It was very hard to form the question and I am sure it is still not clear.
I have a CSV file e.g.: Firstname;Lastname;Adress;product1;product2;product3;product4;
I would like to start replacing ";" with "::". The problem is, I want to start replacing after third semicolon.
I know it can be done in while loop where I check every character, when semicolon occurs I will count +1 and if counter is 3, I will start replacing. But isn't there a way how to do it without a loop?
You can use indexOf(char,fromIndex) method.
Your third semicolon position search can be inlined :
csvLine.indexOf(';', csvLine.indexOf(';', csvLine.indexOf(';') + 1) + 1)
We assume that our csvLine has a least 3 semi-colons...
String csvLine = "Firstname;Lastname;Adress;product1;product2;product3;product4";
//Index of "fromIndex" param is inclusive, that's why we need to add 1
int pos = csvLine.indexOf(';', csvLine.indexOf(';', csvLine.indexOf(';') + 1) + 1);
//Retrieve string from the char after the third semi-colon
String truncatedLine = csvLine.substring(pos + 1);
//Replace ";" by "::" on our substring
truncatedLine = truncatedLine.replaceAll(";", "::");
//Then concat the first part of csvLine with the second
String result = csvLine.substring(0, pos + 1).concat(truncatedLine);
System.out.println(result); //Print => Firstname;Lastname;Adress;product1::product2::product3::product4
Poor input control and performance but we don't have any loops :)
If I have understood what you want, try this.
First search se position of the third semicolon:
String csvContent = "Firstname;Lastname;Adress;product1;product2;product3;product4;";
int i = 0;
int index= 0;
while(i < 4){
index = csvContent.indexOf(';', (index + 1));
i++;
}//index = position of the thrid semicolon
Second, cut your CSV content at the index position.
String tmp1 = csvContent.substring(0, index);
String tmp2 = csvContent.substring(index, csvContent.length());
Thrid, replace all ';' by '::':
tmp2 = tmp2.replaceAll(";", "::");
Finaly, rebuild your file content:
csvContent = tmp1 + tmp2;
int i = 0;
int pos = 0;
while (i < 3) {
pos = string.indexOf(';', pos+1);
i++;
}
String newString = string.substring(0, pos) +";"+ (string.substring(pos + 1, string.length()).replace(";", "::"));
how about a regex solution?
Pattern pattern = Pattern.compile("(.*?;.*?;.*?;)(.*)");
Matcher match = pattern.matcher(str);
if(match.matches()) {
String firstThree = match.group(1);
String rest = match.group(2);
rest = rest.replace(";", "::");
return firstThree + rest;
}
I have a string like a1wwa1xxa1yya1zz.
I would like to get every groups starting with a1 until next a1 excluded.
(In my example, i would be : a1ww, a1xx, a1yyand a1zz
If I use :
Matcher m = Pattern.compile("(a1.*?)a1").matcher("a1wwa1xxa1yya1zz");
while(m.find()) {
String myGroup = m.group(1);
}
myGroup capture 1 group every two groups.
So in my example, I can only capture a1ww and a1yy.
Anyone have a great idea ?
Split is a good solution, but if you want to remain in the regex world, here is a solution:
Matcher m = Pattern.compile("(a1.*?)(?=a1|$)").matcher("a1wwa1xxa1yya1zz");
while (m.find()) {
String myGroup = m.group(1);
System.out.println("> " + myGroup);
}
I used a positive lookahead to ensure the capture is followed by a1, or alternatively by the end of line.
Lookahead are zero-width assertions, ie. they verify a condition without advancing the match cursor, so the string they verify remains available for further testing.
You can use split() method, then append "a1" as a prefix to splitted elements:
String str = "a1wwa1xxa1yya1zz";
String[] parts = str.split("a1");
String[] output = new String[parts.length - 1];
for (int i = 0; i < output.length; i++)
output[i] = "a1" + parts[i + 1];
for (String p : output)
System.out.println(p);
Output:
a1ww
a1xx
a1yy
a1zz
I would use an approach like this:
String str = "a1wwa1xxa1yya1zz";
String[] parts = str.split("a1");
for (int i = 1; i < parts.length; i++) {
String found = "a1" + parts[i];
}
How would I replace a string 10100 with 10010 using the algorithm "replace the last substring 10 with 01."
I tried
s=s.replace(s.substring(a,a+2), "01");
but this returns 01010, replacing both the first and the second substring of "10".
