Way to handle string in Java

Question

I would like to remove all digits that come at the start of a string. I would just like to remove the digits at the start, not at the end or the middle.

Example: "123string67" to "string67"

Is there some function or regex I can use to easily do this?

*"maybe there is an easy regex command"*. There is. What did you try? Or are you asking us to write the regex for you, without doing *any* research on your own? Down-voted for lack of research. — Andreas, Sep 12 '16 at 18:05
http://stackoverflow.com/questions/4643142/regex-to-test-if-string-begins-with-http-or-https similar — Taylor, Sep 12 '16 at 18:05
use this as a hint...http://stackoverflow.com/q/11232801/982161 — ΦXocę 웃 Пepeúpa ツ, Sep 12 '16 at 18:07

score 1 · Answer 1 · answered Sep 12 '16 at 18:50

You can use group capture and ^\d*(.*) regex

Pattern pattern = java.util.regex.Pattern.compile("^\\d*(.*)")
Matcher matcher = pattern.matcher("123string67")
if (matcher.matches()) {
    String truncated = matcher.group(); //string67
}

Here is a regex explained

^    - start of string
\d*  - zero or more digit
 (   - group capture start
  .* - any symbol
 )   - group capture end

score 0 · Answer 2 · answered Sep 12 '16 at 19:03

0

Here's a hint (without regex): (*you still need to consider the case where the string is empty or contains only digits)

int i=0;
while(isDigit(str.charAt(i))) {
    str.deleteCharAt(i);
}

answered Sep 12 '16 at 19:03

PEF

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score 0 · Answer 3 · answered Sep 12 '16 at 19:08

0

You can use for ( i.. To i.string.lenght..) and control with ASCII number and delete digit number so you can replaced your string as you want i hope this helpful for you

answered Sep 12 '16 at 19:08

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Way to handle string in Java

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