如何在 Java 中连接两个字符串？

Question

I am trying to concatenate strings in Java. Why isn't this working?

public class StackOverflowTest {  
    public static void main(String args[]) {
        int theNumber = 42;
        System.out.println("Your number is " . theNumber . "!");
    }
}

Answer 1

You can concatenate Strings using the + operator:

System.out.println("Your number is " + theNumber + "!");

theNumber is implicitly converted to the String "42" .

Answer 2

The concatenation operator in java is + , not .

Read this (including all subsections) before you start. Try to stop thinking the php way ;)

To broaden your view on using strings in Java - the + operator for strings is actually transformed (by the compiler) into something similar to:

new StringBuilder().append("firstString").append("secondString").toString()

Answer 3

There are two basic answers to this question:

1. [simple] Use the + operator (string concatenation). "your number is" + theNumber + "!" (as noted elsewhere)
2. [less simple]: Use StringBuilder (or StringBuffer ).

StringBuilder value;
value.append("your number is");
value.append(theNumber);
value.append("!");

value.toString();

I recommend against stacking operations like this:

new StringBuilder().append("I").append("like to write").append("confusing code");

Edit: starting in java 5 the string concatenation operator is translated into StringBuilder calls by the compiler. Because of this, both methods above are equal.

Note: Spaceisavaluablecommodity,asthissentancedemonstrates.

Caveat: Example 1 below generates multiple StringBuilder instances and is less efficient than example 2 below

Example 1

String Blam = one + two;
Blam += three + four;
Blam += five + six;

Example 2

String Blam = one + two + three + four + five + six;

Answer 4

Out of the box you have 3 ways to inject the value of a variable into a String as you try to achieve:

1. The simplest way

You can simply use the operator + between a String and any object or primitive type, it will automatically concatenate the String and

1. In case of an object, the value of String.valueOf(obj) corresponding to the String " null " if obj is null otherwise the value of obj.toString() .
2. In case of a primitive type, the equivalent of String.valueOf(<primitive-type>) .

Example with a non null object:

Integer theNumber = 42;
System.out.println("Your number is " + theNumber + "!");

Output:

Your number is 42!

Example with a null object:

Integer theNumber = null;
System.out.println("Your number is " + theNumber + "!");

Output:

Your number is null!

Example with a primitive type:

int theNumber = 42;
System.out.println("Your number is " + theNumber + "!");

Output:

Your number is 42!

2. The explicit way and potentially the most efficient one

You can use StringBuilder (or StringBuffer the thread-safe outdated counterpart) to build your String using the append methods.

Example:

int theNumber = 42;
StringBuilder buffer = new StringBuilder()
    .append("Your number is ").append(theNumber).append('!');
System.out.println(buffer.toString()); // or simply System.out.println(buffer)

Output:

Your number is 42!

Behind the scene, this is actually how recent java compilers convert all the String concatenations done with the operator + , the only difference with the previous way is that you have the full control .

Indeed, the compilers will use the default constructor so the default capacity ( 16 ) as they have no idea what would be the final length of the String to build, which means that if the final length is greater than 16 , the capacity will be necessarily extended which has price in term of performances.

So if you know in advance that the size of your final String will be greater than 16 , it will be much more efficient to use this approach to provide a better initial capacity. For instance, in our example we create a String whose length is greater than 16, so for better performances it should be rewritten as next:

Example optimized :

int theNumber = 42;
StringBuilder buffer = new StringBuilder(18)
    .append("Your number is ").append(theNumber).append('!');
System.out.println(buffer)

Output:

Your number is 42!

3. The most readable way

You can use the methods String.format(locale, format, args) or String.format(format, args) that both rely on a Formatter to build your String . This allows you to specify the format of your final String by using place holders that will be replaced by the value of the arguments.

Example:

int theNumber = 42;
System.out.println(String.format("Your number is %d!", theNumber));
// Or if we need to print only we can use printf
System.out.printf("Your number is still %d with printf!%n", theNumber);

Output:

Your number is 42!
Your number is still 42 with printf!

The most interesting aspect with this approach is the fact that we have a clear idea of what will be the final String because it is much more easy to read so it is much more easy to maintain.

Answer 5

The java 8 way:

StringJoiner sj1 = new StringJoiner(", ");
String joined = sj1.add("one").add("two").toString();
// one, two
System.out.println(joined);


StringJoiner sj2 = new StringJoiner(", ","{", "}");
String joined2 = sj2.add("Jake").add("John").add("Carl").toString();
// {Jake, John, Carl}
System.out.println(joined2);

Answer 6

You must be a PHP programmer.

Use a + sign.

System.out.println("Your number is " + theNumber + "!");

Answer 7

For exact concatenation operation of two string please use:

file_names = file_names.concat(file_names1);

In your case use + instead of .

Answer 8

为了获得更好的性能，请使用str1.concat(str2)其中str1和str2是字符串变量。

Answer 9

“+”代替“。”

Answer 10

使用+进行字符串连接。

"Your number is " + theNumber + "!"

Answer 11

This should work

public class StackOverflowTest
{  
    public static void main(String args[])
    {
        int theNumber = 42;
        System.out.println("Your number is " + theNumber + "!");
    }
}

Answer 12

In java concatenate symbol is " + ". If you are trying to concatenate two or three strings while using jdbc then use this:

String u = t1.getString();
String v = t2.getString();
String w = t3.getString();
String X = u + "" + v + "" + w;
st.setString(1, X);

Here "" is used for space only.

Answer 13

String.join( delimiter , stringA , stringB , … )

As of Java 8 and later, we can use String.join .

