someone can help me with code?
How to search word in String text, this word end "." or "," in java
I don't want search like this to find it
String word = "test.";
String wordSerch = "I trying to tasting the Artestem test.";
String word1 = "test,"; // here with ","
String word2 = "test."; // here with "."
String word3 = "test"; //here without
//after i make string array and etc...
if((wordSearch.equalsIgnoreCase(word1))||
(wordSearch.equalsIgnoreCase(word2))||
(wordSearh.equalsIgnoreCase(word3))) {
}
if (wordSearch.contains(gramer))
//it's not working because the word Artestem will contain test too, and I don't need it
You can use the matches(Regex) function with a String
String word = "test.";
boolean check = false;
if (word.matches("\w*[\.,\,]") {
check = true;
}
You can use regex for this
Matcher matcher = Pattern.compile("\\btest\\b").matcher(wordSearch);
if (matcher.find()) {
}
\\b\\b will match only a word. So "Artestem" will not match in this case.
matcher.find() will return true if there is a word test in your sentence and false otherwise.
String stringToSearch = "I trying to tasting the Artestem test. test,";
Pattern p1 = Pattern.compile("test[.,]");
Matcher m = p1.matcher(stringToSearch);
while (m.find())
{
System.out.println(m.group());
}
You can transform your String in an Array divided by words(with "split"), and search on that array , checking the last character of the words(charAt) with the character that you want to find.
String stringtoSearch = "This is a test.";
String whatIwantToFind = ",";
String[] words = stringtoSearch.split("\\s+");
for (String word : words) {
if (whatIwantToFind.equalsignorecas(word.charAt(word.length()-1);)) {
System.out.println("FIND");
}
}
What is a word? E.g.:
Is '5' a word?
Is '漢語' a word, or two words?
Is 'New York' a word, or two words?
Is 'Kraftfahrzeughaftpflichtversicherung' (meaning "automobile liability insurance") a word, or 3 words?
For some languages you can use Pattern.compile("[^\\p{Alnum}\u0301-]+") for split words. Use Pattern#split for this.
I think, you can find word by this pattern:
String notWord = "[^\\p{Alnum}\u0301-]{0,}";
Pattern.compile(notWord + "test" + notWord)`
See also: https://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html
This is my string
1 AP(PEAR + ANC)E (CAN anag)
14 EN(T)ICE (NIECE anag)
15 CHIC ("SHEIK" hom)
The string has an answer hidden,
The first string has an answer APPEARANCE and second string has ENTICE
I must extract those particular answer alone from the strings.
I tried to extract those words from it by
String input = "AP(PEAR + ANC)E (CAN anag)";;
String output = input.substring(0, input.indexOf(' '));
System.out.println(output);
Output:
AP(PEAR
As you can see, there is a space after R so the sub-string stops there and so the output. But Is there any way to read till the character 'E'(ie. end of string 'APPEARANCE') ? I want to stop reading if there is a space and if the next character is '(' .
I have another type of string in the same program "EN(T)ICE (NIECE anag)"
String input = "EN(T)ICE (NIECE anag)";
String output = input.substring(0, input.indexOf(' '));
System.out.println(output);
Output:
EN(T)ICE
There is a space after the character 'E' so it successfully gave the full output. Is there any way to get output like this for the first string. ? Any help would be great !!
Use replaceAll instead of indexOf and substring.
String[] inputs = {
"AP(PEAR + ANC)E (CAN anag)",
"EN(T)ICE (NIECE anag)",
"CHIC (\"SHEIK\" hom)"};
for (String s : inputs) {
String output = s.replaceAll(" \\(.*|[()+\\s]", "");
System.out.println(output);
}
result:
APPEARANCE
ENTICE
CHIC
You can try this splitting the string on the index of " (":
String input1 = "EN(T)ICE (NIECE anag)";
String input2 = "AP(PEAR + ANC)E (CAN anag)";
String input3 = "CHIC (\"SHEIK\" hom)";
System.out.println(extract(input1));
System.out.println(extract(input2));
System.out.println(extract(input3));
public static String extract(String s){
return s.split(" \\(.*")[0]
.replace("(", "")
.replace(")", "")
.replace(" + ", "");
}
Will produce :
ENTICE
APPEARANCE
CHIC
Split on the " (";
like so:
String input = "AP(PEAR + ANC)E (CAN anag)";
System.out.println(input.split(" \\(.*")[0]);
Just replace the String input with new values!
