如何在 Java 中拆分字符串并保留分隔符？
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How can I split a string in Java and retain the delimiters?
提问by yegor256
I have this string (Java 1.5):
我有这个字符串（Java 1.5）：
:alpha;beta:gamma;delta
I need to get an array:
我需要得到一个数组：
{":alpha", ";beta", ":gamma", ";delta"}
What is the most convenient way to do it in Java?
在 Java 中最方便的方法是什么？
回答by Bozho
str.split("(?=[:;])")
This will give you the desired array, only with an empty first item. And:
这将为您提供所需的数组，只有一个空的第一项。和：
str.split("(?=\b[:;])")
This will give the array without the empty first item.
这将给出没有空的第一项的数组。
- The key here is the
(?=X)which is a zero-width positive lookahead (non-capturing construct) (see regex pattern docs). [:;]means "either ; or :"\bis word-boundary - it's there in order not to consider the first:as delimiter (since it is the beginning of the sequence)
- 这里的关键是
(?=X)这是一个零宽度正前瞻（非捕获构造）（请参阅正则表达式模式文档）。 [:;]意思是“要么；要么：”\b是词边界 - 它在那里是为了不将第一个:视为分隔符（因为它是序列的开头）
回答by Fabian Steeg
To keep the separators, you can use a StringTokenizer:
要保留分隔符，您可以使用StringTokenizer：
new StringTokenizer(":alpha;beta:gamma;delta", ":;", true)
That would yield the separators as tokens.
这将产生作为标记的分隔符。
To have them as part of your tokens, you could use String#splitwith lookahead.
要将它们作为令牌的一部分，您可以使用String#splitwith lookahead。
回答by Favonius
You can do this by simply using patterns and matcher class in java regx.
您可以通过在 java regx 中简单地使用模式和匹配器类来做到这一点。
public static String[] mysplit(String text)
{
List<String> s = new ArrayList<String>();
Matcher m = Pattern.compile("(:|;)\w+").matcher(text);
while(m.find()) {
s.add(m.group());
}
return s.toArray(new String[s.size()]);
}
回答by Kanagavelu Sugumar
/**
* @param list an empty String list. used for internal purpose.
* @param str String which has to be processed.
* @return Splited String Array with delimiters.
*/
public String[] split(ArrayList<String> list, String str){
for(int i = str.length()-1 ; i >=0 ; i--){
if(!Character.isLetterOrDigit((str.charAt(i)))) {
list.add(str.substring(i, str.length()));
split(list,str.substring(0,i));
break;
}
}
return list.toArray(new String[list.size()]);
}
回答by Rok T.
This should work with Java 1.5 (Pattern.quote was introduced in Java 1.5).
这应该适用于 Java 1.5（Pattern.quote 是在 Java 1.5 中引入的）。
// Split the string on delimiter, but don't delete the delimiter
private String[] splitStringOnDelimiter(String text, String delimiter, String safeSequence){
// A temporary delimiter must be added as Java split method deletes the delimiter
// for safeSequence use something that doesn't occur in your texts
text=text.replaceAll(Pattern.quote(delimiter), safeSequence+delimiter);
return text.split(Pattern.quote(safeSequence));
}
If first element is the problem:
如果第一个元素是问题：
private String[] splitStringOnDelimiter(String text, String delimiter, String safeSequence){
text=text.replaceAll(Pattern.quote(delimiter), safeSequence+delimiter);
String[] tempArray = text.split(Pattern.quote(safeSequence));
String[] returnArray = new String[tempArray.length-1];
System.arraycopy(tempArray, 1, returnArray, 0, returnArray.length);
return returnArray;
}
E.g., here "a" is the delimiter:
例如，这里的“a”是分隔符：
splitStringOnDelimiter("-asd-asd-g----10-9asdas jadd", "a", "<>")
You get this:
你得到这个：
1.: -
2.: asd-
3.: asd-g----10-9
4.: asd
5.: as j
6.: add
If you in fact want this:
如果你真的想要这个：
1.: -a
2.: sd-a
3.: sd-g----10-9a
4.: sda
5.: s ja
6.: dd
You switch:
你切换：
safeSequence+delimiter
with
和
delimiter+safeSequence
回答by chillysapien
Assuming that you only have a finite set of seperators before the words in your string (eg ;, : etc) you can use the following technique. (apologies for any syntax errors, but its been a while since I used Java)
假设您的字符串中的单词之前只有一组有限的分隔符（例如;、: 等），您可以使用以下技术。（对任何语法错误表示歉意，但自从我使用 Java 以来已经有一段时间了）
String toSplit = ":alpha;beta:gamma;delta "
toSplit = toSplit.replace(":", "~:")
toSplit = toSplit.replace(";", "~;")
//repeat for all you possible seperators
String[] splitStrings = toSplit.split("~")
