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method to take string inside curly braces using split or tokenizer

String s = "author= {insert text here},";
Trying to get the inside of the string, ive looked around but couldn't find a resolution with just split or tokenizer...
so far im doing this
arraySplitBracket = s.trim().split("\\{", 0);
which gives me insert text here},
at array[1] but id like a way to not have } attached
also tried
StringTokenizer st = new StringTokenizer(s, "\\{,\\},");
But it gave me author= as output.

public static void main(String[] args) {
String input="{a c df sdf TDUS^&%^7 }";
String regEx="(.*[{]{1})(.*)([}]{1})";
Matcher matcher = Pattern.compile(regEx).matcher(input);
if(matcher.matches()) {
System.out.println(matcher.group(2));
}
}

You can use \\{([^}]*)\\} Regex to get string between curly braces.
Code Snap :
String str = "{insert text here}";
Pattern p = Pattern.compile("\\{([^}]*)\\}");
Matcher m = p.matcher(str);
while (m.find()) {
System.out.println(m.group(1));
}
Output :
insert text here

String s = "auther ={some text here},";
s = s.substring(s.indexOf("{") + 1); //some text here},
s = s.substring(0, s.indexOf("}"));//some text here
System.out.println(s);

How about taking a substring by excluding the character at arraySplitBracket.length()-1
Something like
arraySplitBracket[1] = arraySplitBracket[1].substring(0,arraySplitBracket.length()-1);
Or use String Class's replaceAll function to replace } ?

How to extract id from url ? Google sheet

I have the follow urls.
https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY/edit#gid=1842172258
https://docs.google.com/a/example.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6PTKTzY0xOM5c6TXY/edit#gid=1842172258
https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY
Foreach url, I need to extract the sheet id: 1mrsetjgfZI2BIypz7SGHMOfHGv6PTKTzY0xOM5c6TXY into a java String.
I am thinking of using split but it can't work with all test cases:
String string = "https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY/edit#gid=1842172258";
String[] parts = string.split("/");
String res = parts[parts.length-2];
Log.d("hello res",res );
How can I that be possible?

You can use regex \/d\/(.*?)(\/|$) (regex demo) to solve your problem, if you look closer you can see that the ID exist between d/ and / or end of line for that you can get every thing between this, check this code demo :
String[] urls = new String[]{
"https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY/edit#gid=1842172258",
"https://docs.google.com/a/example.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6PTKTzY0xOM5c6TXY/edit#gid=1842172258",
"https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY"
};
String regex = "\\/d\\/(.*?)(\\/|$)";
Pattern pattern = Pattern.compile(regex);
for (String url : urls) {
Matcher matcher = pattern.matcher(url);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
}
Outputs
1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY
1mrsetjgfZI2BIypz7SGHMOfHGv6PTKTzY0xOM5c6TXY
1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY

it looks like the id you are looking for always follow "/spreadsheets/d/" if it is the case you can update your code to that
String string = "https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY/edit#gid=1842172258";
String[] parts = string.split("spreadsheets/d/");
String result;
if(parts[1].contains("/")){
String[] parts2 = parts[1].split("/");
result = parts2[0];
}
else{
result=parts[1];
}
System.out.println("hello "+ result);

Using regex
Pattern pattern = Pattern.compile("(?<=\\/d\\/)[^\\/]*");
Matcher matcher = pattern.matcher(url);
System.out.println(matcher.group(1));
Using Java
String result = url.substring(url.indexOf("/d/") + 3);
int slash = result.indexOf("/");
result = slash == -1 ? result
: result.substring(0, slash);
System.out.println(result);

Google use fixed lenght characters for its IDs, in your case they are 44 characters and these are the characters google use: alphanumeric, -, and _ so you can use this regex:
regex = "([\w-]){44}"
match = re.search(regex,url)

complex regular expression in Java

I have a rather complex (to me it seems rather complex) problem that I'm using regular expressions in Java for:
I can get any text string that must be of the format:
M:<some text>:D:<either a url or string>:C:<some more text>:Q:<a number>
I started with a regular expression for extracting the text between the M:/:D:/:C:/:Q: as:
String pattern2 = "(M:|:D:|:C:|:Q:.*?)([a-zA-Z_\\.0-9]+)";
And that works fine if the <either a url or string> is just an alphanumeric string. But it all falls apart when the embedded string is a url of the format:
tcp://someurl.something:port
Can anyone help me adjust the above reg exp to extract the text after :D: to be either a url or a alpha-numeric string?
Here's an example:
public static void main(String[] args) {
String name = "M:myString1:D:tcp://someurl.com:8989:C:myString2:Q:1";
boolean matchFound = false;
ArrayList<String> values = new ArrayList<>();
String pattern2 = "(M:|:D:|:C:|:Q:.*?)([a-zA-Z_\\.0-9]+)";
Matcher m3 = Pattern.compile(pattern2).matcher(name);
while (m3.find()) {
matchFound = true;
String m = m3.group(2);
System.out.println("regex found match: " + m);
values.add(m);
}
}
In the above example, my results would be:
myString1
tcp://someurl.com:8989
myString2
1
And note that the Strings can be of variable length, alphanumeric, but allowing some characters (such as the url format with :// and/or . - characters