Can't you just replace the " + " with nothing, and then split by " " (space)?
Something like this:
private static String convertInputToOutput(final String input) {
String[] splittedArray = input.replaceAll(" \\+ ", "").split(" ");
return splittedArray[0];
}
public static void main(final String[] args) {
System.out.println(convertInputToOutput("AP(PEAR + ANC)E (CAN anag)"));
System.out.println(convertInputToOutput("EN(T)ICE (NIECE anag)"));
System.out.println(convertInputToOutput("CHIC (\"SHEIK\" hom)"));
}
Output:
AP(PEARANC)E
EN(T)ICE
CHIC
If you want output without parenthesis, also use a replaceAll for the parenthesis:
private static String convertInputToOutput(final String input) {
String[] splittedArray = input.replaceAll(" \\+ ", "").split(" ");
return splittedArray[0].replaceAll("(", "").replaceAll(")", "");
}
Output:
APPEARANCE
ENTICE
CHIC
I want to split a string with a delimiter white space. but it should handle quoted strings intelligently. E.g. for a string like
"John Smith" Ted Barry
It should return three strings John Smith, Ted and Barry.
After messing around with it, you can use Regex for this. Run the equivalent of "match all" on:
((?<=("))[\w ]*(?=("(\s|$))))|((?<!")\w+(?!"))
A Java Example:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class Test
{
public static void main(String[] args)
{
String someString = "\"Multiple quote test\" not in quotes \"inside quote\" \"A work in progress\"";
Pattern p = Pattern.compile("((?<=(\"))[\\w ]*(?=(\"(\\s|$))))|((?<!\")\\w+(?!\"))");
Matcher m = p.matcher(someString);
while(m.find()) {
System.out.println("'" + m.group() + "'");
}
}
}
Output:
'Multiple quote test'
'not'
'in'
'quotes'
'inside quote'
'A work in progress'
The regular expression breakdown with the example used above can be viewed here:
http://regex101.com/r/wM6yT9
With all that said, regular expressions should not be the go to solution for everything - I was just having fun. This example has a lot of edge cases such as the handling unicode characters, symbols, etc. You would be better off using a tried and true library for this sort of task. Take a look at the other answers before using this one.
Try this ugly bit of code.
String str = "hello my dear \"John Smith\" where is Ted Barry";
List<String> list = Arrays.asList(str.split("\\s"));
List<String> resultList = new ArrayList<String>();
StringBuilder builder = new StringBuilder();
for(String s : list){
if(s.startsWith("\"")) {
builder.append(s.substring(1)).append(" ");
} else {
resultList.add((s.endsWith("\"")
? builder.append(s.substring(0, s.length() - 1))
: builder.append(s)).toString());
builder.delete(0, builder.length());
}
}
System.out.println(resultList);
well, i made a small snipet that does what you want and some more things. since you did not specify more conditions i did not go through the trouble. i know this is a dirty way and you can probably get better results with something that is already made. but for the fun of programming here is the example:
String example = "hello\"John Smith\" Ted Barry lol\"Basi German\"hello";
int wordQuoteStartIndex=0;
int wordQuoteEndIndex=0;
int wordSpaceStartIndex = 0;
int wordSpaceEndIndex = 0;
boolean foundQuote = false;
for(int index=0;index<example.length();index++) {
if(example.charAt(index)=='\"') {
if(foundQuote==true) {
wordQuoteEndIndex=index+1;
//Print the quoted word
System.out.println(example.substring(wordQuoteStartIndex, wordQuoteEndIndex));//here you can remove quotes by changing to (wordQuoteStartIndex+1, wordQuoteEndIndex-1)
foundQuote=false;
if(index+1<example.length()) {
wordSpaceStartIndex = index+1;
}
}else {
wordSpaceEndIndex=index;
if(wordSpaceStartIndex!=wordSpaceEndIndex) {
//print the word in spaces
System.out.println(example.substring(wordSpaceStartIndex, wordSpaceEndIndex));
}
wordQuoteStartIndex=index;
foundQuote = true;
}
}
if(foundQuote==false) {
if(example.charAt(index)==' ') {
wordSpaceEndIndex = index;
if(wordSpaceStartIndex!=wordSpaceEndIndex) {
//print the word in spaces
System.out.println(example.substring(wordSpaceStartIndex, wordSpaceEndIndex));
}
wordSpaceStartIndex = index+1;
}
if(index==example.length()-1) {
if(example.charAt(index)!='\"') {
//print the word in spaces
System.out.println(example.substring(wordSpaceStartIndex, example.length()));
}
}
}
}
this also checks for words that were not separated with a space after or before the quotes, such as the words "hello" before "John Smith" and after "Basi German".
