I am looking for a solution concerning replacing space digits into another, for example:
"My Example is interesting".replaceAll(" ", "1");
does NOT return "My1Example1is1interesting", but only "My"
I was looking for solutions also on StackOverflow, but usually found "Removing whitespaces from String/URL.." etc.
Scanner s = new Scanner(System.in);
String in = s.next();
in = in.replaceAll("//s+", "1");
System.out.println(in);
The actual issue is the way in which you are reading data. Scanner.next() reads only one word at a time. SO if your print the value which is read, it is actually "My". use nextLine() to read the entire line.
print in and check what it prints. It should print My i.e, only one word and not words separated by space.
Remember that String object is immutable in Java, you should assign the result to a new String:
String res = "My Example is interesting".replaceAll(" ", "1");
Also note that because String#replaceAll accepts a regex as the first parameter, you can improve the regex by having \\s+ which will work on strings that have more than one or more spaces:
String res = "My Example is interesting".replaceAll("\\s+", "1");
Update: After you posted your code, the problem is not with replaceAll, you should use nextLine instead of next because next reads only the first complete token.
You have to assign the result to another variable.
String s = "My Example is interesting";
String result = s.replaceAll(" ", "");
This will remove spaces.
If you wan to replace with another text.
String result = s.replaceAll(" ", "My Text");
Related
How can I delete everything after first empty space in a string which user selects? I was reading this how to remove some words from a string in java. Can this help me in my case?
You can use replaceAll with a regex \s.* which match every thing after space:
String str = "Hello java word!";
str = str.replaceAll("\\s.*", "");
output
Hello
regex demo
Like #Coffeehouse Coder mention in comment, This solution will replace every thing if the input start with space, so if you want to avoid this case, you can trim your input using string.trim() so it can remove the spaces in start and in end.
Assuming that there is no space in the beginning of the string.
Follow these steps-
Split the string at space. It will create an array.
Get the first element of that array.
Hope this helps.
str = "Example string"
String[] _arr = str.split("\\s");
String word = _arr[0];
You need to consider multiple white spaces and space in the beginning before considering the above code.
I am not native to JAVA Programming but have an idea that it has split function for string.
And the reference you cited in the question is bit complex, while you can achieve the desired thing very easily.
P.S. In future if you make a mind to get two words or three, splitting method is better (assuming you have already dealt with multiple white-spaces) else substring is better.
A simple way to do it can be:
System.out.println("Hello world!".split(" ")[0]);
// Taking 'str' as your string
// To remove the first space(s) of the string,
str = str.trim();
int index = str.indexOf(" ");
String word = str.substring(0, index);
This is just one method of many.
str = str.replaceAll("\\s+", " "); // This replaces one or more spaces with one space
String[] words = str.split("\\s");
String first = words[0];
The simplest solution in my opinion would be to just locate the index which the user wants it to be cut off at and then call the substring() method from 0 to the index they wanted. Set that = to a new string and you have the string they want.
If you want to replace the string then just set the original string = to the result of the substring() method.
Link to substring() method: https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#substring(int,%20int)
There are already 5 perfectly good answers, so let me add a sixth one. Variety is the spice of life!
private static final Pattern FIRST_WORD = Pattern.compile("\\S+");
public static String firstWord(CharSequence text) {
Matcher m = FIRST_WORD.matcher(text);
return m.find() ? m.group() : "";
}
Advantages over the .split(...)[0]-type answers:
It directly does exactly what is being asked, i.e. "Find the first sequence of non-space characters." So the self-documentation is more explicit.
It is more efficient when called on multiple strings (e.g. for batch processing a large list of strings) because the regular expression is compiled only once.
It is more space-efficient because it avoids unnecessarily creating a whole array with references to each word when we only need the first.
It works without having to trim the string.
(I know this is probably too late to be of any use to the OP but I'm leaving it here as an alternative solution for future readers.)
This would be more efficient
String str = "Hello world!";
int spaceInd = str.indexOf(' ');
if(spaceInd != -1) {
str = str.substring(0, spaceInd);
}
System.out.println(String.format("[%s]", str));
I found the question here before but somehow I can't see what I'm doing wrong. So I have a given String that looks something like this:
"Some text here\n\nsome more text here"
And I want to remove the linebreaks and display the Text in a TextView. I tried using String.replaceAll:
String newString = oldString.replaceAll("\\n", " ");
But that didn't change anything in the text. However,
oldString.contains("\\n"); returns true. What am I doing wrong?
edit:
I'm sorry, I know, oldString doesn't change. The problem is that, if I print oldString and newString they're exactly the same even though it says that oldString does contain a "\n".