You mention that the format is constant:
M:<some text>:D:<either a url or string>:C:<some more text>:Q:<a number>
Capture groups can do this for you with the pattern:
"M:(.*):D:(.*):C:(.*):Q:(.*)"
Or you can do a String.split() with a pattern of "M:|:D:|:C:|:Q:". However, the split will return an empty element at the first index. Everything else will follow.
public static void main(String[] args) throws Exception {
System.out.println("Regex: ");
String data = "M:<some text>:D:tcp://someurl.something:port:C:<some more text>:Q:<a number>";
Matcher matcher = Pattern.compile("M:(.*):D:(.*):C:(.*):Q:(.*)").matcher(data);
if (matcher.matches()) {
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println(matcher.group(i));
}
}
System.out.println();
System.out.println("String.split(): ");
String[] pieces = data.split("M:|:D:|:C:|:Q:");
for (String piece : pieces) {
System.out.println(piece);
}
}
Results:
Regex:
<some text>
tcp://someurl.something:port
<some more text>
<a number>
String.split():
<some text>
tcp://someurl.something:port
<some more text>
<a number>

To extract the URL/text part you don't need the regular expression. Use
int startPos = input.indexOf(":D:")+":D:".length();
int endPos = input.indexOf(":C:", startPos);
String urlOrText = input.substring(startPos, endPos);

Assuming you need to do some validation along with the parsing:
break the regex into different parts like this:
String m_regex = "[\\w.]+"; //in jsva a . in [] is just a plain dot
String url_regex = "."; //theres a bunch online, pick your favorite.
String d_regex = "(?:" + url_regex + "|\\p{Alnum}+)"; // url or a sequence of alphanumeric characters
String c_regex = "[\\w.]+"; //but i'm assuming you want this to be a bit more strictive. not sure.
String q_regex = "\\d+"; //what sort of number exactly? assuming any string of digits here
String regex = "M:(?<M>" + m_regex + "):"
+ "D:(?<D>" + d_regex + "):"
+ "C:(?<D>" + c_regex + "):"
+ "Q:(?<D>" + q_regex + ")";
Pattern p = Pattern.compile(regex);
Might be a good idea to keep the pattern as a static field somewhere and compile it in a static block so that the temporary regex strings don't overcrowd some class with basically useless fields.
Then you can retrieve each part by its name:
Matcher m = p.matcher( input );
if (m.matches()) {
String m_part = m.group( "M" );
...
String q_part = m.group( "Q" );
}
You can go even a step further by making a RegexGroup interface/objects where each implementing object represents a part of the regex which has a name and the actual regex. Though you definitely lose the simplicity makes it harder to understand it with a quick glance. (I wouldn't do this, just pointing out its possible and has its own benefits)

Java regex not working to find "Ts" followed by "kb"

I have to convert "Tskb" to "TsKB" using java regex whenever it comes as single word. I have written below code which not working.
public class TestBGR {
private static final Pattern s_TsKB = Pattern.compile("/(Ts?.*)(?=.*kb)^(\\w+)$/");
public static void main(String[] args) {
String text = "Tskb";
Matcher matcher = s_TsKB.matcher(text);
StringBuilder builder = new StringBuilder(text);
int offset = 0;
while (matcher.find())
{
String replacement = "KB";
builder.replace(matcher.start() + offset, matcher.end() + offset,
replacement);
offset += replacement.length() - matcher.group().length();
}
System.out.println(builder);
}
}
Here how to find "Ts" followed by "kb" using java regex..?

You can simply do a replaceAll with
(?<=\\bTs)kb\\b
and replace by KB.The lookbehind will make sure kb has Ts before.
See demo.
https://regex101.com/r/fM9lY3/13

