CODEWITHSUNDEEP
regexscalapattern-matching
Kishori
Kishori

Reputation: 151

Search and replace special character in a string in scala

In scala ,I have a string where I need to replace %23 with # , as below:

From https://bucket_name.s3.amazonaws.com/scripts/%23%23%23ENVIRONMENT_NAME%23%23%23/abc/template_abc_windows_%23%23%23ENVIRONMENT_NAME%23%23%23.zip?X-Amz-Security-Token=FQoGZXIvYXdzEOghsfgdghgkjkjjklj

to https://bucket_name.s3.amazonaws.com/scripts/###ENVIRONMENT_NAME###/abc/template_abc_windows_###ENVIRONMENT_NAME###.zip?X-Amz-Security-Token=FQoGZXIvYXdzEOghsfgdghgkjkjjklj

I have used below regex and logic for substitution but I get error as:

java.lang.IllegalStateException: No match found

Code:

val originalURL = "https://bucket_name.s3.amazonaws.com/scripts/%23%23%23ENVIRONMENT_NAME%23%23%23/abc/template_abc_windows_%23%23%23ENVIRONMENT_NAME%23%23%23.zip?X-Amz-Security-Token=FQoGZXIvYXdzEOghsfgdghgkjkjjklj"

  val pattern = Pattern.compile("(https://bucket_name.s3.amazonaws.com/scripts/)((%23){3})([a-zA-Z]+_[a-zA-Z]+)((%23){3})(/abc/template_abc_windows_)((%23){3})([a-zA-Z]+_[a-zA-Z]+)((%23){3})(..*)")
  val matcher = pattern.matcher(originalURL)
  val replacedURL = matcher.group(1)+"###"+ matcher.group(4)+"###"+ matcher.group(7)+"###"+ matcher.group(10)+"###"+matcher.group(13)
  println("*******replacedURL*******  => "+ replacedURL)

Any help is much appreciated.Thank you.

Upvotes: 1

Views: 244

Answers (1)

Krzysztof Atłasik
Krzysztof Atłasik

Reputation: 22635

Maybe you can just use String.replaceAll?

val url = "https://bucket_name.s3.amazonaws.com/scripts/%23%23%23ENVIRONMENT_NAME%23%23%23/abc/template_abc_windows_%23%23%23ENVIRONMENT_NAME%23%23%23.zip?X-Amz-Security-Token=FQoGZXIvYXdzEOghsfgdghgkjkjjklj"
url.replaceAll("%23", "#")

Upvotes: 1

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