CODEWITHSUNDEEP
javasplit
Phillip
Phillip

Reputation: 457

Obtaining the split value after java string split

I have a string that is dynamially generated.

I need to split the string based on the Relational Operator.

For this I can use the split function.

Now I would also like to know that out of the regex mentioned above, based on which Relational Operator was the string actually splitted.

An example, On input 

String sb = "FEES > 200";

applying

List<String> ls =  sb.split(">|>=|<|<=|<>|=");
System.out.println("Splitted Strings: "+s);

will give me the result,

Splitted strings: [FEES ,  200 ]

But expecting result:

Splitted strings: [FEES ,  200 ]
Splitted Relational Operator: >

Upvotes: 10

Views: 747

Answers (5)

mishadoff
mishadoff

Reputation: 10789

I suggest to use regex, which is more flexible in your case.

String sb = "FEES > 200";
Pattern pat = Pattern.compile("(.*?)(>=|<=|<>|=|>|<)(.*)");
Matcher mat = pat.matcher(sb);
if (mat.find()) {
    System.out.println("Groups: " + mat.group(1) + ", " + mat.group(3));
    System.out.println("Operator: " + mat.group(2));
}

Upvotes: 3

The fourth bird
The fourth bird

Reputation: 163467

You could use 3 capturing groups with an alternation for the second group:

(.*?)(>=|<=|<>|>|<)(.*)

Regex demo

Explanation

  • (.*?) Match any character zero or more times non greedy
  • (>=|<=|<>|>|<) Match either >= or <= or <> or > or <
  • (.*) Match any character zero or more times

For example:

String regex = "(.*?)(>=|<=|<>|>|<)(.*)";
String string = "FEES >= 200";            
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(string);
if(matcher.find()) {
    System.out.println("Splitted Relational Operator: " + matcher.group(2));
    System.out.println("Group 1: " + matcher.group(1) + " group 3: " + matcher.group(3));
}

Demo java

Upvotes: 15

drowny
drowny

Reputation: 2147

Use Pattern to do this. 

String sb = "FEES > 200";
Pattern pattern = Pattern.compile("(.*)(>|>=|<|<=|<>|=)(.*)");
Matcher matcher = pattern.matcher(sb);
if (matcher.find()) {
    System.out.println("Grouped params: " + matcher.group(1) + "," + matcher.group(3));
    System.out.println("Split operator: " + matcher.group(2));
}

Note that :

  • matcher.group(0) --> All string
  • matcher.group(1) --> first part of matching
  • matcher.group(2) --> split operator
  • matcher.group(3) --> second part of matching

Upvotes: 1

user8393974
user8393974

Reputation:

You can use replaceAll() if you want exclude number from your string:

replaceAll("\d", ""); - this replace all number on white space

To delete unnecessary words you need to give more information. Because there are different ways of doing things.

Upvotes: 0

Amit Naik
Amit Naik

Reputation: 1073

You can use Pattern matcher to match the delimeters. Example shown below -

public static int indexOf(Pattern pattern, String s) {
        Matcher matcher = pattern.matcher(s);
        return matcher.find() ? matcher.start() : -1;
    }


public static void main(String[] args) {
    String sb = "FEES > 200";
    String[] s =  sb.split(">|>=|<|<=|<>|=");
    System.out.println("Splitted Strings: "+ s);

    int index = indexOf(Pattern.compile(">|>=|<|<=|<>|="), sb);
    System.out.println("del "+ sb.charAt(index));
}

Upvotes: 0

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