A quick and easy way to join array elements with a separator (the opposite of split) in Java
JavaArraysStringJava Problem Overview
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:
-
directly specifying the elements
String joined1 = String.join(",", "a", "b", "c");
-
using arrays
String[] array = new String[] { "a", "b", "c" }; String joined2 = String.join(",", array);
-
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)
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
it's in StringUtils:
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, "");
}