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

另一种方法是将长字符串存储在外部文件中,然后将文件读入字符串。

其他回答

从问题中还不完全清楚作者是否有兴趣处理某种需要有一些动态值的格式化大字符串,但如果是这种情况,像StringTemplate (http://www.stringtemplate.org/)这样的模板引擎可能非常有用。

下面是一个使用StringTemplate的简单代码示例。实际的模板("Hello, < name >")可以从外部纯文本文件加载。模板中的所有缩进都将被保留,不需要转义。

import org.stringtemplate.v4.*;
 
public class Hello {
    public static void main(String[] args) {
        ST hello = new ST("Hello, <name>");
        hello.add("name", "World");
        System.out.println(hello.render());
    }
}

附注:为了可读性和本地化目的,从源代码中删除大块文本总是一个好主意。

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();
}

看到所有的答案,我认为没有人提到比在java的新版本,你可以这样做:

String s = """
    This
    is 
    a
    multiline
    string
    """;
System.out.println(s);

这是它打印的内容:

This
is
a
multiline
string

Java 13及以上版本

Java中现在通过文本块支持多行字符串。在Java 13和14中,这个特性要求你在构建和运行项目时设置——enable-preview选项。在Java 15及以后版本中,由于文本块已成为标准特性,因此不再需要此选项。查看官方的文本块编程指南,了解更多细节。

现在,在Java 13之前,你是这样写查询的:

List<Tuple> posts = entityManager
.createNativeQuery(
    "SELECT *\n" +
    "FROM (\n" +
    "    SELECT *,\n" +
    "           dense_rank() OVER (\n" +
    "               ORDER BY \"p.created_on\", \"p.id\"\n" +
    "           ) rank\n" +
    "    FROM (\n" +
    "        SELECT p.id AS \"p.id\",\n" +
    "               p.created_on AS \"p.created_on\",\n" +
    "               p.title AS \"p.title\",\n" +
    "               pc.id as \"pc.id\",\n" +
    "               pc.created_on AS \"pc.created_on\",\n" +
    "               pc.review AS \"pc.review\",\n" +
    "               pc.post_id AS \"pc.post_id\"\n" +
    "        FROM post p\n" +
    "        LEFT JOIN post_comment pc ON p.id = pc.post_id\n" +
    "        WHERE p.title LIKE :titlePattern\n" +
    "        ORDER BY p.created_on\n" +
    "    ) p_pc\n" +
    ") p_pc_r\n" +
    "WHERE p_pc_r.rank <= :rank\n",
    Tuple.class)
.setParameter("titlePattern", "High-Performance Java Persistence %")
.setParameter("rank", 5)
.getResultList();

多亏了Java 13 Text Blocks,你可以像下面这样重写这个查询:

List<Tuple> posts = entityManager
.createNativeQuery("""
    SELECT *
    FROM (
        SELECT *,
               dense_rank() OVER (
                   ORDER BY "p.created_on", "p.id"
               ) rank
        FROM (
            SELECT p.id AS "p.id",
                   p.created_on AS "p.created_on",
                   p.title AS "p.title",
                   pc.id as "pc.id",
                   pc.created_on AS "pc.created_on",
                   pc.review AS "pc.review",
                   pc.post_id AS "pc.post_id"
            FROM post p
            LEFT JOIN post_comment pc ON p.id = pc.post_id
            WHERE p.title LIKE :titlePattern
            ORDER BY p.created_on
        ) p_pc
    ) p_pc_r
    WHERE p_pc_r.rank <= :rank
    """,
    Tuple.class)
.setParameter("titlePattern", "High-Performance Java Persistence %")
.setParameter("rank", 5)
.getResultList();

可读性强多了,对吧?

支持的想法

IntelliJ IDEA提供了将传统字符串连接块转换为新的多行字符串格式的支持:

Json, html, XML

多行String在编写JSON、HTML或XML时特别有用。

考虑下面的例子,使用String连接来构建一个JSON字符串文字:

entityManager.persist(
    new Book()
    .setId(1L)
    .setIsbn("978-9730228236")
    .setProperties(
        "{" +
        "   \"title\": \"High-Performance Java Persistence\"," +
        "   \"author\": \"Vlad Mihalcea\"," +
        "   \"publisher\": \"Amazon\"," +
        "   \"price\": 44.99," +
        "   \"reviews\": [" +
        "       {" +
        "           \"reviewer\": \"Cristiano\", " +
        "           \"review\": \"Excellent book to understand Java Persistence\", " +
        "           \"date\": \"2017-11-14\", " +
        "           \"rating\": 5" +
        "       }," +
        "       {" +
        "           \"reviewer\": \"T.W\", " +
        "           \"review\": \"The best JPA ORM book out there\", " +
        "           \"date\": \"2019-01-27\", " +
        "           \"rating\": 5" +
        "       }," +
        "       {" +
        "           \"reviewer\": \"Shaikh\", " +
        "           \"review\": \"The most informative book\", " +
        "           \"date\": \"2016-12-24\", " +
        "           \"rating\": 4" +
        "       }" +
        "   ]" +
        "}"
    )
);

由于转义字符和大量的双引号和加号,您几乎无法阅读JSON。

使用Java文本块,JSON对象可以这样写:

entityManager.persist(
    new Book()
    .setId(1L)
    .setIsbn("978-9730228236")
    .setProperties("""
        {
           "title": "High-Performance Java Persistence",
           "author": "Vlad Mihalcea",
           "publisher": "Amazon",
           "price": 44.99,
           "reviews": [
               {
                   "reviewer": "Cristiano",
                   "review": "Excellent book to understand Java Persistence",
                   "date": "2017-11-14",
                   "rating": 5
               },
               {
                   "reviewer": "T.W",
                   "review": "The best JPA ORM book out there",
                   "date": "2019-01-27",
                   "rating": 5
               },
               {
                   "reviewer": "Shaikh",
                   "review": "The most informative book",
                   "date": "2016-12-24",
                   "rating": 4
               }
           ]
        }
        """
    )
);

自从我在2004年使用c#以来,我一直希望在Java中有这个功能,现在我们终于有了它。

如果你像我一样喜欢谷歌的番石榴,它可以提供一个相当干净的表示和一个很好的,简单的方法来不硬编码你的换行符:

String out = Joiner.on(newline).join(ImmutableList.of(
    "line1",
    "line2",
    "line3"));