I am trying to get an element from my string which I have attained through a getSelectedValue().toString from a JList.
It returns [1] testString
What I am trying to do it only get the 1 from the string. Is there a way to only get that element from the string or remove all else from the string?
I have tried:
String longstring = Customer_list.getSelectedValue().toString();
int index = shortstring.indexOf(']');
String firstPart = myStr.substring(0, index);
You have many ways to do it, for example
String#replaceAllString#substringSee below code to use all methods.
import java.util.*;
import java.lang.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
class Test {
public static void main(String args[]) {
String[] data = { "[1] test", " [2] [3] text ", " just some text " };
for (String s : data) {
String r0 = null;
Matcher matcher = Pattern.compile("\\[(.*?)\\]").matcher(s);
if (matcher.find()) {
r0 = matcher.group(1);
}
System.out.print(r0 + " ");
}
System.out.println();
for (String s : data) {
String r1 = null;
r1 = s.replaceAll(".*\\[|\\].*", "");
System.out.print(r1 + " ");
}
System.out.println();
for (String s : data) {
String r2 = null;
int i = s.indexOf("[");
int j = s.indexOf("]");
if (i != -1 && j != -1) {
r2 = s.substring(i + 1, j);
}
System.out.print(r2 + " ");
}
System.out.println();
}
}
However results may vary, for example String#replaceAll will give you wrong results when input is not what you expecting.
1 2 null
1 3 just some text
1 2 null
What worked best for me is the String#replace(charSequence, charSequence) combined with String#substring(int,int)
I have done as followed:
String longstring = Customer_list.getSelectedValue().toString();
String shortstring = longstring.substring(0, longstring.indexOf("]") + 1);
String shota = shortstring.replace("[", "");
String shortb = shota.replace("]", "");
My string has been shortened, and the [ and ] have been removed thereafter in 2 steps.
If you know each input string begins with a number in square brackets followed by a SPACE character, then I would simply split the input by SPACE character, take the first part, and replace the square brackets with nothing. That leaves characters to be parsed as digits.
String input = "[1] testString";
String[] parts = input.split( " " );
String part = parts[0];
String digits = part.replace( "[" , "" ).replace( "]" , "" ) ;
int number = Integer.parseInt( digits ) ;
See this code run at Ideone.com.
1
Briefer:
String digits = input.split( " " )[0].replace( "[" , "" ).replace( "]" , "" ) ;
int number = Integer.parseInt( digits ) ;
Briefest:
int number = Integer.parseInt( input.split( " " )[0].replace( "[" , "" ).replace( "]" , "" ) ) ;
Yes, or more simply: getSelectedValue().replaceAll("\\[(.*?)\\]", "$1")
String#substring(int)defines the substring from the beginning index to the end. This will just give you everything after the[1]. if you useString#substring(int,int), which is defined as create a substring from posxto excluse posythen you could do it like this:longstring.substring(0, longstring.indexOf("]") + 1).String#replace(charSequence, charSequence)method. The first parameter should be either"["or"]", your second could be a empty String as"". Just call it twice after each other and the brackets should be gone.