Caveat: You must pass all String or CharSequence objects. So your int variable 42 does not work directly. One alternative is using an object rather than primitive, and then calling toString .

Integer theNumber = 42;
String output = 
    String                                                   // `String` class in Java 8 and later gained the new `join` method.
    .join(                                                   // Static method on the `String` class. 
        "" ,                                                 // Delimiter.
        "Your number is " , theNumber.toString() , "!" ) ;   // A series of `String` or `CharSequence` objects that you want to join.
    )                                                        // Returns a `String` object of all the objects joined together separated by the delimiter.
;

Dump to console.

System.out.println( output ) ;

See this code run live at IdeOne.com .

Answer 14

在 Java 中，连接符号是“+”，而不是“.”。

Answer 15

"+" not "."

But be careful with String concatenation. Here's a link introducing some thoughts from IBM DeveloperWorks .

Answer 16

You can concatenate Strings using the + operator:

String a="hello ";
String b="world.";
System.out.println(a+b);

Output:

hello world.

That's it

Answer 17

So from the able answer's you might have got the answer for why your snippet is not working. Now I'll add my suggestions on how to do it effectively. This article is a good place where the author speaks about different way to concatenate the string and also given the time comparison results between various results.

Different ways by which Strings could be concatenated in Java

1. By using + operator (20 + "")
2. By using concat method in String class
3. Using StringBuffer
4. By using StringBuilder

Method 1:

This is a non-recommended way of doing. Why? When you use it with integers and characters you should be explicitly very conscious of transforming the integer to toString() before appending the string or else it would treat the characters to ASCI int's and would perform addition on the top.

String temp = "" + 200 + 'B';

//This is translated internally into,

new StringBuilder().append( "" ).append( 200 ).append('B').toString();

Method 2:

This is the inner concat method's implementation

 public String concat(String str) { int olen = str.length(); if (olen == 0) { return this; } if (coder() == str.coder()) { byte[] val = this.value; byte[] oval = str.value; int len = val.length + oval.length; byte[] buf = Arrays.copyOf(val, len); System.arraycopy(oval, 0, buf, val.length, oval.length); return new String(buf, coder); } int len = length(); byte[] buf = StringUTF16.newBytesFor(len + olen); getBytes(buf, 0, UTF16); str.getBytes(buf, len, UTF16); return new String(buf, UTF16); }

This creates a new buffer each time and copies the old content to the newly allocated buffer. So, this is would be too slow when you do it on more Strings.

Method 3:

This is thread safe and comparatively fast compared to (1) and (2). This uses StringBuilder internally and when it allocates new memory for the buffer (say it's current size is 10) it would increment it's 2*size + 2 (which is 22). So when the array becomes bigger and bigger this would really perform better as it need not allocate buffer size each and every time for every append call.

 private int newCapacity(int minCapacity) { // overflow-conscious code int oldCapacity = value.length >> coder; int newCapacity = (oldCapacity << 1) + 2; if (newCapacity - minCapacity < 0) { newCapacity = minCapacity; } int SAFE_BOUND = MAX_ARRAY_SIZE >> coder; return (newCapacity <= 0 || SAFE_BOUND - newCapacity < 0) ? hugeCapacity(minCapacity) : newCapacity; } private int hugeCapacity(int minCapacity) { int SAFE_BOUND = MAX_ARRAY_SIZE >> coder; int UNSAFE_BOUND = Integer.MAX_VALUE >> coder; if (UNSAFE_BOUND - minCapacity < 0) { // overflow throw new OutOfMemoryError(); } return (minCapacity > SAFE_BOUND) ? minCapacity : SAFE_BOUND; }

Method 4

StringBuilder would be the fastest one for String concatenation since it's not thread safe . Unless you are very sure that your class which uses this is single ton I would highly recommend not to use this one.

In short, use StringBuffer until you are not sure that your code could be used by multiple threads. If you are damn sure, that your class is singleton then go ahead with StringBuilder for concatenation.

Answer 18

First method: You could use "+" sign for concatenating strings, but this always happens in print. Another way: The String class includes a method for concatenating two strings: string1.concat(string2);

Answer 19

import com.google.common.base.Joiner;

String delimiter = "";
Joiner.on(delimiter).join(Lists.newArrayList("Your number is ", 47, "!"));

This may be overkill to answer the op's question, but it is good to know about for more complex join operations. This stackoverflow question ranks highly in general google searches in this area, so good to know.

Answer 20

you can use stringbuffer, stringbuilder, and as everyone before me mentioned, "+". I'm not sure how fast "+" is (I think it is the fastest for shorter strings), but for longer I think builder and buffer are about equal (builder is slightly faster because it's not synchronized).

Answer 21

here is an example to read and concatenate 2 string without using 3rd variable:

public class Demo {
    public static void main(String args[]) throws Exception  {
        InputStreamReader r=new InputStreamReader(System.in);     
        BufferedReader br = new BufferedReader(r);
        System.out.println("enter your first string");
        String str1 = br.readLine();
        System.out.println("enter your second string");
        String str2 = br.readLine();
        System.out.println("concatenated string is:" + str1 + str2);
    }
}

Answer 22

There are multiple ways to do so, but Oracle and IBM say that using + , is a bad practice, because essentially every time you concatenate String, you end up creating additional objects in memory. It will utilize extra space in JVM, and your program may be out of space, or slow down.

Using StringBuilder or StringBuffer is best way to go with it. Please look at Nicolas Fillato's comment above for example related to StringBuffer .

String first = "I eat";  String second = "all the rats."; 
System.out.println(first+second);

Answer 23

Using "+" symbol u can concatenate strings.

String a="I"; 
String b="Love."; 
String c="Java.";
System.out.println(a+b+c);