I'm trying to write a function that extracts each word from a sentence that contains a certain substring e.g. Looking for 'Po' in 'Porky Pork Chop' will return Porky Pork.
I've tested my regex on regexpal but the Java code doesn't seem to work. What am I doing wrong?
private static String foo()
{
String searchTerm = "Pizza";
String text = "Cheese Pizza";
String sPattern = "(?i)\b("+searchTerm+"(.+?)?)\b";
Pattern pattern = Pattern.compile ( sPattern );
Matcher matcher = pattern.matcher ( text );
if(matcher.find ())
{
String result = "-";
for(int i=0;i < matcher.groupCount ();i++)
{
result+= matcher.group ( i ) + " ";
}
return result.trim ();
}else
{
System.out.println("No Luck");
}
}
In Java to pass \b word boundaries to regex engine you need to write it as \\b. \b represents backspace in String object.
Judging by your example you want to return all words that contains your substring. To do this don't use for(int i=0;i < matcher.groupCount ();i++) but while(matcher.find()) since group count will iterate over all groups in single match, not over all matches.
In case your string can contain some special characters you probably should use Pattern.quote(searchTerm)
In your code you are trying to find "Pizza" in "Cheese Pizza" so I assume that you also want to find strings that same as searched substring. Although your regex will work fine for it, you can change your last part (.+?)?) to \\w* and also add \\w* at start if substring should also be matched in the middle of word (not only at start).
So your code can look like
private static String foo() {
String searchTerm = "Pizza";
String text = "Cheese Pizza, Other Pizzas";
String sPattern = "(?i)\\b\\w*" + Pattern.quote(searchTerm) + "\\w*\\b";
StringBuilder result = new StringBuilder("-").append(searchTerm).append(": ");
Pattern pattern = Pattern.compile(sPattern);
Matcher matcher = pattern.matcher(text);
while (matcher.find()) {
result.append(matcher.group()).append(' ');
}
return result.toString().trim();
}
While the regex approach is certainly a valid method, I find it easier to think through when you split the words up by whitespace. This can be done with String's split method.
public List<String> doIt(final String inputString, final String term) {
final List<String> output = new ArrayList<String>();
final String[] parts = input.split("\\s+");
for(final String part : parts) {
if(part.indexOf(term) > 0) {
output.add(part);
}
}
return output;
}
Of course it is worth nothing that doing this will effectively be doing two passes through your input String. The first pass to find the characters that are whitespace to split on, and the second pass looking through each split word for your substring.
If one pass is necessary though, the regex path is better.
I find nicholas.hauschild's answer to be the best.
However if you really wanted to use regex, you could do it as such:
String searchTerm = "Pizza";
String text = "Cheese Pizza";
Pattern pattern = Pattern.compile("\\b" + Pattern.quote(searchTerm)
+ "\\b", Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(text);
while (matcher.find()) {
System.out.println(matcher.group());
}
Output:
Pizza
The pattern should have been
String sPattern = "(?i)\\b("+searchTerm+"(?:.+?)?)\\b";
You want to capture the whole (pizza)string.?: ensures you don't capture a part of the string twice.
Try this pattern:
String searchTerm = "Po";
String text = "Porky Pork Chop oPod zzz llPo";
Pattern p = Pattern.compile("\\p{Alpha}+" + substring + "|\\p{Alpha}+" + substring + "\\p{Alpha}+|" + substring + "\\p{Alpha}+");
Matcher m = p.matcher(myString);
while(m.find()) {
System.out.println(">> " + m.group());
}
Ok, I give you a pattern in raw style (not java style, you must double escape yourself):
(?i)\b[a-z]*po[a-z]*\b
And that's all.
Related
I have several strings in the rough form:
[some text] [some number] [some more text]
I want to extract the text in [some number] using the Java Regex classes.
I know roughly what regular expression I want to use (though all suggestions are welcome). What I'm really interested in are the Java calls to take the regex string and use it on the source data to produce the value of [some number].
EDIT: I should add that I'm only interested in a single [some number] (basically, the first instance). The source strings are short and I'm not going to be looking for multiple occurrences of [some number].
Full example:
private static final Pattern p = Pattern.compile("^([a-zA-Z]+)([0-9]+)(.*)");
public static void main(String[] args) {
// create matcher for pattern p and given string
Matcher m = p.matcher("Testing123Testing");
// if an occurrence if a pattern was found in a given string...
if (m.find()) {
// ...then you can use group() methods.
