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String s = "abc//jason:1234567#123.123.213.212/";
I want to replace all the substring before and after ":" delimiter with "......."
I want my final output to be :
"abc//.....:.......#123.123.213:212/"
I tried doing this since there is a second : in the string it gets messed up, is it there better way to be able to get my output:
String [] headersplit;
headersplit = s.split(":");
If you want to locate only symbols between "//" and "#" then algorithm is simple, provided that mention symbols are compulsory.
public class Main {
public static void main(String[] args) {
String s = "abc//jason:1234567#123.123.213.212/";
System.out.println(replaceSensitiveInfo(s));
}
static String replaceSensitiveInfo(String src) {
int slashes = src.indexOf("//");
int colon = src.indexOf(":", slashes);
int at = src.indexOf("#", colon);
StringBuilder sb = new StringBuilder(src);
sb.replace(slashes + 2, colon, ".".repeat(colon - slashes - 2));
sb.replace(colon + 1, at, ".".repeat(at - colon - 1));
return sb.toString();
}
}
Not the best way but it works for your example and should work for others:
String s = "abc//jason:1234567#123.123.213:212/";
String result = replaceSensitiveInfo(s);
private String replaceSensitiveInfo(String info){
StringBuilder sb = new StringBuilder(info);
String substitute = ".";
int start = sb.indexOf("//") + 2;
int end = sb.indexOf(":");
String firstReplace = substitute.repeat(end - start);
sb.replace(start, end, firstReplace);
int start2 = sb.indexOf(":") + 1;
int end2 = sb.indexOf("#");
String secondReplace = substitute.repeat(end2 - start2);
sb.replace(start2, end2, secondReplace);
return sb.toString();
}
i need replace json mapper generated string sequences as fallow:
:"{ -> :{
}"} -> }}
how would look pattern for that ?
update: full string example
{"method":"createInvoice","params":"{"btcDue":null,"btcPaid":null,
"btcPrice":null,"currency":"PLN","currentTime":null,
"exceptionStatus":null,"expirationTime":null,
"guid":"99250130","id":null,"invoiceTime":null,
"paymentUrls":null,"price":1.23,"rate":null,
"status":null,"transactions":null,"url":null
}"}
but assume we will have more instances to replace like 2 :)
to clarify: android String methods
public String replace(CharSequence target, CharSequence replacement) {
String replacementStr = replacement.toString();
String targetStr = target.toString();
// Special case when target == "".
// .. cut
// This is the "regular" case.
int lastMatch = 0;
StringBuilder sb = null;
for (;;) {
int currentMatch = indexOf(this, targetStr, lastMatch);
if (currentMatch == -1) {
break;
}
if (sb == null) {
sb = new StringBuilder(count);
}
sb.append(this, lastMatch, currentMatch);
sb.append(replacementStr);
lastMatch = currentMatch + targetStr.count;
}
if (sb != null) {
sb.append(this, lastMatch, count);
return sb.toString();
} else {
return this;
}
}
public String replaceAll(String regex, String replacement) {
return Pattern.compile(regex).matcher(this).replaceAll(replacement);
}
Nothing difficult about it:
String result = "{\"method\":\"createInvoice\",\"params\":\"{\"btcDue\":null,\"btcPaid\":null,\"btcPrice\":null,\"currency\":\"PLN\",\"currentTime\":null,\"exceptionStatus\":null,\"expirationTime\":null,\"guid\":\"99250130\",\"id\":null,\"invoiceTime\":null,\"paymentUrls\":null,\"price\":1.23,\"rate\":null,\"status\":null,\"transactions\":null,\"url\":null}\"}"
.replace(":\"{", ":{")
.replace("}\"}", "}}");
System.out.println(result);
You can match all quotes surrounded by :+{ and }+} using the lookahead/lookbehind construct:
(?<=})"(?=})|(?<=:)"(?={)
Pass this to replaceAll to remove the quotes (demo).
Search by
\"
Then replace it to ""
Original Text
:"{
}"}
Result
:{
}}
JAVA CODE
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "\\\"";
final String string = ":\"{ -> :{ \n"
+ "}\"} -> }} ";
final String subst = "";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
// The substituted value will be contained in the result variable
final String result = matcher.replaceAll(subst);
System.out.println("Substitution result: " + result);
I wanna do something like this!
