来自Perl,我肯定错过了在源代码中创建多行字符串的“here-document”方法:

$string = <<"EOF"  # create a three-line string
text
text
text
EOF

在Java中,当我从头开始连接多行字符串时,我必须在每一行上使用繁琐的引号和加号。

有什么更好的选择吗?在属性文件中定义我的字符串?

编辑:有两个答案说StringBuilder.append()比加号更可取。谁能详细解释一下他们为什么这么想?在我看来,这一点也不可取。我正在寻找一种方法来解决多行字符串不是一级语言结构这一事实,这意味着我绝对不想用方法调用取代一级语言结构(字符串连接与加号)。

编辑:为了进一步澄清我的问题,我根本不关心性能。我关心的是可维护性和设计问题。


当前回答

当使用一长串的+时,只会创建一个StringBuilder,除非在编译时确定String,在这种情况下不使用StringBuilder !

StringBuilder唯一更高效的情况是使用多个语句构造String。

String a = "a\n";
String b = "b\n";
String c = "c\n";
String d = "d\n";

String abcd = a + b + c + d;
System.out.println(abcd);

String abcd2 = "a\n" +
        "b\n" +
        "c\n" +
        "d\n";
System.out.println(abcd2);

注意:只创建了一个StringBuilder。

  Code:
   0:   ldc     #2; //String a\n
   2:   astore_1
   3:   ldc     #3; //String b\n
   5:   astore_2
   6:   ldc     #4; //String c\n
   8:   astore_3
   9:   ldc     #5; //String d\n
   11:  astore  4
   13:  new     #6; //class java/lang/StringBuilder
   16:  dup
   17:  invokespecial   #7; //Method java/lang/StringBuilder."<init>":()V
   20:  aload_1
   21:  invokevirtual   #8; //Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
   24:  aload_2
   25:  invokevirtual   #8; //Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
   28:  aload_3
   29:  invokevirtual   #8; //Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
   32:  aload   4
   34:  invokevirtual   #8; //Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
   37:  invokevirtual   #9; //Method java/lang/StringBuilder.toString:()Ljava/lang/String;
   40:  astore  5
   42:  getstatic       #10; //Field java/lang/System.out:Ljava/io/PrintStream;
   45:  aload   5
   47:  invokevirtual   #11; //Method java/io/PrintStream.println:(Ljava/lang/String;)V
   50:  ldc     #12; //String a\nb\nc\nd\n
   52:  astore  6
   54:  getstatic       #10; //Field java/lang/System.out:Ljava/io/PrintStream;
   57:  aload   6
   59:  invokevirtual   #11; //Method java/io/PrintStream.println:(Ljava/lang/String;)V
   62:  return

为了进一步澄清我的问题,我根本不关心性能。我关心的是可维护性和设计问题。

尽你所能让它清晰简单。

其他回答

String newline = System.getProperty ("line.separator");
string1 + newline + string2 + newline + string3

但是,最好的替代方法是使用String.format

String multilineString = String.format("%s\n%s\n%s\n",line1,line2,line3);

一个小技巧。使用这个,我在动态创建的HTML页面中注入javascript

StringBuilder builder = new StringBuilder();

public String getString()
{
    return builder.toString();
}
private DropdownContent _(String a)
{
    builder.append(a);
    return this;
}

public String funct_showhide()
{
   return
    _("function slidedown_showHide(boxId)").
    _("{").
    _("if(!slidedown_direction[boxId])slidedown_direction[boxId] = 1;").
    _("if(!slideDownInitHeight[boxId])slideDownInitHeight[boxId] = 0;").
    _("if(slideDownInitHeight[boxId]==0)slidedown_direction[boxId]=slidedownSpeed; ").
    _("else slidedown_direction[boxId] = slidedownSpeed*-1;").
    _("slidedownContentBox = document.getElementById(boxId);").
    _("var subDivs = slidedownContentBox.getElementsByTagName('DIV');").
    _("for(var no=0;no<subDivs.length;no++){").
    _(" if(subDivs[no].className=='dhtmlgoodies_content')slidedownContent = subDivs[no];").
    _("}").
    _("contentHeight = slidedownContent.offsetHeight;").
    _("slidedownContentBox.style.visibility='visible';").
    _("slidedownActive = true;").
    _("slidedown_showHide_start(slidedownContentBox,slidedownContent);").
    _("}").getString();

}

使用JDK/12早期访问构建# 12,现在可以在Java中使用多行字符串,如下所示:

String multiLine = `First line
    Second line with indentation
Third line
and so on...`; // the formatting as desired
System.out.println(multiLine);

这将导致以下输出:

第一行 第二行有缩进 第三行 等等……

编辑: 推迟至java 13

你可以在一个单独的方法中连接你的追加:

public static String multilineString(String... lines){
   StringBuilder sb = new StringBuilder();
   for(String s : lines){
     sb.append(s);
     sb.append ('\n');
   }
   return sb.toString();
}

无论哪种方式,都更喜欢StringBuilder而不是加号符号。

Java 13预览:

Text Blocks Come to Java. Java 13 delivers long-awaited multiline string by Mala Gupta With text blocks, Java 13 is making it easier for you to work with multiline string literals. You no longer need to escape the special characters in string literals or use concatenation operators for values that span multiple lines. Text block is defined using three double quotes (""") as the opening and closing delimiters. The opening delimiter can be followed by zero or more white spaces and a line terminator.

例子:

 String s1 = """
 text
 text
 text
 """;