JavaScript有Array.join()

js>["Bill","Bob","Steve"].join(" and ")
Bill and Bob and Steve

Java有这样的东西吗?我知道我可以用StringBuilder自己拼凑一些东西:

static public String join(List<String> list, String conjunction)
{
   StringBuilder sb = new StringBuilder();
   boolean first = true;
   for (String item : list)
   {
      if (first)
         first = false;
      else
         sb.append(conjunction);
      sb.append(item);
   }
   return sb.toString();
}

. .但是如果像这样的东西已经是JDK的一部分,那么这样做就没有意义了。


当前回答

如果您正在使用Eclipse Collections(以前的GS Collections),则可以使用makeString()方法。

List<String> list = Arrays.asList("Bill", "Bob", "Steve");

String string = ListAdapter.adapt(list).makeString(" and ");

Assert.assertEquals("Bill and Bob and Steve", string);

如果您可以将List转换为Eclipse Collections类型,那么您就可以摆脱适配器。

MutableList<String> list = Lists.mutable.with("Bill", "Bob", "Steve");
String string = list.makeString(" and ");

如果您只想要一个逗号分隔的字符串,您可以使用不接受参数的makeString()版本。

Assert.assertEquals(
    "Bill, Bob, Steve", 
    Lists.mutable.with("Bill", "Bob", "Steve").makeString());

注意:我是Eclipse Collections的提交者。

其他回答

用java 1.8的流可以使用,

import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;

List<String> list = Arrays.asList("Bill","Bob","Steve").
String str = list.stream().collect(Collectors.joining(" and "));

String.join

在Java 8中,你不需要任何第三方库就可以做到这一点。

如果你想加入一个字符串集合,你可以使用String.join()方法:

List<String> list = Arrays.asList("foo", "bar", "baz");
String joined = String.join(" and ", list); // "foo and bar and baz"

Collectors.joining

如果你有一个非String类型的Collection,你可以使用Stream API来连接Collector:

List<Person> list = Arrays.asList(
  new Person("John", "Smith"),
  new Person("Anna", "Martinez"),
  new Person("Paul", "Watson ")
);

String joinedFirstNames = list.stream()
  .map(Person::getFirstName)
  .collect(Collectors.joining(", ")); // "John, Anna, Paul"

StringJoiner类可能也很有用。

试试这个:

java.util.Arrays.toString(anArray).replaceAll(", ", ",")
                .replaceFirst("^\\[","").replaceFirst("\\]$","");

实现它的一个正统方法是定义一个新函数:

public static String join(String joinStr, String... strings) {
    if (strings == null || strings.length == 0) {
        return "";
    } else if (strings.length == 1) {
        return strings[0];
    } else {
        StringBuilder sb = new StringBuilder(strings.length * 1 + strings[0].length());
        sb.append(strings[0]);
        for (int i = 1; i < strings.length; i++) {
            sb.append(joinStr).append(strings[i]);
        }
        return sb.toString();
    }
}

示例:

String[] array = new String[] { "7, 7, 7", "Bill", "Bob", "Steve",
        "[Bill]", "1,2,3", "Apple ][","~,~" };

String joined;
joined = join(" and ","7, 7, 7", "Bill", "Bob", "Steve", "[Bill]", "1,2,3", "Apple ][","~,~");
joined = join(" and ", array); // same result

System.out.println(joined);

输出:

7,7,7和比尔,鲍勃,史蒂夫和[比尔],1,2,3和苹果][和~,~

如果您正在使用Eclipse Collections(以前的GS Collections),则可以使用makeString()方法。

List<String> list = Arrays.asList("Bill", "Bob", "Steve");

String string = ListAdapter.adapt(list).makeString(" and ");

Assert.assertEquals("Bill and Bob and Steve", string);

如果您可以将List转换为Eclipse Collections类型,那么您就可以摆脱适配器。

MutableList<String> list = Lists.mutable.with("Bill", "Bob", "Steve");
String string = list.makeString(" and ");

如果您只想要一个逗号分隔的字符串,您可以使用不接受参数的makeString()版本。

Assert.assertEquals(
    "Bill, Bob, Steve", 
    Lists.mutable.with("Bill", "Bob", "Steve").makeString());

注意:我是Eclipse Collections的提交者。