The approach that I've taken has evolved since Java 1.0 to provide readability and maintain reasonable options for backward-compatibility with older Java versions, while also providing method signatures that are drop-in replacements for those from apache commons-lang. For performance reasons, I can see some possible objections to the use of Arrays.asList but I prefer helper methods that have sensible defaults without duplicating the one method that performs the actual work. This approach provides appropriate entry points to a reliable method that does not require array/list conversions prior to calling.

Java版本兼容性的可能变化包括用StringBuffer (Java 1.0)代替StringBuilder (Java 1.5)，切换出Java 1.5迭代器，并从集合(Java 1.2)中删除通用通配符(Java 1.5)。如果想进一步提高向后兼容性，可以删除使用Collection的方法，并将逻辑移到基于数组的方法中。

public static String join(String[] values)
{
    return join(values, ',');
}

public static String join(String[] values, char delimiter)
{
    return join(Arrays.asList(values), String.valueOf(delimiter));
}

// To match Apache commons-lang: StringUtils.join(values, delimiter)
public static String join(String[] values, String delimiter)
{
    return join(Arrays.asList(values), delimiter);
}

public static String join(Collection<?> values)
{
    return join(values, ',');
}

public static String join(Collection<?> values, char delimiter)
{
    return join(values, String.valueOf(delimiter));
}

public static String join(Collection<?> values, String delimiter)
{
    if (values == null)
    {
        return new String();
    }

    StringBuffer strbuf = new StringBuffer();

    boolean first = true;

    for (Object value : values)
    {
        if (!first) { strbuf.append(delimiter); } else { first = false; }
        strbuf.append(value.toString());
    }

    return strbuf.toString();
}

所有这些其他答案都包括运行时开销……比如使用ArrayList.toString(). replaceall(…)，这是非常浪费的。

我会给出零开销的最优算法; 它看起来不像其他选项那么漂亮，但在内部，这是它们都在做的事情(在成堆的其他隐藏检查、多个数组分配和其他杂事之后)。

因为您已经知道您正在处理字符串，所以您可以通过手动执行所有操作来节省大量的数组分配。这并不漂亮，但如果跟踪其他实现所执行的实际方法调用，您将看到它的运行时开销可能最少。

public static String join(String separator, String ... values) {
  if (values.length==0)return "";//need at least one element
  //all string operations use a new array, so minimize all calls possible
  char[] sep = separator.toCharArray();

  // determine final size and normalize nulls
  int totalSize = (values.length - 1) * sep.length;// separator size
  for (int i = 0; i < values.length; i++) {
    if (values[i] == null)
      values[i] = "";
    else
      totalSize += values[i].length();
  }

  //exact size; no bounds checks or resizes
  char[] joined = new char[totalSize];
  int pos = 0;
  //note, we are iterating all the elements except the last one
  for (int i = 0, end = values.length-1; i < end; i++) {
    System.arraycopy(values[i].toCharArray(), 0, 
      joined, pos, values[i].length());
    pos += values[i].length();
    System.arraycopy(sep, 0, joined, pos, sep.length);
    pos += sep.length;
  }
  //now, add the last element; 
  //this is why we checked values.length == 0 off the hop
  System.arraycopy(values[values.length-1].toCharArray(), 0,
    joined, pos, values[values.length-1].length());

  return new String(joined);
}

你可以用正则表达式来使用replace和replaceAll。

String[] strings = {"a", "b", "c"};

String result = Arrays.asList(strings).toString().replaceAll("(^\\[|\\]$)", "").replace(", ", ",");

因为Arrays.asList(). tostring()生成:"[a, b, c]"，我们执行replaceAll来删除第一个和最后一个括号，然后(可选地)您可以更改"，"序列为"，"(您的新分隔符)。

精简版(字符数更少):

String[] strings = {"a", "b", "c"};

String result = ("" + Arrays.asList(strings)).replaceAll("(^.|.$)", "").replace(", ", "," );

正则表达式非常强大，特别是字符串方法“replaceFirst”和“replaceAll”。试一试吧。

更简单的是，你可以使用数组，所以你会得到一个字符串，数组的值由"，"隔开

String concat = Arrays.toString(myArray);

所以你会得到这样的结果:concat = "[a,b,c]"

更新

然后，您可以使用Jeff建议的子字符串来消除括号

concat = concat.substring(1, concat.length() -1);

所以你得到concat = "a,b,c"

如果你想使用Kotlin:

 val concat = myArray.joinToString(separator = ",") //"a,b,c"

如果你在Android上，你可以使用TextUtils。加入(分隔符,令牌)

这个选项是快速和清晰的:

  public static String join(String separator, String... values) {
    StringBuilder sb = new StringBuilder(128);
    int end = 0;
    for (String s : values) {
      if (s != null) {
        sb.append(s);
        end = sb.length();
        sb.append(separator);
      }
    }
    return sb.substring(0, end);
  }