A quick and easy way to join array elements with a separator (the opposite of split) in Java

JavaArraysString

Java Problem Overview


See Related .NET question

I'm looking for a quick and easy way to do exactly the opposite of split so that it will cause ["a","b","c"] to become "a,b,c"

Iterating through an array requires either adding a condition (if this is not the last element, add the seperator) or using substring to remove the last separator.

I'm sure there is a certified, efficient way to do it (Apache Commons?)

How do you prefer doing it in your projects?

Java Solutions


Solution 1 - Java

Using Java 8 you can do this in a very clean way:

String.join(delimiter, elements);

This works in three ways:

  1. directly specifying the elements

    String joined1 = String.join(",", "a", "b", "c");

  2. using arrays

    String[] array = new String[] { "a", "b", "c" }; String joined2 = String.join(",", array);

  3. using iterables

    List list = Arrays.asList(array); String joined3 = String.join(",", list);

Solution 2 - Java

If you're on Android you can TextUtils.join(delimiter, tokens)

Solution 3 - Java

I prefer Guava over Apache StringUtils for this particular problem:

Joiner.on(separator).join(array)

Compared to StringUtils, the Joiner API has a fluent design and is a bit more flexible, e.g. null elements may be skipped or replaced by a placeholder. Also, Joiner has a feature for joining maps with a separator between key and value.

Solution 4 - Java

Apache Commons Lang does indeed have a StringUtils.join method which will connect String arrays together with a specified separator.

For example:

String[] s = new String[] {"a", "b", "c"};
String joined = StringUtils.join(s, ",");  // "a,b,c"

However, I suspect that, as you mention, there must be some kind of conditional or substring processing in the actual implementation of the above mentioned method.

If I were to perform the String joining and didn't have any other reasons to use Commons Lang, I would probably roll my own to reduce the number of dependencies to external libraries.

Solution 5 - Java

A fast and simple solution without any 3rd party includes.

public static String strJoin(String[] aArr, String sSep) {
	StringBuilder sbStr = new StringBuilder();
	for (int i = 0, il = aArr.length; i < il; i++) {
		if (i > 0)
			sbStr.append(sSep);
		sbStr.append(aArr[i]);
	}
	return sbStr.toString();
}

Solution 6 - Java

"I'm sure there is a certified, efficient way to do it (Apache Commons?)"

yes, apparenty it's

StringUtils.join(array, separator)

http://www.java2s.com/Code/JavaAPI/org.apache.commons.lang/StringUtilsjoinObjectarrayStringseparator.htm

Solution 7 - Java

With Java 1.8 there is a new StringJoiner class - so no need for Guava or Apache Commons:

String str = new StringJoiner(",").add("a").add("b").add("c").toString();

Or using a collection directly with the new stream api:

String str = Arrays.asList("a", "b", "c").stream().collect(Collectors.joining(","));

Solution 8 - Java

Even easier you can just use Arrays, so you will get a String with the values of the array separated by a ","

String concat = Arrays.toString(myArray);

so you will end up with this: concat = "[a,b,c]"

Update

You can then get rid of the brackets using a sub-string as suggested by Jeff

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

so you end up with concat = "a,b,c"

if you want to use Kotlin:

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

Solution 9 - Java

You can use replace and replaceAll with regular expressions.

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

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

Because Arrays.asList().toString() produces: "[a, b, c]", we do a replaceAll to remove the first and last brackets and then (optionally) you can change the ", " sequence for "," (your new separator).

A stripped version (fewer chars):

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

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

Regular expressions are very powerful, specially String methods "replaceFirst" and "replaceAll". Give them a try.

Solution 10 - Java

All of these other answers include runtime overhead... like using ArrayList.toString().replaceAll(...) which are very wasteful.

I will give you the optimal algorithm with zero overhead; it doesn't look as pretty as the other options, but internally, this is what they are all doing (after piles of other hidden checks, multiple array allocation and other crud).

Since you already know you are dealing with strings, you can save a bunch of array allocations by performing everything manually. This isn't pretty, but if you trace the actual method calls made by the other implementations, you'll see it has the least runtime overhead possible.

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);
}

Solution 11 - Java

Solution 12 - Java

This options is fast and clear:

  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);
  }

Solution 13 - Java

This small function always comes in handy.

public static String join(String[] strings, int startIndex, String separator) {
    StringBuffer sb = new StringBuffer();
    for (int i=startIndex; i < strings.length; i++) {
        if (i != startIndex) sb.append(separator);
        sb.append(strings[i]);
    }
    return sb.toString();
}

Solution 14 - Java

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.

Possible variations for Java version compatibility include substituting StringBuffer (Java 1.0) for StringBuilder (Java 1.5), switching out the Java 1.5 iterator and removing the generic wildcard (Java 1.5) from the Collection (Java 1.2). If you want to take backward compatibility a step or two further, delete the methods that use Collection and move the logic into the array-based method.

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();
}

Solution 15 - Java

public String join(String[] str, String separator){
	String retval = "";
	for (String s: str){ retval+= separator + s;}
	return retval.replaceFirst(separator, "");
}

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