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

这个问题有两个答案:

In you want to stick to pure Java, with Java 14 being released in March 2020, you can leverage the JEP 368 - Text Blocks, in Second Preview mode. Actually the feature is in preview mode in other releases (at least 13 has it). I created and example set here. While this feature is useful, it can be easily abused. Remember that Java requires compilation - having large character arrays in your code can be an easy way to shoot yourself in the leg (if you want a quick change, you will need recompilation - that toolset might not be available to the guy operating your application).

根据我的经验,建议在配置文件中保留大字符串(通常是应用程序操作员可以/应该在运行时更改的字符串)。

总结:负责地使用文本块:)。

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);

这是你不应该在不考虑它在做什么的情况下使用的东西。但对于一次性脚本,我已经成功地使用了这个方法:

例子:

    System.out.println(S(/*
This is a CRAZY " ' ' " multiline string with all sorts of strange 
   characters!
*/));

代码:

// From: http://blog.efftinge.de/2008/10/multi-line-string-literals-in-java.html
// Takes a comment (/**/) and turns everything inside the comment to a string that is returned from S()
public static String S() {
    StackTraceElement element = new RuntimeException().getStackTrace()[1];
    String name = element.getClassName().replace('.', '/') + ".java";
    StringBuilder sb = new StringBuilder();
    String line = null;
    InputStream in = classLoader.getResourceAsStream(name);
    String s = convertStreamToString(in, element.getLineNumber());
    return s.substring(s.indexOf("/*")+2, s.indexOf("*/"));
}

// From http://www.kodejava.org/examples/266.html
private static String convertStreamToString(InputStream is, int lineNum) {
    /*
     * To convert the InputStream to String we use the BufferedReader.readLine()
     * method. We iterate until the BufferedReader return null which means
     * there's no more data to read. Each line will appended to a StringBuilder
     * and returned as String.
     */
    BufferedReader reader = new BufferedReader(new InputStreamReader(is));
    StringBuilder sb = new StringBuilder();

    String line = null; int i = 1;
    try {
        while ((line = reader.readLine()) != null) {
            if (i++ >= lineNum) {
                sb.append(line + "\n");
            }
        }
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        try {
            is.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    return sb.toString();
}

由于Java(还)不支持多行字符串,目前唯一的方法是使用前面提到的技术之一来破解它。我使用上面提到的一些技巧构建了下面的Python脚本:

import sys
import string
import os

print 'new String('
for line in sys.stdin:
    one = string.replace(line, '"', '\\"').rstrip(os.linesep)
    print '  + "' + one + ' "'
print ')'

把它放在一个名为javastringify.py的文件中,把你的字符串放在一个名为mystring.txt的文件中,然后像下面这样运行:

cat mystring.txt | python javastringify.py

然后,您可以复制输出并将其粘贴到编辑器中。

修改这需要处理任何特殊情况,但这是为我的需要。希望这能有所帮助!

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
 """;

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。