when the string is modified to "John Smith" Ted Barry the output is three strings,
1) "John Smith"
2) Ted
3) Barry
The string in the example is hello"John Smith" Ted Barry lol"Basi German"hello and prints
1)hello
2)"John Smith"
3)Ted
4)Barry
5)lol
6)"Basi German"
7)hello
Hope it helps
This is my own version, clean up from http://pastebin.com/aZngu65y (posted in the comment).
It can take care of Unicode. It will clean up all excessive spaces (even in quote) - this can be good or bad depending on the need. No support for escaped quote.
private static String[] parse(String param) {
String[] output;
param = param.replaceAll("\"", " \" ").trim();
String[] fragments = param.split("\\s+");
int curr = 0;
boolean matched = fragments[curr].matches("[^\"]*");
if (matched) curr++;
for (int i = 1; i < fragments.length; i++) {
if (!matched)
fragments[curr] = fragments[curr] + " " + fragments[i];
if (!fragments[curr].matches("(\"[^\"]*\"|[^\"]*)"))
matched = false;
else {
matched = true;
if (fragments[curr].matches("\"[^\"]*\""))
fragments[curr] = fragments[curr].substring(1, fragments[curr].length() - 1).trim();
if (fragments[curr].length() != 0)
curr++;
if (i + 1 < fragments.length)
fragments[curr] = fragments[i + 1];
}
}
if (matched) {
return Arrays.copyOf(fragments, curr);
}
return null; // Parameter failure (double-quotes do not match up properly).
}
Sample input for comparison:
"sdfskjf" sdfjkhsd "hfrif ehref" "fksdfj sdkfj fkdsjf" sdf sfssd
asjdhj sdf ffhj "fdsf fsdjh"
日本語 中文 "Tiếng Việt" "English"
dsfsd
sdf " s dfs fsd f " sd f fs df fdssf "日本語 中文"
"" "" ""
" sdfsfds " "f fsdf
(2nd line is empty, 3rd line is spaces, last line is malformed).
Please judge with your own expected output, since it may varies, but the baseline is that, the 1st case should return [sdfskjf, sdfjkhsd, hfrif ehref, fksdfj sdkfj fkdsjf, sdf, sfssd].
commons-lang has a StrTokenizer class to do this for you, and there is also java-csv library.
Example with StrTokenizer:
String params = "\"John Smith\" Ted Barry"
// Initialize tokenizer with input string, delimiter character, quote character
StrTokenizer tokenizer = new StrTokenizer(params, ' ', '"');
for (String token : tokenizer.getTokenArray()) {
System.out.println(token);
}
Output:
John Smith
Ted
Barry
Looking for quick, simple way in Java to change this string
" hello there "
to something that looks like this
"hello there"
where I replace all those multiple spaces with a single space, except I also want the one or more spaces at the beginning of string to be gone.
Something like this gets me partly there
String mytext = " hello there ";
mytext = mytext.replaceAll("( )+", " ");
but not quite.
Try this:
String after = before.trim().replaceAll(" +", " ");
See also
String.trim()
Returns a copy of the string, with leading and trailing whitespace omitted.
regular-expressions.info/Repetition
No trim() regex
It's also possible to do this with just one replaceAll, but this is much less readable than the trim() solution. Nonetheless, it's provided here just to show what regex can do:
String[] tests = {
" x ", // [x]
" 1 2 3 ", // [1 2 3]
"", // []
" ", // []
};
for (String test : tests) {
System.out.format("[%s]%n",
test.replaceAll("^ +| +$|( )+", "$1")
);
}
There are 3 alternates:
^_+ : any sequence of spaces at the beginning of the string
Match and replace with $1, which captures the empty string
_+$ : any sequence of spaces at the end of the string
Match and replace with $1, which captures the empty string
(_)+ : any sequence of spaces that matches none of the above, meaning it's in the middle
Match and replace with $1, which captures a single space
See also
regular-expressions.info/Anchors
You just need a:
replaceAll("\\s{2,}", " ").trim();
where you match one or more spaces and replace them with a single space and then trim whitespaces at the beginning and end (you could actually invert by first trimming and then matching to make the regex quicker as someone pointed out).
