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
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~，~
其他回答
谷歌的Guava API也有.join()，尽管(从其他回复中应该很明显)，Apache Commons在这里几乎是标准的。
你可以这样做:
String aToString = java.util.Arrays.toString(anArray);
// Do not need to do this if you are OK with '[' and ']'
aToString = aToString.substring(1, aToString.length() - 1);
或者单行语句(仅当不需要'['和']'时)
String aToString = java.util.Arrays.toString(anArray).substring(1).replaceAll("\\]$", "");希望这能有所帮助。
用java 1.8的流可以使用，
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
List<String> list = Arrays.asList("Bill","Bob","Steve").
String str = list.stream().collect(Collectors.joining(" and "));
如果您正在使用Eclipse Collections(以前的GS Collections)，则可以使用makeString()方法。
List<String> list = Arrays.asList("Bill", "Bob", "Steve");
String string = ListAdapter.adapt(list).makeString(" and ");
Assert.assertEquals("Bill and Bob and Steve", string);
如果您可以将List转换为Eclipse Collections类型，那么您就可以摆脱适配器。
MutableList<String> list = Lists.mutable.with("Bill", "Bob", "Steve");
String string = list.makeString(" and ");
如果您只想要一个逗号分隔的字符串，您可以使用不接受参数的makeString()版本。
Assert.assertEquals(
"Bill, Bob, Steve",
Lists.mutable.with("Bill", "Bob", "Steve").makeString());
注意:我是Eclipse Collections的提交者。
在纯JDK中有一种有趣的方法:
String[] array = new String[] { "Bill", "Bob", "Steve","[Bill]","1,2,3","Apple ][" };
String join = " and ";
String joined = Arrays.toString(array).replaceAll(", ", join)
.replaceAll("(^\\[)|(\\]$)", "");
System.out.println(joined);
输出:
比尔、鲍勃、史蒂夫、[比尔]、1,2,3和苹果][
一个不太完美也不太有趣的方式!
String[] array = new String[] { "7, 7, 7","Bill", "Bob", "Steve", "[Bill]",
"1,2,3", "Apple ][" };
String join = " and ";
for (int i = 0; i < array.length; i++) array[i] = array[i].replaceAll(", ", "~,~");
String joined = Arrays.toString(array).replaceAll(", ", join)
.replaceAll("(^\\[)|(\\]$)", "").replaceAll("~,~", ", ");
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][
