英文:
How do I split a string into sentences while maintaining all the spaces in Java?
问题
我尝试了以下正则表达式：
sentences = sb.toString().split("(?<=[a-z])*\\.\\s*");
我正在使用一个字符串构建器 sb，将其转换为字符串，然后使用 split 函数。正则表达式检查在'.'之前的0个或多个字符以及在'.'之后的0个或多个空格。
然而，它对以下输入不起作用：
Hello World. Shipped to U.S on Friday.We are here .Good input
但我需要保留"We are here"之前的空格。
期望的输出是：
Hello World
Shipped to U.S on Friday
We are here
Good input
英文:
I tried the following regex:
sentences = sb.toString().split("(?<=[a-z])*\\.\\s*");
I am using a stringBuilder sb and converting it to string and then using a split function<br>
The regex checks for 0 or more characters before '.' and 0 or more spaces after the '.'<br>
However, it doesn't work for the following input<be>
Hello World. Shipped to U.S on Friday.We are here .Good input
But I need to keep the space before We are here
Req Output
Hello World
Shipped to U.S on Friday
We are here
Good input
答案1
得分: 3
使用这个正则表达式：([^\.]+)(\.|$)*?
您可以阅读有关组匹配并在此处查看完整的匹配：https://regex101.com/r/yV9GES/5
编辑：已更新评论中的答案链接。
英文:
use this regex: ([^\.]+)(\.|$)*?
you can read about group matchers and see the full matches here : https://regex101.com/r/yV9GES/5
edit: updated the link for answer in the comment.
答案2
得分: 1
使用\\.拆分你的字符串，即在.上拆分。
演示：
import java.util.Arrays;
public class Main {
public static void main(String[] args) {
System.out.println(Arrays.toString("Hello World. We are here .Good input.".split("\\.")));
}
}
输出：
[Hello World, We are here , Good input]
英文:
Split your string using \\. i.e. on .
Demo:
import java.util.Arrays;
public class Main {
public static void main(String[] args) {
System.out.println(Arrays.toString("Hello World. We are here .Good input.".split("\\.")));
}
}
Output:
[Hello World, We are here , Good input]
答案3
得分: 1
为什么要使用正则表达式？
你可以简单地使用 indexOf 和 substring
public List<String> splitOnDot(String input) {
List<String> result = new ArrayList<>();
int idx;
while ((idx = input.indexOf('.')) != -1) {
result.add(input.substring(0, idx));
input = input.substring(idx + 1);
}
return result;
}
成功的测试：
@Test
public void test1() {
assertThat(splitOnDot("Hello World. We are here .Good input.")).contains("Hello World", " We are here ", "Good input");
}
英文:
Why do you have to use a RegEx?
You can simply use indexOf and substring
public List<String> splitOnDot(String input) {
List<String> result = new ArrayList<>();
int idx;
while ((idx = input.indexOf('.')) != -1) {
result.add(input.substring(0, idx));
input = input.substring(idx + 1);
}
return result;
}
Successful test:
@Test
public void test1() {
assertThat(splitOnDot("Hello World. We are here .Good input.")).contains("Hello World", " We are here ", "Good input");
}
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