在java中按字符串拆分字符串
我从网上得到一个字符串，如下所示：
Latest Episode@04x22^Killing Your Number^May/15/2009 然后我需要在不同的变量中存储04x22 ， Killing Your Number和May/15/2009 04x22 ，但它不起作用。
String[] all = inputLine.split("@"); String[] need = all[1].split("^"); show.setNextNr(need[0]); show.setNextTitle(need[1]); show.setNextDate(need[2]);
现在它只存储NextNr ，包含整个字符串
04x22^Killing Your Number^May/15/2009
哪里不对？
String.split(String regex)
参数是一个regualr表达式， ^具有特殊含义; “锚定到开始”
你需要这样做：
String[] need = all[1].split("\\^");
通过逃避^你说“我的意思是字符’^’”
如果您有分隔符但不知道它是否包含特殊字符，则可以使用以下方法
String[] parts = Pattern.compile(separator, Pattern.LITERAL).split(text);
使用番石榴，你可以优雅而快速地做到：
private static final Splitter RECORD_SPLITTER = Splitter.on(CharMatcher.anyOf("@^")).trimResults().omitEmptyStrings(); ... Iterator splitLine = Iterables.skip(RECORD_SPLITTER.split(inputLine), 1).iterator(); show.setNextNr(splitLine.next()); show.setNextTitle(splitLine.next()); show.setNextDate(splitLine.next());
public static String[] split(String string, char separator) { int count = 1; for (int index = 0; index < string.length(); index++) if (string.charAt(index) == separator) count++; String parts[] = new String[count]; int partIndex = 0; int startIndex = 0; for (int index = 0; index < string.length(); index++) if (string.charAt(index) == separator) { parts[partIndex++] = string.substring(startIndex, index); startIndex = index + 1; } parts[partIndex++] = string.substring(startIndex); return parts; }
String input = "Latest Episode@04x22^Killing Your Number^May/15/2009"; //split will work for both @ and ^ String splitArr[] = input.split("[@\\^]"); /*The output will be, [Latest Episode, 04x22, Killing Your Number, May/15/2009] */ System.out.println(Arrays.asList(splitArr));