"a" represents s.lastindexOf("10");
Here's a simple and extensible function you can use. First its use/output and then its code.
String original = "10100";
String toFind = "10";
String toReplace = "01";
int ocurrence = 2;
String replaced = replaceNthOcurrence(original, toFind, toReplace, ocurrence);
System.out.println(replaced); // Output: "10010"
original = "This and This and This";
toFind = "This";
toReplace = "That";
ocurrence = 3;
replaced = replaceNthOcurrence(original, toFind, toReplace, ocurrence);
System.out.println(replaced); // Output: "This and This and That"
Function code:
public static String replaceNthOcurrence(String str, String toFind, String toReplace, int ocurrence) {
Pattern p = Pattern.compile(Pattern.quote(toFind));
Matcher m = p.matcher(str);
StringBuffer sb = new StringBuffer(str);
int i = 0;
while (m.find()) {
if (++i == ocurrence) { sb.replace(m.start(), m.end(), toReplace); break; }
}
return sb.toString();
}
If you want to access the last two indices of a string, then you can use: -
str.substring(str.length() - 2);
This gives you string from index str.length() - 2 to the last character, which is exactly the last two character.
Now, you can replace the last two indices with whatever string you want.
UPDATE: -
Of you want to access the last occurrence of a character or substring, you can use String#lastIndexOf method: -
str.lastIndexOf("10");
Ok, you can try this code: -
String str = "10100";
int fromIndex = str.lastIndexOf("10");
str = str.substring(0, fromIndex) + "01" + str.substring(fromIndex + 2);
System.out.println(str);
10100 with 10010
String result = "10100".substring(0, 2) + "10010".substring(2, 4) + "10100".substring(4, 5);
You can get the last index of a character or substring using string's lastIndexOf method. See the documentation link below for how to use it.
http://docs.oracle.com/javase/1.4.2/docs/api/java/lang/String.html#lastIndexOf(java.lang.String)
Once you know the index of your substring, you can get the substring of all characters before that index, and the substring of all characters after the last character in your search string, and concatenate.
This is a little drawn out, and I didn't actually run it (so I might have a syntax error), but it gives you the point of what I'm trying to convey at least. You could do this all in one line if you want, but it wouldn't illustrate the point as well.
string s = "10100";
string searchString = "10";
string replacementString = "01";
string charsBeforeSearchString = s.substring(0, s.lastIndexOf(searchString) - 1);
string charsAfterSearchString = s.substring(s.lastIndexIf(searchString) + 2);
s = charsBeforeSearchString + replacementString + charsAfterSearchString;
The easiest way:
String input = "10100";
String result = Pattern.compile("(10)(?!.*10.*)").matcher(input).replaceAll("01");
System.out.println(result);
String text = "select ename from emp";
I want to know the space index after the word from. How to do it?
If you're specifically looking for the index of the first space after the word "from", you can use:
text.substring(text.indexOf("from")).indexOf(' ');
If you're trying to do something more general, than you'll need to give a bit more information. But the indexOf() method will probably be very useful to you.
Edit: This should actually be
text.indexOf(' ', text.indexOf("from"));
The first version returns the index relative to "from", whereas the second returns the index relative to the original string. (thanks #jpm)
This loop will find all space characters in the given string:
int index = text.indexOf(' ');
while (index >= 0) {
System.out.println(index);
index = text.indexOf(' ', index + 1);
}
The very basic answer might look something like...
String text = "select ename from emp";
text = text.toLowerCase();
if (text.contains("from ")) {
int index = text.indexOf("from ") + "from".length();
System.out.println("Found space # " + index);
System.out.println(text.substring(index));
} else {
System.out.println(text + " does not contain `from `");
}
Or you could use some regular expression (this is a rather poor example, but hay)
Pattern pattern = Pattern.compile("from ");
Matcher matcher = pattern.matcher(text);
String match = null;
int endIndex = -1;
if (matcher.find()) {
endIndex = matcher.end();
}
if (endIndex > -1) {
endIndex--;
System.out.println("Found space # " + endIndex);
System.out.println(text.substring(endIndex));
} else {
System.out.println(text + " does not contain `from `");
}
To find the index of each space you could do something like...
Pattern pattern = Pattern.compile(" ");
Matcher matcher = pattern.matcher(text);
String match = null;
while (matcher.find()) {
System.out.println(matcher.start());
}
Which will output
6
12
17
use indexOf() method. Hope u got the answer
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;
}