System.out.println(m.group(0)); // whole matched expression
System.out.println(m.group(1)); // first expression from round brackets (Testing)
System.out.println(m.group(2)); // second one (123)
System.out.println(m.group(3)); // third one (Testing)
}
}
Since you're looking for the first number, you can use such regexp:
^\D+(\d+).*
and m.group(1) will return you the first number. Note that signed numbers can contain a minus sign:
^\D+(-?\d+).*
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Regex1 {
public static void main(String[]args) {
Pattern p = Pattern.compile("\\d+");
Matcher m = p.matcher("hello1234goodboy789very2345");
while(m.find()) {
System.out.println(m.group());
}
}
}
Output:
1234
789
2345
Allain basically has the java code, so you can use that. However, his expression only matches if your numbers are only preceded by a stream of word characters.
"(\\d+)"
should be able to find the first string of digits. You don't need to specify what's before it, if you're sure that it's going to be the first string of digits. Likewise, there is no use to specify what's after it, unless you want that. If you just want the number, and are sure that it will be the first string of one or more digits then that's all you need.
If you expect it to be offset by spaces, it will make it even more distinct to specify
"\\s+(\\d+)\\s+"
might be better.
If you need all three parts, this will do:
"(\\D+)(\\d+)(.*)"
EDIT The Expressions given by Allain and Jack suggest that you need to specify some subset of non-digits in order to capture digits. If you tell the regex engine you're looking for \d then it's going to ignore everything before the digits. If J or A's expression fits your pattern, then the whole match equals the input string. And there's no reason to specify it. It probably slows a clean match down, if it isn't totally ignored.
In addition to Pattern, the Java String class also has several methods that can work with regular expressions, in your case the code will be:
"ab123abc".replaceFirst("\\D*(\\d*).*", "$1")
where \\D is a non-digit character.
In Java 1.4 and up:
String input = "...";
Matcher matcher = Pattern.compile("[^0-9]+([0-9]+)[^0-9]+").matcher(input);
if (matcher.find()) {
String someNumberStr = matcher.group(1);
// if you need this to be an int:
int someNumberInt = Integer.parseInt(someNumberStr);
}
This function collect all matching sequences from string. In this example it takes all email addresses from string.
static final String EMAIL_PATTERN = "[_A-Za-z0-9-\\+]+(\\.[_A-Za-z0-9-]+)*#"
+ "[A-Za-z0-9-]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})";
public List<String> getAllEmails(String message) {
List<String> result = null;
Matcher matcher = Pattern.compile(EMAIL_PATTERN).matcher(message);
if (matcher.find()) {
result = new ArrayList<String>();
result.add(matcher.group());
while (matcher.find()) {
result.add(matcher.group());
}
}
return result;
}
For message = "adf#gmail.com, <another#osiem.osiem>>>> lalala#aaa.pl" it will create List of 3 elements.
Try doing something like this:
Pattern p = Pattern.compile("^.+(\\d+).+");
Matcher m = p.matcher("Testing123Testing");
if (m.find()) {
System.out.println(m.group(1));
}
Simple Solution
// Regexplanation:
// ^ beginning of line
// \\D+ 1+ non-digit characters
// (\\d+) 1+ digit characters in a capture group
// .* 0+ any character
String regexStr = "^\\D+(\\d+).*";
// Compile the regex String into a Pattern
Pattern p = Pattern.compile(regexStr);
// Create a matcher with the input String
Matcher m = p.matcher(inputStr);
// If we find a match
if (m.find()) {
// Get the String from the first capture group
String someDigits = m.group(1);
// ...do something with someDigits
}
Solution in a Util Class
public class MyUtil {
private static Pattern pattern = Pattern.compile("^\\D+(\\d+).*");
private static Matcher matcher = pattern.matcher("");
// Assumptions: inputStr is a non-null String
public static String extractFirstNumber(String inputStr){
// Reset the matcher with a new input String
matcher.reset(inputStr);
// Check if there's a match
if(matcher.find()){
// Return the number (in the first capture group)
return matcher.group(1);
}else{
// Return some default value, if there is no match
return null;
}
}
}
...
