一个简单的选择是使用SciTE (http://www.scintilla.org/SciTEDownload.html)这样的编辑器编辑java代码，它允许您对文本进行WRAP，以便容易地查看和编辑长字符串。如果你需要转义字符，你只需输入它们。通过关闭“换行”选项，您可以检查字符串是否仍然是一个很长的单行字符串。当然，如果不是，编译器也会告诉你。

Eclipse或NetBeans是否支持编辑器中的文本包装，我不知道，因为它们有太多的选项。但如果没有，这将是一个很好的补充。

我还没有看到的另一个答案是java.io.PrintWriter。

StringWriter stringWriter = new StringWriter();
PrintWriter writer = new PrintWriter(stringWriter);
writer.println("It was the best of times, it was the worst of times");
writer.println("it was the age of wisdom, it was the age of foolishness,");
writer.println("it was the epoch of belief, it was the epoch of incredulity,");
writer.println("it was the season of Light, it was the season of Darkness,");
writer.println("it was the spring of hope, it was the winter of despair,");
writer.println("we had everything before us, we had nothing before us");
String string = stringWriter.toString();

此外，java.io.BufferedWriter有一个newLine()方法的事实也未被提及。

我所知道的唯一方法是用加号连接多行

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。

我建议使用ThomasP所建议的实用程序，然后将其链接到构建过程中。仍然存在一个外部文件以包含文本，但在运行时不读取该文件。 工作流程如下:

构建一个“文本文件到java代码”实用程序并检查到版本控制 在每次构建时，针对资源文件运行实用程序以创建修改后的java源 Java源代码包含类似于 类TextBlock{…后面跟着一个从资源文件自动生成的静态字符串 用剩下的代码构建生成的java文件

使用这个库

https://github.com/alessio-santacroce/multiline-string-literals

可以这样写

System.out.println(newString(/*
      Wow, we finally have
      multiline strings in
      Java! HOOO!
*/));

很好很简单，但是只适用于单元测试