depends on the situation that you have but assuming that the string can be repeated at most twice and not three or more times you could find the length of the entire string, find the halfway point and compare each index after the halfway point with the matching beginning index. If the string can be repeated more than once you will need a more complicated algorithm that would first determine how many times the string is repeated and then finds the starting index of each repeat and truncates all index's from the beginning of the first repeat onward. If you can provide some more context into what possible scenarios you expect to handle we can start putting together some ideas.
//Doing it in Java 8
String str1 = "I am am am a good Good coder";
String[] arrStr = str1.split(" ");
String[] element = new String[1];
return Arrays.stream(arrStr).filter(str1 -> {
if (!str1.equalsIgnoreCase(element[0])) {
element[0] = str1;
return true;
}return false;
}).collect(Collectors.joining(" "));
I am looking to remove parts of a string if it ends in a certain string.
An example would be to take this string: "am.sunrise.ios#2x.png"
And remove the #2x.png so it looks like: "am.sunrise.ios"
How would I go about checking to see if the end of a string contains "#2x.png" and remove it?
You could check the lastIndexOf, and if it exists in the string, use substring to remove it:
String str = "am.sunrise.ios#2x.png";
String search = "#2x.png";
int index = str.lastIndexOf(search);
if (index > 0) {
str = str.substring(0, index);
}
Assuming you have a string initialized as String file = "am.sunrise.ios#2x.png";.
if(file.endsWith("#2x.png"))
file = file.substring(0, file.lastIndexOf("#2x.png"));
The endsWith(String) method returns a boolean determining if the string has a certain suffix. Depending on that you can replace the string with a substring of itself starting with the first character and ending before the index of the character that you are trying to remove.
private static String removeSuffixIfExists(String key, String suffix) {
return key.endswith(suffix)
? key.substring(0, key.length() - suffix.length())
: key;
}
}
String suffix = "#2x.png";
String key = "am.sunrise.ios#2x.png";
String output = removeSuffixIfExists(key, suffix);
public static void main(String [] args){
String word = "am.sunrise.ios#2x.png";
word = word.replace("#2x.png", "");
System.out.println(word);
}
If you want to generally remove entire content of string from # till end you can use
yourString = yourString.replaceAll("#.*","");
where #.* is regex (regular expression) representing substring starting with # and having any character after it (represented by .) zero or more times (represented by *).
In case there will be no #xxx part your string will be unchanged.
If you want to change only this particular substring #2x.png (and not substirng like #3x.png) while making sure that it is placed at end of your string you can use
yourString = yourString.replaceAll("#2x\\.png$","");
where
$ represents end of string
\\. represents . literal (we need to escape it since like shown earlier . is metacharacter representing any character)
Since I was trying to do this on an ArrayList of items similarly styled I ended up using the following code:
for (int image = 0; image < IBImages.size(); image++) {
IBImages.set(image, IBImages.get(image).split("~")[0].split("#")[0].split(".png")[0]);
}
If I have a list of images with the names
[am.sunrise.ios.png, am.sunrise.ios#2x.png, am.sunrise.ios#3x.png, am.sunrise.ios~ipad.png, am.sunrise.ios~ipad#2x.png]
This allows me to split the string into 2 parts.
For example, "am.sunrise.ios~ipad.png" will be split into "am.sunrise.ios" and "~ipad.png" if I split on "~". I can just get the first part back by referencing [0]. Therefore I get what I'm looking for in one line of code.
Note that image is "am.sunrise.ios~ipad.png"
You could use String.split():
public static void main(String [] args){
String word = "am.sunrise.ios#2x.png";
String[] parts = word.split("#");
if (parts.length == 2) {
System.out.println("looks like user#host...");
System.out.println("User: " + parts[0]);
System.out.println("Host: " + parts[1]);
}
}
Then you haven an array of Strings, where the first element contains the part before "#" and the second element the part after the "#".
Combining the answers 1 and 2:
String str = "am.sunrise.ios#2x.png";
String search = "#2x.png";
if (str.endsWith(search)) {
str = str.substring(0, str.lastIndexOf(search));
}
I'm a computer science student learning Java, and as an exercise, we're doing a permutation algorhythm.
Now, i'm stuck at a point where i need to search for a natural number within a String full of numbers, splitted by a comma:
String myString = "0,1,2,10,14,";
The problem is i'm using...
myString.contains(String.valueOf(anInteger);
...to check for the presence of a specific number. This works for numbers from 0 to 9, but when looking for a more-than-1-digit number, the program does not recognize it as a natural number.
In other words, and as an example: "14" is not the integer 14, its just a string with an "1", and a "4"; so, if i run...
String myString = "0,1,2,10,14,";
if (myString.contains(myString.valueOf(4))) { doSomething(); }
...the "if" statement will be true, since the integer "4" is present in the string, as part of the natural number "14".
At this point, i've been searching through StackOverflow and other pages for a solution, and learnt i should use Pattern and Matcher.
My question is: what's the best way to do use them?
Relevant part of my code:
for (int i = 0; i<r; i++)
{
if (!act.contains(String.valueOf(i)))
{
...
}
...
}
I use this method several times in my code, so an exact substitution would be nice.
Thank you all in advance!
You only need a method call to matches():
if (myString.matches(".*\\b" + anInteger + "\\b.*"))
// string contains the number
This works using by creating a regex that has a word boundary (\b) at either end of the target number. The leading and trailing .* are required because matches() must match the whole string to return true.
Look into how to split a String into an array of String. So:
String[] splitStrings = myString.split(",")
ArrayList<Integer> parsedInts = new ArrayList<Integer>();
for (String str : splitStrings) {
parsedInts.add(Integer.parseInt(str));
}
then in your for loop:
if (parsedInts.contains(i)) {
// body
}
Something like this:
String myString = "0,1,2,10,14,";
String[] split = myString.split(",");
for (String string : split) {
int num = Integer.parseInt(string);
if (num == 4) {
System.out.println(num);
// ...
}
}
String myString = "0,1,2,10,14,2323232";
String[] allList = myString.split(",");
for (String string : allList) {
if(string.matches("[0-9]*"))
{
System.out.println("Its number with value "+string);
}
}
I think you need to pick all the numbers in the given string and find the permutation.
I think you need to Tokenize the given string with the Comma Separator.
When I do such program, I divide my logic to parse the String and write the logic in another method. Below is the snippet
String myString = "0,1,2,10,14,";
StringTokenizer st2 = new StringTokenizer(myString , ",");
while (st2.hasMoreElements()) {
doSomething(st2.nextElement());
}