To test this out quickly try:
System.out.println(new String(" hello there ").trim().replaceAll("\\s{2,}", " "));
and it will return:
"hello there"
Use the Apache commons StringUtils.normalizeSpace(String str) method. See docs here
This worked perfectly for me : sValue = sValue.trim().replaceAll("\\s+", " ");
trim() method removes the leading and trailing spaces and using replaceAll("regex", "string to replace") method with regex "\s+" matches more than one space and will replace it with a single space
myText = myText.trim().replaceAll("\\s+"," ");
The following code will compact any whitespace between words and remove any at the string's beginning and end
String input = "\n\n\n a string with many spaces, \n"+
" a \t tab and a newline\n\n";
String output = input.trim().replaceAll("\\s+", " ");
System.out.println(output);
This will output a string with many spaces, a tab and a newline
Note that any non-printable characters including spaces, tabs and newlines will be compacted or removed
For more information see the respective documentation:
String#trim() method
String#replaceAll(String regex, String replacement) method
For information about Java's regular expression implementation see the documentation of the Pattern class
"[ ]{2,}"
This will match more than one space.
String mytext = " hello there ";
//without trim -> " hello there"
//with trim -> "hello there"
mytext = mytext.trim().replaceAll("[ ]{2,}", " ");
System.out.println(mytext);
OUTPUT:
hello there
To eliminate spaces at the beginning and at the end of the String, use String#trim() method. And then use your mytext.replaceAll("( )+", " ").
You can first use String.trim(), and then apply the regex replace command on the result.
Try this one.
Sample Code
String str = " hello there ";
System.out.println(str.replaceAll("( +)"," ").trim());
OUTPUT
hello there
First it will replace all the spaces with single space. Than we have to supposed to do trim String because Starting of the String and End of the String it will replace the all space with single space if String has spaces at Starting of the String and End of the String So we need to trim them. Than you get your desired String.
String blogName = "how to do in java . com";
String nameWithProperSpacing = blogName.replaceAll("\\\s+", " ");
trim()
Removes only the leading & trailing spaces.
From Java Doc,
"Returns a string whose value is this string, with any leading and trailing whitespace removed."
System.out.println(" D ev Dum my ".trim());
"D ev Dum my"
replace(), replaceAll()
Replaces all the empty strings in the word,
System.out.println(" D ev Dum my ".replace(" ",""));
System.out.println(" D ev Dum my ".replaceAll(" ",""));
System.out.println(" D ev Dum my ".replaceAll("\\s+",""));
Output:
"DevDummy"
"DevDummy"
"DevDummy"
Note: "\s+" is the regular expression similar to the empty space character.
Reference : https://www.codedjava.com/2018/06/replace-all-spaces-in-string-trim.html
In Kotlin it would look like this
val input = "\n\n\n a string with many spaces, \n"
val cleanedInput = input.trim().replace(Regex("(\\s)+"), " ")
A lot of correct answers been provided so far and I see lot of upvotes. However, the mentioned ways will work but not really optimized or not really readable.
I recently came across the solution which every developer will like.
String nameWithProperSpacing = StringUtils.normalizeSpace( stringWithLotOfSpaces );
You are done.
This is readable solution.
You could use lookarounds also.
test.replaceAll("^ +| +$|(?<= ) ", "");
OR
test.replaceAll("^ +| +$| (?= )", "")
<space>(?= ) matches a space character which is followed by another space character. So in consecutive spaces, it would match all the spaces except the last because it isn't followed by a space character. This leaving you a single space for consecutive spaces after the removal operation.
Example:
String[] tests = {
" x ", // [x]
" 1 2 3 ", // [1 2 3]
"", // []
" ", // []
};
for (String test : tests) {
System.out.format("[%s]%n",
test.replaceAll("^ +| +$| (?= )", "")
);
}
See String.replaceAll.
Use the regex "\s" and replace with " ".
Then use String.trim.
String str = " hello world"
reduce spaces first
str = str.trim().replaceAll(" +", " ");
capitalize the first letter and lowercase everything else
str = str.substring(0,1).toUpperCase() +str.substring(1,str.length()).toLowerCase();
you should do it like this
String mytext = " hello there ";
mytext = mytext.replaceAll("( +)", " ");
put + inside round brackets.
