我写了这个(我用它来bean和利用toString，所以不要写Collection<String>):

public static String join(Collection<?> col, String delim) {
    StringBuilder sb = new StringBuilder();
    Iterator<?> iter = col.iterator();
    if (iter.hasNext())
        sb.append(iter.next().toString());
    while (iter.hasNext()) {
        sb.append(delim);
        sb.append(iter.next().toString());
    }
    return sb.toString();
}

但是JSP不支持Collection，所以对于TLD我写了:

public static String join(List<?> list, String delim) {
    int len = list.size();
    if (len == 0)
        return "";
    StringBuilder sb = new StringBuilder(list.get(0).toString());
    for (int i = 1; i < len; i++) {
        sb.append(delim);
        sb.append(list.get(i).toString());
    }
    return sb.toString();
}

并放入。tld文件:

<?xml version="1.0" encoding="UTF-8"?>
<taglib version="2.1" xmlns="http://java.sun.com/xml/ns/javaee"
    <function>
        <name>join</name>
        <function-class>com.core.util.ReportUtil</function-class>
        <function-signature>java.lang.String join(java.util.List, java.lang.String)</function-signature>
    </function>
</taglib>

并在JSP文件中使用它:

<%@taglib prefix="funnyFmt" uri="tag:com.core.util,2013:funnyFmt"%>
${funnyFmt:join(books, ", ")}

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。

所有对Apache Commons的引用都很好(这是大多数人使用的)，但我认为与Guava相当的Joiner具有更好的API。

你可以使用简单的连接

Joiner.on(" and ").join(names)

但也很容易处理空值:

Joiner.on(" and ").skipNulls().join(names);

or

Joiner.on(" and ").useForNull("[unknown]").join(names);

和(就我而言，它比common -lang更有用)，处理map的能力:

Map<String, Integer> ages = .....;
String foo = Joiner.on(", ").withKeyValueSeparator(" is ").join(ages);
// Outputs:
// Bill is 25, Joe is 30, Betty is 35

这对于调试等非常有用。

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~，~

使用Java .util. stringjoiner的Java 8解决方案

Java 8有一个StringJoiner类。但您仍然需要编写一些样板文件，因为它是Java。

StringJoiner sj = new StringJoiner(" and ", "" , "");
String[] names = {"Bill", "Bob", "Steve"};
for (String name : names) {
   sj.add(name);
}
System.out.println(sj);

另一个解，它是另一个答案的变体

public static String concatStringsWSep(Iterable<String> strings, String separator) {
    Iterator<String> it = strings.iterator();
    if( !it.hasNext() ) return "";
    StringBuilder sb = new StringBuilder(it.next());
    while( it.hasNext()) {
        sb.append(separator).append(it.next());
    }
    return sb.toString();                           
}