英文:
Split a String by multiple delimiters in java
问题
以下是翻译好的内容：
什么是通过多个分隔符拆分字符串的可能方法？StringTokenizer能帮助我实现这一点吗？
String str="list1|10456103|10456102|10456121#list2|10456105|10456122";
String str="list1|10513846#list2|";
String str3="list1#list2|10509855";
String str4="list2|10481812|";
String str5="list1|10396496|";
String str6="list1#list2|";
现在，我应该能够仅提取长整型值：
对于Str1，Finallist=[10456103,10456102,10456121,10456105,10456122]
对于Str2，Finallist=[10513846]
对于Str3，Finallist=[10509855]
对于Str4，Finallist=[10481812]
对于Str5，Finallist=[10396496]
对于Str6，Finallist=[]
英文:
what is the possible way to split a String by multiple delimiters? Is StringTokenizer can help me out to achieve this?
String str="list1|10456103|10456102|10456121#list2|10456105|10456122";
String str="list1|10513846#list2|";
String str3="list1#list2|10509855";
String str4="list2|10481812|";
String str5="list1|10396496|";
String str6="list1#list2|";
So now I should be able to extract only the long values :
For Str1 Finallist=[10456103,10456102,10456121,10456105,10456122]
For Str2 Finallist=[10513846]
For Str3 Finallist=[10509855]
For Str4 Finallist=[10481812]
For Str5 Finallist=[10396496]
For Str6 Finallist[]
答案1
得分: 1
你可以使用 Java 中的字符串 split 方法来将它们分割，然后检查是否为数字。
该字符串通过 #、| 或 , 进行分割，可以出现一次或多次。
然后对分割后的字符串进行数字检测，如下所示：
public static void main(String []args){
String str = "list1|10456103|10456102|10456121#list2|10456105|10456122";
String arr[] = str.split("[|,#]+");
for (String s : arr) {
try {
int num = Integer.parseInt(s);
System.out.println(num + " 是一个数字"); // 添加到列表
} catch (Exception err) {
System.out.println(s + " 不是一个数字"); // 不添加到列表
}
}
}
英文:
you can split them using the split method for String in java, then check if it's numeric or not.
the string is split by having # or | or , once or more than once.
then the split strings are tested to be numeric or not, as so:
public static void main(String []args){
String str="list1|10456103|10456102|10456121#list2|10456105|10456122";
String arr[] = str.split("[|,#]+");
for(String s: arr){
try{
int num=Integer.parseInt(s);
System.out.println(num + " is a number"); //add to list
}catch(Exception err) {
System.out.println(s + " is not a number"); //don't add to list
}
}
}
答案2
得分: 1
以下是翻译好的部分：
为了您的考虑，您还可以将它们即时转换为列表的映射以供以后处理。
它的工作方式如下：
- 将每个字符串以
[|#]分割 - 仅筛选匹配
\\d+的数字 - 转换为长整型
- 并保存到列表中
LinkedHashMap 只是保留了最终映射中的处理顺序。实际上，这并不是必需的。
String[] strings = { str1, str2, str3, str4, str5, str6 };
Map<String, List<Long>> map = IntStream
.range(0, strings.length)
.boxed()
.collect(Collectors.toMap(i -> "str" + (i + 1),
i -> Arrays.stream(strings[i].split("[|#]"))
.filter(str -> str.matches("\\d+"))
.map(Long::valueOf)
.collect(Collectors.toList()),
(a, b) -> a, LinkedHashMap::new));
map.entrySet().forEach(System.out::println);
输出结果如下：
str1=[10456103, 10456102, 10456121, 10456105, 10456122]
str2=[10513846]
str3=[10509855]
str4=[10481812]
str5=[10396496]
str6=[]
英文:
For your consideration you can also convert them to a map of lists on the fly for later processing.
It works as follows:
- Split each String on
[|#] - filter on numbers only matching on
\\d+ - convert to a long
- and save to a list
The LinkedHashMap simply preserves processing order in the final map. It is not really required for this to work.
String[] strings = { str1, str2,str3,str4,str5,str6 };
Map<String, List<Long>> map = IntStream
.range(0, strings.length)
.boxed()
.collect(Collectors.toMap(i->"str"+(i+1),
i->Arrays.stream(strings[i].split("[|#]"))
.filter(str->str.matches("\\d+"))
.map(Long::valueOf)
.collect(Collectors.toList()),
(a,b)->a, LinkedHashMap::new));
map.entrySet().forEach(System.out::println);
Prints
str1=[10456103, 10456102, 10456121, 10456105, 10456122]
str2=[10513846]
str3=[10509855]
str4=[10481812]
str5=[10396496]
str6=[]
</details>
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