// Use the util function and print out the result
String firstNum = MyUtil.extractFirstNumber("Testing4234Things");
System.out.println(firstNum);
Look you can do it using StringTokenizer
String str = "as:"+123+"as:"+234+"as:"+345;
StringTokenizer st = new StringTokenizer(str,"as:");
while(st.hasMoreTokens())
{
String k = st.nextToken(); // you will get first numeric data i.e 123
int kk = Integer.parseInt(k);
System.out.println("k string token in integer " + kk);
String k1 = st.nextToken(); // you will get second numeric data i.e 234
int kk1 = Integer.parseInt(k1);
System.out.println("new string k1 token in integer :" + kk1);
String k2 = st.nextToken(); // you will get third numeric data i.e 345
int kk2 = Integer.parseInt(k2);
System.out.println("k2 string token is in integer : " + kk2);
}
Since we are taking these numeric data into three different variables we can use this data anywhere in the code (for further use)
How about [^\\d]*([0-9]+[\\s]*[.,]{0,1}[\\s]*[0-9]*).* I think it would take care of numbers with fractional part.
I included white spaces and included , as possible separator.
I'm trying to get the numbers out of a string including floats and taking into account that the user might make a mistake and include white spaces while typing the number.
Sometimes you can use simple .split("REGEXP") method available in java.lang.String. For example:
String input = "first,second,third";
//To retrieve 'first'
input.split(",")[0]
//second
input.split(",")[1]
//third
input.split(",")[2]
if you are reading from file then this can help you
try{
InputStream inputStream = (InputStream) mnpMainBean.getUploadedBulk().getInputStream();
BufferedReader br = new BufferedReader(new InputStreamReader(inputStream));
String line;
//Ref:03
while ((line = br.readLine()) != null) {
if (line.matches("[A-Z],\\d,(\\d*,){2}(\\s*\\d*\\|\\d*:)+")) {
String[] splitRecord = line.split(",");
//do something
}
else{
br.close();
//error
return;
}
}
br.close();
}
}
catch (IOException ioExpception){
logger.logDebug("Exception " + ioExpception.getStackTrace());
}
Pattern p = Pattern.compile("(\\D+)(\\d+)(.*)");
Matcher m = p.matcher("this is your number:1234 thank you");
if (m.find()) {
String someNumberStr = m.group(2);
int someNumberInt = Integer.parseInt(someNumberStr);
}
For example, replace "HOW do I replace different how in the same sentence by using Matcher?" with "LOL do I replace different lol in the same sentence?"
If HOW is all caps, replace it with LOL. Otherwise, replace it with lol.
I only know how to find them:
String source = "HOW do I replace different how in the same " +
"sentence by using Matcher?"
Pattern pattern = Pattern.compile(how, Pattern.CASE_INSENSITIVE);
Matcher m = pattern.matcher(source);
while (m.find()) {
if(m.group.match("^[A-Z]*$"))
System.out.println("I am uppercase");
else
System.out.println("I am lowercase");
}
But I don't know how to replace them by using matcher and pattern.
Here's one way to achieve your goal: (not necessarily the most efficient, but it works and is simply understood)
String source = "HOW do I replace different how in the same sentence by using Matcher?";
String[] split = source.replaceAll("HOW", "LOL").split(" ");
String newSource = "";
for(int i = 0; i < split.length; i++) {
String at = split[i];
if(at.equalsIgnoreCase("how")) at = "lol";
newSource+= " " + at;
}
newSource.substring(1, newSource.length());
//The output string is newSource
Replace all uppercase, then iterate over each word and replace the remaining "how"s with "lol". That substring at the end is simply to remove the extra space.
I came up with a really dumb solution:
String result = source;
result = result.replaceAll(old_Word, new_Word);
result = result.replaceAll(old_Word.toUpperCase(),
newWord.toUpperCase());
I have this alphabet: {'faa','fa','af'}
and I have this string: "faaf"
I have this regex: "(faa|fa|af)*" which helps me match the string with the alphabet.
How do I make Java trim my string into: {fa,af}, which is the correct way to write the string: "faaf" based on my alphabet?
here is my code:
String regex = "(faa|fa|af)*";
String str = "faaf";
boolean isMatch = Pattern.matches(regex, str);
if(isMatch)
{
//trim the string
while(str.length()!=0)
{
Pattern pattern = Pattern.compile("^(faa|fa|af)(faa|fa|af)*$");
Matcher mc = pattern.matcher(str);
if (mc.find())
{
String l =mc.group(1);
alphabet.add(l);
str = str.substring(l.length());
System.out.println("\n"+ l);
}
}
}
Thanks to Aaron who helped me with this problem.
You need a loop.