java.lang.StringIndexOutOfBoundsException: from java.util.regex.Matcher

I am trying to use regex to remove nbsp; from my string . Following is the program.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class MyTest {
private static final StringBuffer testRegex =
new StringBuffer("<FONT style=\"BACKGROUND-COLOR: #ff6600\">Test</font></p><br><p>" +
"<FONT style=\"BACKGROUND-COLOR: #ff6600\">Test</font></p><br><p>" +
"<FONT style=\"BACKGROUND-COLOR: #ff6600\">Test</font>" +
"<BLOCKQUOTE style=\"MARGIN-RIGHT: 0px\" dir=ltr><br><p>Test</p><strong>" +
"<FONT color=#333333>TestTest</font></strong></p><br><p>Test</p></blockquote>" +
"<br><p>TestTest</p><br><BLOCKQUOTE style=\"MARGIN-RIGHT: 0px\" dir=ltr><br><p>" +
"<FONT style=\"BACKGROUND-COLOR: #ffcc66\">TestTestTestTestTest</font><br>" +
"<p>TestTestTestTest</p></blockquote><br><p>" +
"<FONT style=\"BACKGROUND-COLOR: #003333\">TestTestTest</font></p><p>" +
"<FONT style=\"BACKGROUND-COLOR: #003399\">TestTest</font></p><p> </p>");
//"This is test<P>Tag Tag</P>";
public static void main(String[] args) {
System.out.println("***Testing***");
String temp = checkRegex(testRegex);
System.out.println("***FINAL = "+temp);
}
private static String checkRegex(StringBuffer sample){
Pattern pattern = Pattern.compile("<[^>]+? [^<]+?>");
Matcher matcher = pattern.matcher(sample);
while (matcher.find()) {
int start = matcher.start();
int end = matcher.end();
String group = matcher.group();
System.out.println("start = "+start+" end = "+end+"" +"***GROUP = "+group);
String substring = sample.substring(start, end);
System.out.println(" Substring = "+substring);
String replacedSubString = substring.replaceAll(" "," ");
System.out.println("Replaced Substring = "+replacedSubString);
sample.replace(start, end, replacedSubString);
System.out.println(" NEW SAMPLE = "+sample);
}
System.out.println("********WHILE OVER ********");
return sample.toString();
}
}
I am getting java.lang.StringIndexOutOfBoundsException at line while (matcher.find()). I am currently using java Pattern and Matcher to find nbsp; and the replace it with " ". Does anyone know what causes this ? What should I do to remove the extra nbsp; from my string ?
Thanks

Use matcher.reset(); after sample.replace(start, end, replacedSubString);
This is because when you replace the string sample, the end would point to an invalid position.So,you need to use matcher.reset(); after every replace.
For example if start is 0 and end is 5 and when you replace with ,the end would point to an invalid position and then find method would throw a StringIndexOutOfBoundsException exception if end points to position outside the string length.
If string is huge,reset can cause a major performance bottleneck because reset would again start matching from beginning.You can instead use
matcher.region(start,sample.length());
This would start matching from the last matched position!

You need to create a new StringBuffer to hold the replaced string, then use appendReplacement(StringBuffer sb, String replacement) and appendTail(StringBuffer sb) methods in Matcher class to do the replacement. There is probably way to do this in-place, but the approach above is the most straight-forward way to do this.
This is your checkRegex method re-written:
private static String checkRegex(String inputString){
Pattern pattern = Pattern.compile("<[^>]+? [^<]+?>");
Matcher matcher = pattern.matcher(inputString);
// Create a new StringBuffer to hold the string after replacement
StringBuffer replacedString = new StringBuffer();
while (matcher.find()) {
// matcher.group() returns the substring that matches the whole regex
String substring = matcher.group();
System.out.println(" Substring = "+substring);
String replacedSubstring = substring.replaceAll(" "," ");
System.out.println("Replaced Substring = "+replacedSubstring);
// appendReplacement is a clean approach to append the text which comes
// before a match, and append the replacement text for the matched text
// Note that appendReplacement will interpret $ in the replacement string
// with special meaning (for referring to text matched by capturing group).
// Matcher.quoteReplacement is necessary to provide a literal string as
// replacement
matcher.appendReplacement(replacedString, Matcher.quoteReplacement(replacedSubstring));
System.out.println(" NEW SAMPLE = "+replacedString);
}
// appendTail is used to append the text after the last match to the
// replaced string.
matcher.appendTail(replacedString);
System.out.println("********WHILE OVER ********");
return replacedString.toString();
}

// change the group and it is source string is automatically updated
There is no way what so ever to change any string in Java, so what you're asking for is impossible.
To remove or replace a pattern with a string can be achieved with a call like
someString = someString.replaceAll(toReplace, replacement);
To transform the matched substring, as seems to be indicated by your line
m.group().replaceAll("something","");
The best solution is probably to use a StringBuffer for the result
Matcher.appendReplacement and Matcher.appendTail.
Example:
String regex = "ipsum";
String sourceString = "lorem ipsum dolor sit";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(sourceString);
StringBuffer sb = new StringBuffer();
while (m.find()) {
// For example: transform match to upper case
String replacement = m.group().toUpperCase();
m.appendReplacement(sb, replacement);
}
m.appendTail(sb);
sourceString = sb.toString();
System.out.println(sourceString); // "lorem IPSUM dolor sit"