String str = " this is string ";
str = str.replaceAll("\\s+", " ").trim();
This worked for me
scan= filter(scan, " [\\s]+", " ");
scan= sac.trim();
where filter is following function and scan is the input string:
public String filter(String scan, String regex, String replace) {
StringBuffer sb = new StringBuffer();
Pattern pt = Pattern.compile(regex);
Matcher m = pt.matcher(scan);
while (m.find()) {
m.appendReplacement(sb, replace);
}
m.appendTail(sb);
return sb.toString();
}
The simplest method for removing white space anywhere in the string.
public String removeWhiteSpaces(String returnString){
returnString = returnString.trim().replaceAll("^ +| +$|( )+", " ");
return returnString;
}
check this...
public static void main(String[] args) {
String s = "A B C D E F G\tH I\rJ\nK\tL";
System.out.println("Current : "+s);
System.out.println("Single Space : "+singleSpace(s));
System.out.println("Space count : "+spaceCount(s));
System.out.format("Replace all = %s", s.replaceAll("\\s+", ""));
// Example where it uses the most.
String s = "My name is yashwanth . M";
String s2 = "My nameis yashwanth.M";
System.out.println("Normal : "+s.equals(s2));
System.out.println("Replace : "+s.replaceAll("\\s+", "").equals(s2.replaceAll("\\s+", "")));
}
If String contains only single-space then replace() will not-replace,
If spaces are more than one, Then replace() action performs and removes spacess.
public static String singleSpace(String str){
return str.replaceAll(" +| +|\t|\r|\n","");
}
To count the number of spaces in a String.
public static String spaceCount(String str){
int i = 0;
while(str.indexOf(" ") > -1){
//str = str.replaceFirst(" ", ""+(i++));
str = str.replaceFirst(Pattern.quote(" "), ""+(i++));
}
return str;
}
Pattern.quote("?") returns literal pattern String.
My method before I found the second answer using regex as a better solution. Maybe someone needs this code.
private String replaceMultipleSpacesFromString(String s){
if(s.length() == 0 ) return "";
int timesSpace = 0;
String res = "";
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if(c == ' '){
timesSpace++;
if(timesSpace < 2)
res += c;
}else{
res += c;
timesSpace = 0;
}
}
return res.trim();
}
Stream version, filters spaces and tabs.
Stream.of(str.split("[ \\t]")).filter(s -> s.length() > 0).collect(Collectors.joining(" "))
I know replaceAll method is much easier but I wanted to post this as well.
public static String removeExtraSpace(String input) {
input= input.trim();
ArrayList <String> x= new ArrayList<>(Arrays.asList(input.split("")));
for(int i=0; i<x.size()-1;i++) {
if(x.get(i).equals(" ") && x.get(i+1).equals(" ")) {
x.remove(i);
i--;
}
}
String word="";
for(String each: x)
word+=each;
return word;
}
String myText = " Hello World ";
myText = myText.trim().replace(/ +(?= )/g,'');
// Output: "Hello World"
string.replaceAll("\s+", " ");
If you already use Guava (v. 19+) in your project you may want to use this:
CharMatcher.whitespace().trimAndCollapseFrom(input, ' ');
or, if you need to remove exactly SPACE symbol ( or U+0020, see more whitespaces) use:
CharMatcher.anyOf(" ").trimAndCollapseFrom(input, ' ');
public class RemoveExtraSpacesEfficient {
public static void main(String[] args) {
String s = "my name is mr space ";
char[] charArray = s.toCharArray();
char prev = s.charAt(0);
for (int i = 0; i < charArray.length; i++) {
char cur = charArray[i];
if (cur == ' ' && prev == ' ') {
} else {
System.out.print(cur);
}
prev = cur;
}
}
}
The above solution is the algorithm with the complexity of O(n) without using any java function.
Please use below code
package com.myjava.string;
import java.util.StringTokenizer;
public class MyStrRemoveMultSpaces {
public static void main(String a[]){
String str = "String With Multiple Spaces";
StringTokenizer st = new StringTokenizer(str, " ");
StringBuffer sb = new StringBuffer();
while(st.hasMoreElements()){
sb.append(st.nextElement()).append(" ");
}
System.out.println(sb.toString().trim());
}
}