String.substring（）OR运算符

2 回答

• -1

  indexOf 不够复杂，无法做到这一点 . 但是，您可以使用正则表达式来获取子字符串：

  Pattern p = Pattern.compile("start(.*?)(?:x|y)");
  Matcher m = p.matcher(str);
  if (m.find()) {
      String sub = m.group(1);
  }
  

  Demo.

  回复于 2024-05-05T10:10:26+08:00
• 4

  解：

  -> "startthexstrayingx".replaceAll ("start([^xy]*(x|y)).*", "$1") 
  |  Expression value is: "thex"
  |    assigned to temporary variable $72 of type String
  

  第一组从“开始”开始并捕获，直到满足第一个x或y .

  使用String start的锚点，它将只捕获一个匹配项，而不管名称replaceAll：

  -> "startstartthexstrayingstartAgainandynotyy".replaceAll ("^start([^xy]*(x|y)).*", "$1") 
  |  Expression value is: "startthex"
  |    assigned to temporary variable $77 of type String
  

  Demo

  从字面上看问题，模式必须是 "^s(tart[^xy]*(x|y)).*" 但我怀疑这不是故意的 .

  但不是那样的：

  -> String str = "startthex|ystray"
  |  Modified variable str of type String with initial value "startthex|ystray"
  |    Update overwrote variable str
  
  -> str.substring(str.indexOf("start")+1, str.indexOf("x|y"))
  |  Expression value is: "tartthe"
  |    assigned to temporary variable $62 of type String
  
  -> String str = "startthexstray"
  |  Modified variable str of type String with initial value "startthexstray"
  |    Update overwrote variable str
  
  -> str.substring(str.indexOf("start")+1, str.indexOf("x|y"))
  |  java.lang.StringIndexOutOfBoundsException thrown: begin 1, end -1, length 14
  |        at String.checkBoundsBeginEnd (String.java:3119)
  |        at String.substring (String.java:1907)
  |        at (#98:1)
  

  回复于 2024-05-05T10:10:26+08:00