So i am only left with the website part of the string. I was having problems with the quotations within the string.
/////////////////////This is what i read into a string.
///<td width="118"><a href="research.html" class="navText style10 style12">
///////I wanna be able to parse this so i am only left with research.html
//I sometimes also get a string that contains:
//<font size="3">University of Calgary</font></div>
//From this string i wanna keep http://www.ucalgary.ca
What I got so far doesnt always work for every case. I would appreciate your help!! My code is
public class Parse
{
public static void main(String[] args)
{
String h = "<a href=\"http://www.departmentofmedicine.com/policy.htm\">";
int n = getIndexOf(h, '"', 0);
String[] a = h.substring(n).split(">");
String url = a[0].replaceAll("\"", "");
//String value = a[1].replaceAll("</a", "");
System.out.println(url + " " );
}
public static int getIndexOf(String str, char c, int n)
{
int pos = str.indexOf(c, 0);
while (n-- > 0 && pos != -1)
{
pos = str.indexOf(c, pos + 1);
}
return pos;
}
}
I would give Pattern and Matcher a try like this:
String s = "<a href=\"http://www.departmentofmedicine.com/policy.htm\">";
Pattern p = Pattern.compile(".*href=\"([^\"]*).*");
Matcher m = p.matcher(s);
if(m.matches()) {
System.out.println(m.group(1));
}
Little code:
String h = "http://www.departmentofmedicine.com/policy.htm\">";
String url = h.substring(h.indexOf("http")).replace("\">", "");
System.out.println(url);
Output will be:
http://www.departmentofmedicine.com/policy.htm
Tested on my machine.
Also post what are possible cases. So that I can tell you better solution.
Solution for all three posibilities:
//String h1 = "<a href=\"http://www.departmentofmedicine.com/policy.htm\">";
//String h1 = `"<font size=\"3\">University of Calgary</font>";
String h1="<td width=\"118\"><a href=\"research.html\" class=\"navText style10 style12\">";`
String url = h1.substring(h1.indexOf("href=\"") + "href=\"".length()).substring(0, h1.substring(h1.indexOf("href=\"") + "href=\"".length()).indexOf("\""));
System.out.println(url);
Uncomment String h1; object one by one and check your requirements.
Above code is giving output:
research.html
http://www.departmentofmedicine.com/policy.htm
ucalgary.ca
I have a string,
String s = "test string (67)";
I want to get the no 67 which is the string between ( and ).
Can anyone please tell me how to do this?
There's probably a really neat RegExp, but I'm noob in that area, so instead...
String s = "test string (67)";
s = s.substring(s.indexOf("(") + 1);
s = s.substring(0, s.indexOf(")"));
System.out.println(s);
A very useful solution to this issue which doesn't require from you to do the indexOf is using Apache Commons libraries.
StringUtils.substringBetween(s, "(", ")");
This method will allow you even handle even if there multiple occurrences of the closing string which wont be easy by looking for indexOf closing string.
You can download this library from here:
https://mvnrepository.com/artifact/org.apache.commons/commons-lang3/3.4
Try it like this
String s="test string(67)";
String requiredString = s.substring(s.indexOf("(") + 1, s.indexOf(")"));
The method's signature for substring is:
s.substring(int start, int end);
By using regular expression :
String s = "test string (67)";
Pattern p = Pattern.compile("\\(.*?\\)");
Matcher m = p.matcher(s);
if(m.find())
System.out.println(m.group().subSequence(1, m.group().length()-1));
Java supports Regular Expressions, but they're kind of cumbersome if you actually want to use them to extract matches. I think the easiest way to get at the string you want in your example is to just use the Regular Expression support in the String class's replaceAll method:
String x = "test string (67)".replaceAll(".*\\(|\\).*", "");
// x is now the String "67"
This simply deletes everything up-to-and-including the first (, and the same for the ) and everything thereafter. This just leaves the stuff between the parenthesis.
However, the result of this is still a String. If you want an integer result instead then you need to do another conversion:
int n = Integer.parseInt(x);
// n is now the integer 67
In a single line, I suggest:
String input = "test string (67)";
input = input.subString(input.indexOf("(")+1, input.lastIndexOf(")"));
System.out.println(input);`
You could use apache common library's StringUtils to do this.
import org.apache.commons.lang3.StringUtils;
...
String s = "test string (67)";
s = StringUtils.substringBetween(s, "(", ")");
....
Test String test string (67) from which you need to get the String which is nested in-between two Strings.
String str = "test string (67) and (77)", open = "(", close = ")";
Listed some possible ways: Simple Generic Solution:
String subStr = str.substring(str.indexOf( open ) + 1, str.indexOf( close ));
System.out.format("String[%s] Parsed IntValue[%d]\n", subStr, Integer.parseInt( subStr ));
Apache Software Foundation commons.lang3.
StringUtils class substringBetween() function gets the String that is nested in between two Strings. Only the first match is returned.
String substringBetween = StringUtils.substringBetween(subStr, open, close);
System.out.println("Commons Lang3 : "+ substringBetween);
Replaces the given String, with the String which is nested in between two Strings. #395
Pattern with Regular-Expressions: (\()(.*?)(\)).*
The Dot Matches (Almost) Any Character
.? = .{0,1}, .* = .{0,}, .+ = .{1,}
String patternMatch = patternMatch(generateRegex(open, close), str);
System.out.println("Regular expression Value : "+ patternMatch);
Regular-Expression with the utility class RegexUtils and some functions.
Pattern.DOTALL: Matches any character, including a line terminator.
Pattern.MULTILINE: Matches entire String from the start^ till end$ of the input sequence.
public static String generateRegex(String open, String close) {
return "(" + RegexUtils.escapeQuotes(open) + ")(.*?)(" + RegexUtils.escapeQuotes(close) + ").*";
}
public static String patternMatch(String regex, CharSequence string) {
final Pattern pattern = Pattern.compile(regex, Pattern.DOTALL);
final Matcher matcher = pattern .matcher(string);
String returnGroupValue = null;
if (matcher.find()) { // while() { Pattern.MULTILINE }
System.out.println("Full match: " + matcher.group(0));
System.out.format("Character Index [Start:End]«[%d:%d]\n",matcher.start(),matcher.end());
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
if( i == 2 ) returnGroupValue = matcher.group( 2 );
}
}
return returnGroupValue;
}
String s = "test string (67)";
int start = 0; // '(' position in string
int end = 0; // ')' position in string
for(int i = 0; i < s.length(); i++) {
if(s.charAt(i) == '(') // Looking for '(' position in string
start = i;
else if(s.charAt(i) == ')') // Looking for ')' position in string
end = i;
}
String number = s.substring(start+1, end); // you take value between start and end
String result = s.substring(s.indexOf("(") + 1, s.indexOf(")"));
public String getStringBetweenTwoChars(String input, String startChar, String endChar) {
try {
int start = input.indexOf(startChar);
if (start != -1) {
int end = input.indexOf(endChar, start + startChar.length());
if (end != -1) {
return input.substring(start + startChar.length(), end);
}
}
} catch (Exception e) {
e.printStackTrace();
}
return input; // return null; || return "" ;
}
Usage :
String input = "test string (67)";
String startChar = "(";
String endChar = ")";
String output = getStringBetweenTwoChars(input, startChar, endChar);
System.out.println(output);
// Output: "67"
Another way of doing using split method
public static void main(String[] args) {
String s = "test string (67)";
String[] ss;
ss= s.split("\\(");
ss = ss[1].split("\\)");
System.out.println(ss[0]);
}
Use Pattern and Matcher
public class Chk {
public static void main(String[] args) {
String s = "test string (67)";
ArrayList<String> arL = new ArrayList<String>();
ArrayList<String> inL = new ArrayList<String>();
Pattern pat = Pattern.compile("\\(\\w+\\)");
Matcher mat = pat.matcher(s);
while (mat.find()) {
arL.add(mat.group());
System.out.println(mat.group());
}
for (String sx : arL) {
Pattern p = Pattern.compile("(\\w+)");
Matcher m = p.matcher(sx);
while (m.find()) {
inL.add(m.group());
System.out.println(m.group());
}
}
System.out.println(inL);
}
}
The "generic" way of doing this is to parse the string from the start, throwing away all the characters before the first bracket, recording the characters after the first bracket, and throwing away the characters after the second bracket.