This is my code:
Log.d(TAG, "contains: " + str.contains("\\n"));
Log.d(TAG, "old: " + str);
str = str.replaceAll("\\n", " ");
Log.d(TAG, "new: " + str);
And this is what I get:
contains: true
old: Vorgang nicht möglich\n\nBitte Karte entnehmen
new: Vorgang nicht möglich\n\nBitte Karte entnehmen
UPDATE
Thanks to Shivanshu Verma, I tried str.replace("\\n"" "); instead of str.replaceAll("\\n", " "); and that works! Does anybody know, why I can't use replaceAll() here?
Strings in Java are immutable and as such the new string with the replacements is stored in newString, not oldString.
EDIT
I see now that your issue was not actually related to Java String immutability but rather the difference between replace() and replaceAll(). The difference between these is that replaceAll() takes in a regex as the first argument, which will then replace any matches with the second argument, whereas replace() simply takes in a CharSequence (of which String is an implementation) and will replace exact matches with the second argument.
In your case, I think your original String had the newline characters escaped:
String str = "Vorgang nicht möglich\\n\\nBitte Karte entnehmen";
which meant that the String didn't actually contain newline characters at all; it contained literally "\n". This would mean that:
str.replaceAll("\\n", " ");
will resolve the first argument to a regex and replace newline characters (of which there were none), and:
str.replace("\\n", " ");
will replace exact matches of "\n". It's also worth noting that as others have pointed out contains() also doesn't take in a regex, which is why running:
oldString.contains("\\n");
returned true.
Your code works perfectly fine.
This test will pass without any error:
#Test
public void testReplaceAll() {
String newString = "line1\nline2\nline3".replaceAll("\\n", " ");
assertThat(newString).isEqualTo("line1 line2 line3");
assertThat(newString).doesNotContain("\\n");
}
try to Replace() method instead of ReplaceAll()
String newString = oldString.replace("\\n", " ");
may be it work for u
String newString = oldString.replace("\n", " ");
The question is we have to split the string and write how many words we have.
Scanner in = new Scanner(System.in);
String st = in.nextLine();
String[] tokens = st.split("[\\W]+");
When I gave the input as a new line and printed the no. of tokens .I have got the answer as one.But i want it as zero.What should i do? Here the delimiters are all the symbols.
Short answer: To get the tokens in str (determined by whitespace separators), you can do the following:
String str = ... //some string
str = str.trim() + " "; //modify the string for the reasons described below
String[] tokens = str.split("\\s+");
Longer answer:
First of all, the argument to split() is the delimiter - in this case one or more whitespace characters, which is "\\s+".
If you look carefully at the Javadoc of String#split(String, int) (which is what String#split(String) calls), you will see why it behaves like this.
If the expression does not match any part of the input then the resulting array has just one element, namely this string.
This is why "".split("\\s+") would return an array with one empty string [""], so you need to append the space to avoid this. " ".split("\\s+") returns an empty array with 0 elements, as you want.
When there is a positive-width match at the beginning of this string then an empty leading substring is included at the beginning of the resulting array.
This is why " a".split("\\s+") would return ["", "a"], so you need to trim() the string first to remove whitespace from the beginning.
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.
Since String#split(String) calls String#split(String, int) with the limit argument of zero, you can add whitespace to the end of the string without changing the number of words (because trailing empty strings will be discarded).
UPDATE:
If the delimiter is "\\W+", it's slightly different because you can't use trim() for that:
String str = ...
str = str.replaceAll("^\\W+", "") + " ";
String[] tokens = str.split("\\W+");
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String line = null;
while (!(line = in.nextLine()).isEmpty()) {
//logic
}
System.out.print("Empty Line");
}
output
Empty Line
I am trying to split a string according to a certain set of delimiters.
My delimiters are: ,"():;.!? single spaces or multiple spaces.
This is the code i'm currently using,
String[] arrayOfWords= inputString.split("[\\s{2,}\\,\"\\(\\)\\:\\;\\.\\!\\?-]+");
which works fine for most cases but i'm have a problem when the the first word is surrounded by quotation marks. For example
String inputString = "\"Word\" some more text.";
Is giving me this output
arrayOfWords[0] = ""
arrayOfWords[0] = "Word"
arrayOfWords[1] = "some"
arrayOfWords[2] = "more"
arrayOfWords[3] = "text"
I want the output to give me an array with
arrayOfWords[0] = "Word"
arrayOfWords[1] = "some"
arrayOfWords[2] = "more"
arrayOfWords[3] = "text"
This code has been working fine when quotation marks are used in the middle of the sentence, I'm not sure what the trouble is when it's at the beginning.