Pattern pattern = Pattern.compile(regex + "*");
LinkedList<String> parts = new LinkedList<>();
while (!str.isEmpty()) {
Matcher m = pattern.matcher(str);
if (!m.matches()) { // In the first loop step.
break;
}
parts.addFirst(m.group(1)); // The last repetition matching group.
str = str.substring(0, m.start(1));
}
String result = parts.stream().collect(Collectors.joining(", ", "{", "}"));
This utilizes that a match (X)+ will yield in m.group(1) the last occurrence's value of X.
Unfortunately the regex module does not provide a bored-open matches, such as the overloaded replaceAll with a lambda working on a single MatchResult.
Note that matches applies to the entire string.
I have string with multiple {!XXX} phrases. For example:
Kumar gaurav {!str1} is just {!str2}, adasdas {!str3}
I need to replace all {!str} values with corresponding str, how to replace all {!str} from my string?
You can use a Pattern and Matcher, which provides you the means to query the string for a unknown number of elements, in combination with a regular expression of \{!str\d\} which will allow you to break the text down based on the tags
For example...
String text = "All that {!str1} is {!str2}";
Map<String, String> values = new HashMap<>(25);
values.put("{!str1}", "glitters");
values.put("{!str2}", "gold");
Pattern p = Pattern.compile("\\{!str\\d\\}");
Matcher matcher = p.matcher(text);
while (matcher.find()) {
String match = matcher.group();
text = text.replaceAll("\\" + match, values.get(match));
}
System.out.println(text);
Which outputs
All that glitters is gold
You could also use something like...
int previousStart = 0;
StringBuilder sb = new StringBuilder();
while (matcher.find()) {
String match = matcher.group();
int start = matcher.start();
int end = matcher.end();
sb.append(text.substring(previousStart, start));
sb.append(values.get(match));
previousStart = end;
}
if (previousStart < text.length()) {
sb.append(text.substring(previousStart));
}
Which does away with the String creation in a loop and relies more on the position of the match to cut the original text around the tokens, which makes me happier ;)
use this regex, simple
String string="hello world {!hello}";
string=string.replaceAll("\\{!(.*?)\\}", "replace");
System.out.println(string); //this will print (hello world replace)
I have an array input like this which is an email id in reverse order along with some data:
MOC.OOHAY#ABC.PQRqwertySDdd
MOC.OOHAY#AB.JKLasDDbfn
MOC.OOHAY#XZ.JKGposDDbfn
I want my output to come as
MOC.OOHAY#ABC.PQR
MOC.OOHAY#AB.JKL
MOC.OOHAY#XZ.JKG
How should I filter the string since there is no pattern?
There is a pattern, and that is any upper case character which is followed either by another upper case letter, a period or else the # character.
Translated, this would become something like this:
String[] input = new String[]{"MOC.OOHAY#ABC.PQRqwertySDdd","MOC.OOHAY#AB.JKLasDDbfn" , "MOC.OOHAY#XZ.JKGposDDbfn"};
Pattern p = Pattern.compile("([A-Z.]+#[A-Z.]+)");
for(String string : input)
{
Matcher matcher = p.matcher(string);
if(matcher.find())
System.out.println(matcher.group(1));
}
Yields:
MOC.OOHAY#ABC.PQR
MOC.OOHAY#AB.JKL
MOC.OOHAY#XZ.JKG
Why do you think there is no pattern?
You clearly want to get the string till you find a lowercase letter.
You can use the regex (^[^a-z]+) to match it and extract.
Regex Demo
Simply split on [a-z], with limit 2:
String s1 = "MOC.OOHAY#ABC.PQRqwertySDdd";
String s2 = "MOC.OOHAY#AB.JKLasDDbfn";
String s3 = "MOC.OOHAY#XZ.JKGposDDbfn";
System.out.println(s1.split("[a-z]", 2)[0]);
System.out.println(s2.split("[a-z]", 2)[0]);
System.out.println(s3.split("[a-z]", 2)[0]);
Demo.
You can do it like this:
String arr[] = { "MOC.OOHAY#ABC.PQRqwertySDdd", "MOC.OOHAY#AB.JKLasDDbfn", "MOC.OOHAY#XZ.JKGposDDbfn" };
for (String test : arr) {
Pattern p = Pattern.compile("[A-Z]*\\.[A-Z]*#[A-Z]*\\.[A-Z.]*");
Matcher m = p.matcher(test);
if (m.find()) {
System.out.println(m.group());
}
}