[java] Java; String replace (using regular expressions)?

As part of a project for school, I need to replace a string from the form:

5 * x^3 - 6 * x^1 + 1

to something like:

5x<sup>3</sup> - 6x<sup>1</sup> + 1

I believe this can be done with regular expressions, but I don't know how to do it yet.

Can you lend me a hand?

P.S. The actual assignment is to implement a Polynomial Processing Java application, and I'm using this to pass polynomial.toString() from the model to the view, and I want do display it using html tags in a pretty way.

private String removeScript(String content) {
    Pattern p = Pattern.compile("<script[^>]*>(.*?)</script>",
            Pattern.DOTALL | Pattern.CASE_INSENSITIVE);
    return p.matcher(content).replaceAll("");
}

String input = "hello I'm a java dev" +
"no job experience needed" +
"senior software engineer" +
"java job available for senior software engineer";

String fixedInput = input.replaceAll("(java|job|senior)", "<b>$1</b>");

Take a look at antlr4. It will get you much farther along in creating a tree structure than regular expressions alone.

https://github.com/antlr/grammars-v4/tree/master/calculator (calculator.g4 contains the grammar you need)

In a nutshell, you define the grammar to parse an expression, use antlr to generate java code, and add callbacks to handle evaluation when the tree is being built.

import java.util.regex.PatternSyntaxException;

// (:?\d+) \* x\^(:?\d+)
// 
// Options: ^ and $ match at line breaks
// 
// Match the regular expression below and capture its match into backreference number 1 «(:?\d+)»
//    Match the character “:” literally «:?»
//       Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
//    Match a single digit 0..9 «\d+»
//       Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
// Match the character “ ” literally « »
// Match the character “*” literally «\*»
// Match the characters “ x” literally « x»
// Match the character “^” literally «\^»
// Match the regular expression below and capture its match into backreference number 2 «(:?\d+)»
//    Match the character “:” literally «:?»
//       Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
//    Match a single digit 0..9 «\d+»
//       Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
try {
    String resultString = subjectString.replaceAll("(?m)(:?\\d+) \\* x\\^(:?\\d+)", "$1x<sup>$2</sup>");
} catch (PatternSyntaxException ex) {
    // Syntax error in the regular expression
} catch (IllegalArgumentException ex) {
    // Syntax error in the replacement text (unescaped $ signs?)
} catch (IndexOutOfBoundsException ex) {
    // Non-existent backreference used the replacement text
}

You'll want to look into capturing in regex to handle wrapping the 3 in ^3.

"5 * x^3 - 6 * x^1 + 1".replaceAll("\\W*\\*\\W*","").replaceAll("\\^(\\d+)","<sup>$1</sup>");

please note that joining both replacements in a single regex/replacement would be a bad choice because more general expressions such as x^3 - 6 * x would fail.

Try this, may not be the best way. but it works

String str = "5 * x^3 - 6 * x^1 + 1";
str = str.replaceAll("(?x)(\\d+)(\\s+?\\*?\\s+?)(\\w+?)(\\^+?)(\\d+?)", "$1$3<sup>$5</sup>");
System.out.println(str);

str.replaceAll("\\^([0-9]+)", "<sup>$1</sup>");

class Replacement 
{
    public static void main(String args[])
    {
        String Main = "5 * x^3 - 6 * x^1 + 1";
        String replaced = Main.replaceAll("(?m)(:?\\d+) \\* x\\^(:?\\d+)", "$1x<sup>$2</sup>");
        System.out.println(replaced);
    }
}

If this is for any general math expression and parenthetical expressions are allowed, it will be very difficult (perhaps impossible) to do this with regular expressions.

If the only replacements are the ones you showed, it's not that hard to do. First strip out *'s, then use capturing like Can Berk Güder showed to handle the ^'s.

Try this:

String str = "5 * x^3 - 6 * x^1 + 1";
String replacedStr = str.replaceAll("\\^(\\d+)", "<sup>\$1</sup>");

Be sure to import java.util.regex.

What is your polynomial? If you're "processing" it, I'm envisioning some sort of tree of sub-expressions being generated at some point, and would think that it would be much simpler to use that to generate your string than to re-parse the raw expression with a regex.

Just throwing a different way of thinking out there. I'm not sure what else is going on in your app.