I'm sure there's a regex library or something to do it though.
The least generic way I found to do this with Regex and Pattern / Matcher classes:
String text = "test string (67)";
String START = "\\("; // A literal "(" character in regex
String END = "\\)"; // A literal ")" character in regex
// Captures the word(s) between the above two character(s)
String pattern = START + "(\w+)" + END;
Pattern pattern = Pattern.compile(pattern);
Matcher matcher = pattern.matcher(text);
while(matcher.find()) {
System.out.println(matcher.group()
.replace(START, "").replace(END, ""));
}
This may help for more complex regex problems where you want to get the text between two set of characters.
The other possible solution is to use lastIndexOf where it will look for character or String from backward.
In my scenario, I had following String and I had to extract <<UserName>>
1QAJK-WKJSH_MyApplication_Extract_<<UserName>>.arc
So, indexOf and StringUtils.substringBetween was not helpful as they start looking for character from beginning.
So, I used lastIndexOf
String str = "1QAJK-WKJSH_MyApplication_Extract_<<UserName>>.arc";
String userName = str.substring(str.lastIndexOf("_") + 1, str.lastIndexOf("."));
And, it gives me
<<UserName>>
String s = "test string (67)";
System.out.println(s.substring(s.indexOf("(")+1,s.indexOf(")")));
Something like this:
public static String innerSubString(String txt, char prefix, char suffix) {
if(txt != null && txt.length() > 1) {
int start = 0, end = 0;
char token;
for(int i = 0; i < txt.length(); i++) {
token = txt.charAt(i);
if(token == prefix)
start = i;
else if(token == suffix)
end = i;
}
if(start + 1 < end)
return txt.substring(start+1, end);
}
return null;
}
This is a simple use \D+ regex and job done.
This select all chars except digits, no need to complicate
/\D+/
it will return original string if no match regex
var iAm67 = "test string (67)".replaceFirst("test string \\((.*)\\)", "$1");
add matches to the code
String str = "test string (67)";
String regx = "test string \\((.*)\\)";
if (str.matches(regx)) {
var iAm67 = str.replaceFirst(regx, "$1");
}
---EDIT---
i use https://www.freeformatter.com/java-regex-tester.html#ad-output to test regex.
turn out it's better to add ? after * for less match. something like this:
String str = "test string (67)(69)";
String regx1 = "test string \\((.*)\\).*";
String regx2 = "test string \\((.*?)\\).*";
String ans1 = str.replaceFirst(regx1, "$1");
String ans2 = str.replaceFirst(regx2, "$1");
System.out.println("ans1:"+ans1+"\nans2:"+ans2);
// ans1:67)(69
// ans2:67
String s = "(69)";
System.out.println(s.substring(s.lastIndexOf('(')+1,s.lastIndexOf(')')));
Little extension to top (MadProgrammer) answer
public static String getTextBetween(final String wholeString, final String str1, String str2){
String s = wholeString.substring(wholeString.indexOf(str1) + str1.length());
s = s.substring(0, s.indexOf(str2));
return s;
}
So say I have a string called x that = "Hello world". I want to somehow make it so that it will flip those two words and instead display "world Hello". I am not very good with loops or arrays and obviously am a beginner. Could I accomplish this somehow by splitting my string? If so, how? If not, how could I do this? Help would be appreciated, thanks!
1) split string into String array on space.
String myArray[] = x.split(" ");
2) Create new string with words in reverse order from array.
String newString = myArray[1] + " " + myArray[0];
Bonus points for using a StringBuilder instead of concatenation.
String abc = "Hello world";
String cba = abc.replace( "Hello world", "world Hello" );
abc = "This is a longer string. Hello world. My String";
cba = abc.replace( "Hello world", "world Hello" );
If you want, you can explode your string as well:
String[] pieces = abc.split(" ");
for( int i=0; i<pieces.length-1; ++i )
if( pieces[i]=="Hello" && pieces[i+1]=="world" ) swap(pieces[i], pieces[i+1]);
There are many other ways you can do it too. Be careful for capitalization. You can use .toUpperCase() in your if statements and then make your matching conditionals uppercase, but leave the results with their original capitalization, etc.
