Question

在scala中，我有一个字符串，需要在其中用%23替换#，如下所示：

来自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

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

我在正则表达式和逻辑下面使用了替换，但出现错误：

java.lang.IllegalStateException: No match found

代码：

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)

非常感谢您的帮助。谢谢。

Answer 1

也许您可以只使用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", "#")

在Scala中搜索并替换字符串中的特殊字符

1 个答案: