Java: Print specific substring based on characters appearing before, and end if meeting letters after?

Question

I have an input string, let's say it's 111.112.111.0.0 > 111.33.44.55.11 pxs ikf len 3348 , it might also sometimes be input via the stream as 111.112.111.0.0 > 111.33.44.55.11 pxs ikf len 3348 BBBTTT.

How do I capture the numbers occuring after "len" in the via substring? Using system.out.println(stringStr.substring(stringStr.lastIndexOf(" ") + 1) will only print 3348 if it's the last part of the string above, but if BBBTTT appears, it will not work.

How do I print the numbers of a substring specifically after "len" but not including anything but the numbers that follow(might be 0-9999, not more), so not any characters following?

This is a candidate for RegularExpressions. Learn about it.

Heri
– Heri

2016-10-11 18:38:34 +00:00
Commented Oct 11, 2016 at 18:38 — Heri
– Heri, Commented Oct 11, 2016 at 18:38

JitterbugChew · Accepted Answer · 2016-10-11 20:01:15Z

3

Use RegEx. The pattern would be something like "len ([\d]+)"

String inputString = "111.112.111.0.0 > 111.33.44.55.11 pxs ikf len 3348 BBBTTT";
String regEx = "len ([\\d]+)";
String regExMatch = "";

Pattern p = Pattern.compile(regEx);
Matcher m = p.matcher(inputString);
if (m.find()) {
    regExMatch = m.group(1); // save the first capture group
}
System.out.println(regExMatch);

edited Oct 11, 2016 at 20:01

answered Oct 11, 2016 at 18:43

JitterbugChew

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2 Comments

cbll Over a year ago

Can't I use this directly in the system.out.println line? regex that is

JitterbugChew Over a year ago

@cbll you could either store it to a variable or create a separate function and have it return just the string. If you HAD to, I guess you could put it all in a single line, but it's not good practice to do so and it would be difficult to read.

Raul Vasi · Accepted Answer · 2016-10-11 18:47:02Z

2

You can use split method for this.

        String s = "111.112.111.0.0 > 111.33.44.55.11 pxs ikf len 3348 BBBTTT";
    s = s.split("len")[1];
    s = s.replaceAll("\\D+", "");
    System.out.println(s);

or just

s = s.split("len")[1].replaceAll("\\D+", "");

edited Oct 11, 2016 at 18:47

answered Oct 11, 2016 at 18:41

Raul Vasi

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3 Comments

cbll Over a year ago

I didnt express myself clearly. I want only the 1-4 digits, not including BBBTTT

Nicolas Filotto Over a year ago

still not good if you have 111.112.111.0.0 > 111.33.44.55.11 pxs ikf len 3348 BBBTT18T, you will get 334818 instead of 3348

Raul Vasi Over a year ago

then use split again s = s.split("len")[1]; and s = s.split(" ")[1]; will split from the next whitespace

GhostCat · Accepted Answer · 2016-10-11 18:42:48Z

1

You could be using regular expressions here, but there are other ways:

Determine the index of "len" in your string
Add 4 to that - that is where your number starts
Do a substring from there

That string will either contain "just digigits"; or digits followed by a space. So you can do another indexOf() call to figure if there is a space; and if so, throw away everything after that space, too; like:

String numberWithChars = "3348 BBBTTT";
int indexOfSpace = numberWithChars.indexOf(" ");
if  (indexOfSpace > 0) {
   String pureNumber = numberWithChars.substring(0, indexOfSpace);
} else {
   String pureNumber = numberWithChars;
}

Hope that helps.

answered Oct 11, 2016 at 18:42

GhostCat

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Comments

Wiktor Stribiżew · Accepted Answer · 2016-10-11 19:43:23Z

1

You may use a replaceFirst with a not-so-simple regex:

.*\blen\s+(\d+).*

See the regex demo

Details:

.* - any 0+ chars other than linebreak symbols, as many as possible
\blen - a whole word len (\b is a word boundary)
\s+ - 1+ whitespaces
(\d+) - Group 1 capturing 1+ digits
.* - any 0+ chars other than linebreak symbols.

The $1 is a backereference to the contents captured into Group 1.

If your string can contain linebreaks, add (?s) at the start of the regex.

See Java demo:

String s  = "111.112.111.0.0 > 111.33.44.55.11 pxs ikf len 3348";
System.out.println(s.replaceFirst(".*\\blen\\s+(\\d+).*", "$1"));

answered Oct 11, 2016 at 19:43

Wiktor Stribiżew

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Comments

Tony · Accepted Answer · 2016-10-11 21:22:22Z

1

Example with your two strings using a regex :

        String s1 = "111.112.111.0.0 > 111.33.44.55.11 pxs ikf len 3348";
        String s2 = "111.112.111.0.0 > 111.33.44.55.11 pxs ikf len 3349 BBBTTT";

        Pattern pattern = Pattern.compile(".*len (\\d*).*");
        Matcher m1 = pattern.matcher(s1);
        Matcher m2 = pattern.matcher(s2);
        if (m1.matches()) {
            System.out.println(m1.group(1));
        }
        if (m2.matches()) {
            System.out.println(m2.group(1));
        }

You could put this in a separate method :

public class ExtractLen {
private static final Pattern LEN_PATTERN = Pattern.compile(".*len (\\d*).*");

public String extractLen(String s) {
    Matcher m1 = LEN_PATTERN.matcher(s);
    if (m1.matches()) {
        return m1.group(1);
    }
    return null;
}}

edited Oct 11, 2016 at 21:22

answered Oct 11, 2016 at 18:50

Tony

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2 Comments

cbll Over a year ago

Can't I implement it directly in the system.out.println as a condition, like the substring?

Tony Over a year ago

@cbll You could extract it to a class with method like the updated answer

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