Here's the solution:
import java.util.*;
public class ReverseWords {
public String reverseWords(String phrase) {
List<String> wordList = Arrays.asList(phrase.split("[ ]"));
Collections.reverse(wordList);
StringBuilder sbReverseString = new StringBuilder();
for(String word: wordList) {
sbReverseString.append(word + " ");
}
return sbReverseString.substring(0, sbReverseString.length() - 1);
}
}
The above solution was coded by me, for Google Code Jam and is also blogged here: Reverse Words - GCJ 2010
Just use this method, call it and pass the string that you want to split out
static String reverseWords(String str) {
// Specifying the pattern to be searched
Pattern pattern = Pattern.compile("\\s");
// splitting String str with a pattern
// (i.e )splitting the string whenever their
// is whitespace and store in temp array.
String[] temp = pattern.split(str);
String result = "";
// Iterate over the temp array and store
// the string in reverse order.
for (int i = 0; i < temp.length; i++) {
if (i == temp.length - 1) {
result = temp[i] + result;
} else {
result = " " + temp[i] + result;
}
}
return result;
}
Depending on your exact requirements, you may want to split on other forms of whitespace (tabs, multiple spaces, etc.):
static Pattern p = Pattern.compile("(\\S+)(\\s+)(\\S+)");
public String flipWords(String in)
{
Matcher m = p.matcher(in);
if (m.matches()) {
// reverse the groups we found
return m.group(3) + m.group(2) + m.group(1);
} else {
return in;
}
}
If you want to get more complex see the docs for Pattern http://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html
Try something as follows:
String input = "how is this";
List<String> words = Arrays.asList(input.split(" "));
Collections.reverse(words);
String result = "";
for(String word : words) {
if(!result.isEmpty()) {
result += " ";
}
result += word;
}
System.out.println(result);
Output:
this is how
Too much?
private static final Pattern WORD = Pattern.compile("^(\\p{L}+)");
private static final Pattern NUMBER = Pattern.compile("^(\\p{N}+)");
private static final Pattern SPACE = Pattern.compile("^(\\p{Z}+)");
public static String reverseWords(final String text) {
final StringBuilder sb = new StringBuilder(text.length());
final Matcher wordMatcher = WORD.matcher(text);
final Matcher numberMatcher = NUMBER.matcher(text);
final Matcher spaceMatcher = SPACE.matcher(text);
int offset = 0;
while (offset < text.length()) {
wordMatcher.region(offset, text.length());
numberMatcher.region(offset, text.length());
spaceMatcher.region(offset, text.length());
if (wordMatcher.find()) {
final String word = wordMatcher.group();
sb.insert(0, reverseCamelCase(word));
offset = wordMatcher.end();
} else if (numberMatcher.find()) {
sb.insert(0, numberMatcher.group());
offset = numberMatcher.end();
} else if (spaceMatcher.find()) {
sb.insert(0, spaceMatcher.group(0));
offset = spaceMatcher.end();
} else {
sb.insert(0, text.charAt(offset++));
}
}
return sb.toString();
}
private static final Pattern CASE_REVERSAL = Pattern
.compile("(\\p{Lu})(\\p{Ll}*)(\\p{Ll})$");
private static String reverseCamelCase(final String word) {
final StringBuilder sb = new StringBuilder(word.length());
final Matcher caseReversalMatcher = CASE_REVERSAL.matcher(word);
int wordEndOffset = word.length();
while (wordEndOffset > 0 && caseReversalMatcher.find()) {
sb.insert(0, caseReversalMatcher.group(3).toUpperCase());
sb.insert(0, caseReversalMatcher.group(2));
sb.insert(0, caseReversalMatcher.group(1).toLowerCase());
wordEndOffset = caseReversalMatcher.start();
caseReversalMatcher.region(0, wordEndOffset);
}
sb.insert(0, word.substring(0, wordEndOffset));
return sb.toString();
}