EDIT: I just realized I have same problem when any of the delimiters are used as the first character of the string
Unfortunately you wont be able to remove this empty first element using only split. You should probably remove first elements from your string that match your delimiters and split after it. Also your regex seems to be incorrect because
by adding {2,} inside [...] you are in making { 2 , and } characters delimiters,
you don't need to escape rest of your delimiters (note that you don't have to escape - only because it is at end of character class [] so he cant be used as range operator).
Try maybe this way
String regexDelimiters = "[\\s,\"():;.!?\\-]+";
String inputString = "\"Word\" some more text.";
String[] arrayOfWords = inputString.replaceAll(
"^" + regexDelimiters,"").split(regexDelimiters);
for (String s : arrayOfWords)
System.out.println("'" + s + "'");
output:
'Word'
'some'
'more'
'text'
A delimiter is interpreted as separating the strings on either side of it, thus the empty string on its left is added to the result as well as the string to its right ("Word"). To prevent this, you should first strip any leading delimiters, as described here:
How to prevent java.lang.String.split() from creating a leading empty string?
So in short form you would have:
String delim = "[\\s,\"():;.!?\\-]+";
String[] arrayOfWords = inputString.replaceFirst("^" + delim, "").split(delim);
Edit: Looking at Pshemo's answer, I realize he is correct regarding your regex. Inside the brackets it's unnecessary to specify the number of space characters, as they will be caught be the + operator.
Pretty basic question for someone who knows.
Instead of getting from
"This is my text.
And here is a new line"
To:
"This is my text. And here is a new line"
I get:
"This is my text.And here is a new line.
Any idea why?
L.replaceAll("[\\\t|\\\n|\\\r]","\\\s");
I think I found the culprit.
On the next line I do the following:
L.replaceAll( "[^a-zA-Z0-9|^!|^?|^.|^\\s]", "");
And this seems to be causing my issue.
Any idea why?
I am obviously trying to do the following: remove all non-chars, and remove all new lines.
\s is a shortcut for whitespace characters in regex. It has no meaning in a string. ==> You can't use it in your replacement string. There you need to put exactly the character(s) that you want to insert. If this is a space just use " " as replacement.
The other thing is: Why do you use 3 backslashes as escape sequence? Two are enough in Java. And you don't need a | (alternation operator) in a character class.
L.replaceAll("[\\t\\n\\r]+"," ");
Remark
L is not changed. If you want to have a result you need to do
String result = L.replaceAll("[\\t\\n\\r]+"," ");
Test code:
String in = "This is my text.\n\nAnd here is a new line";
System.out.println(in);
String out = in.replaceAll("[\\t\\n\\r]+"," ");
System.out.println(out);
The new line separator is different for different OS-es - '\r\n' for Windows and '\n' for Linux.
To be safe, you can use regex pattern \R - the linebreak matcher introduced with Java 8:
String inlinedText = text.replaceAll("\\R", " ");
Try
L.replaceAll("(\\t|\\r?\\n)+", " ");
Depending on the system a linefeed is either \r\n or just \n.
I found this.
String newString = string.replaceAll("\n", " ");
Although, as you have a double line, you will get a double space. I guess you could then do another replace all to replace double spaces with a single one.
If that doesn't work try doing:
string.replaceAll(System.getProperty("line.separator"), " ");
If I create lines in "string" by using "\n" I had to use "\n" in the regex. If I used System.getProperty() I had to use that.
Your regex is good altough I would replace it with the empty string
String resultString = subjectString.replaceAll("[\t\n\r]", "");
You expect a space between "text." and "And" right?
I get that space when I try the regex by copying your sample
"This is my text. "
So all is well here. Maybe if you just replace it with the empty string it will work. I don't know why you replace it with \s. And the alternation | is not necessary in a character class.
You May use first split and rejoin it using white space.
it will work sure.
String[] Larray = L.split("[\\n]+");
L = "";
for(int i = 0; i<Larray.lengh; i++){
L = L+" "+Larray[i];
}
This should take care of space, tab and newline:
data = data.replaceAll("[ \t\n\r]*", " ");