Kotlin - How to correctly concatenate a String

I agree with the accepted answer above but it is only good for known string values. For dynamic string values here is my suggestion.

// A list may come from an API JSON like
{
   "names": [
      "Person 1",
      "Person 2",
      "Person 3",
         ...
      "Person N"
   ]
}
var listOfNames = mutableListOf<String>() 

val stringOfNames = listOfNames.joinToString(", ") 
// ", " <- a separator for the strings, could be any string that you want

// Posible result
// Person 1, Person 2, Person 3, ..., Person N

This is useful for concatenating list of strings with separator.

Try this, I think this is a natively way to concatenate strings in Kotlin:

val result = buildString{
    append("a")
    append("b")
}

println(result)

// you will see "ab" in console.

Similar to @Rhusfer answer I wrote this. In case you have a group of EditTexts and want to concatenate their values, you can write:

listOf(edit_1, edit_2, edit_3, edit_4).joinToString(separator = "") { it.text.toString() }

If you want to concatenate Map, use this:

map.entries.joinToString(separator = ", ")

To concatenate Bundle, use

bundle.keySet().joinToString(", ") { key -> "$key=${bundle[key]}" }

It sorts keys in alphabetical order.

Example:

val map: MutableMap<String, Any> = mutableMapOf("price" to 20.5)
map += "arrange" to 0
map += "title" to "Night cream"
println(map.entries.joinToString(separator = ", "))

// price=20.5, arrange=0, title=Night cream

val bundle = bundleOf("price" to 20.5)
bundle.putAll(bundleOf("arrange" to 0))
bundle.putAll(bundleOf("title" to "Night cream"))
val bundleString =
    bundle.keySet().joinToString(", ") { key -> "$key=${bundle[key]}" }
println(bundleString)

// arrange=0, price=20.5, title=Night cream

There are various way to concatenate strings in kotlin Example -

a = "Hello" , b= "World"

1. Using + operator a+b

2. Using plus() operator

   a.plus(b)

Note - + is internally converted to .plus() method only

In above 2 methods, a new string object is created as strings are immutable. if we want to modify the existing string, we can use StringBuilder

StringBuilder str = StringBuilder("Hello").append("World")

The most simplest way so far which will add separator and exclude the empty/null strings from concatenation:

val finalString = listOf(a, b, c)
    .filterNot { it.isNullOrBlank() }
    .joinToString(separator = " ")

yourString += "newString"

This way you can concatenate a string

I suggest if you have a limited and predefined set of values then the most efficient and readable approach is to use the String Template (It uses String Builder to perform concatination).

val a = "Hello"
val b = "World"
val c = "$a  ${b.toUpperCase()} !"

println(c) //prints: Hello  WORLD !

On the other hand if you have a collection of values then use joinToString method.

val res = (1..100).joinToString(",")
println(res) //prints: 1,2,3,...,100

I believe that some suggested solutions on this post are are not efficient. Like using plus or + or creating a collection for a limited set of entris and then applying joinToString on them.

If you have an object and want to concatenate two values of an object like

data class Person(
val firstName: String,
val lastName: String
)

Then simply following won't work

val c = "$person.firstName $person.lastName"

Correct way in this case will be

"${person.firstName} ${person.lastName}"

If you want any string in between concatenated values, just write there without any helper symbol. For example if I want "Name is " and then a hyphen in between first and last name then

 "Name is ${person.firstName}-${person.lastName}"

In Kotlin we can join string array using joinToString()

val tags=arrayOf("hi","bye")

val finalString=tags.joinToString (separator = ","){ "#$it" }

Result is :

#hi,#bye

if list coming from server

var tags = mutableListOf<Tags>() // list from server

val finalString=tags.joinToString (separator = "-"){ "#${it.tagname}" }

Result is :

#hi-#bye