如何在 Java 中格式化字符串

How to format strings in Java

提问by katit

Primitive question, but how do I format strings like this:

原始问题，但我如何格式化这样的字符串：

"Step {1} of {2}"

“{2} 的第 {1} 步”

by substituting variables using Java? In C# it's easy.

通过使用Java替换变量？在 C# 中很容易。

采纳答案by ataylor

In addition to String.format, also take a look java.text.MessageFormat. The format less terse and a bit closer to the C# example you've provided and you can use it for parsing as well.

除了String.format，也看看java.text.MessageFormat。格式更简洁，更接近您提供的 C# 示例，您也可以将其用于解析。

For example:

例如：

int someNumber = 42;
String someString = "foobar";
Object[] args = {new Long(someNumber), someString};
MessageFormat fmt = new MessageFormat("String is \"{1}\", number is {0}.");
System.out.println(fmt.format(args));

A nicer example takes advantage of the varargs and autoboxing improvements in Java 1.5 and turns the above into a one-liner:

一个更好的例子利用了 Java 1.5 中的可变参数和自动装箱改进，并将上面的内容变成了一个单行：

MessageFormat.format("String is \"{1}\", number is {0}.", 42, "foobar");

MessageFormatis a little bit nicer for doing i18nized plurals with the choice modifier. To specify a message that correctly uses the singular form when a variable is 1 and plural otherwise, you can do something like this:

MessageFormat用选择修饰符做 i18nized 复数会更好一点。要指定在变量为 1 时正确使用单数形式的消息，否则可以执行以下操作：

String formatString = "there were {0} {0,choice,0#objects|1#object|1<objects}";
MessageFormat fmt = new MessageFormat(formatString);
fmt.format(new Object[] { new Long(numberOfObjects) });

回答by Martin T?rnwall

Take a look at String.format. Note, however, that it takes format specifiers similar to those of C's printf family of functions -- for example:

看看String.format。但是请注意，它采用类似于 C 的 printf 系列函数的格式说明符——例如：

String.format("Hello %s, %d", "world", 42);

Would return "Hello world, 42". You may find thishelpful when learning about the format specifiers. Andy Thomas-Cramer was kind enough to leave thislink in a comment below, which appears to point to the official spec. The most commonly used ones are:

将返回“Hello world, 42”。在了解格式说明符时，您可能会发现这很有帮助。安迪·托马斯-克莱默 (Andy Thomas-Cramer)在下面的评论中留下了这个链接，这似乎指向了官方规范。最常用的有：

• %s - insert a string
• %d - insert a signed integer (decimal)
• %f - insert a real number, standard notation

• %s - 插入一个字符串
• %d - 插入一个有符号整数（十进制）
• %f - 插入一个实数，标准符号

This is radically different from C#, which uses positional references with an optional format specifier. That means that you can't do things like:

这与 C# 完全不同，C# 使用带有可选格式说明符的位置引用。这意味着您不能执行以下操作：

String.format("The {0} is repeated again: {0}", "word");

... without actually repeating the parameter passed to printf/format.(see The Scrum Meister's comment below)

...实际上没有重复传递给 printf/format 的参数。（请参阅下面的 Scrum 大师的评论）

---

---

If you just want to print the result directly, you may find System.out.printf (PrintStream.printf) to your liking.

如果您只想直接打印结果，您可以根据自己的喜好找到 System.out.printf ( PrintStream.printf)。

回答by Ryan Amos

If you choose not to use String.format, the other option is the + binary operator

如果您选择不使用 String.format，另一个选项是 + 二元运算符

String str = "Step " + a + " of " + b;

This is the equivalent of

这相当于

new StringBuilder("Step ").append(String.valueOf(1)).append(" of ").append(String.valueOf(2));

new StringBuilder("Step ").append(String.valueOf(1)).append(" of ").append(String.valueOf(2));

Whichever you use is your choice. StringBuilder is faster, but the speed difference is marginal. I prefer to use the +operator (which does a StringBuilder.append(String.valueOf(X)))and find it easier to read.

无论您使用哪种，都是您的选择。StringBuilder 更快，但速度差异很小。我更喜欢使用+运算符（它执行 aStringBuilder.append(String.valueOf(X)))并发现它更易于阅读。

回答by Ved Prakash

public class StringFormat {

    public static void main(String[] args) {
            Scanner sc=new Scanner(System.in);
            System.out.println("================================");
            for(int i=0;i<3;i++){
                String s1=sc.next();
                int x=sc.nextInt();
                System.out.println(String.format("%-15s%03d",s1,x));
            }
            System.out.println("================================");

    }
}

outpot "================================"
ved15space123 ved15space123 ved15space123 "================================

输出点“================================”
ved15space123 ved15space123 ved15space123 “============ ======================

Java solution

Java解决方案

• The "-" is used to left indent

• The "15" makes the String's minimum length it takes up be 15

• The "s" (which is a few characters after %) will be substituted by our String
• The 0 pads our integer with 0s on the left
• The 3 makes our integer be a minimum length of 3

• “-”用于左缩进

• “15”使字符串的最小长度为 15

• “s”（% 后的几个字符）将被我们的字符串替换
• 0 在左边用 0 填充我们的整数
• 3 使我们的整数最小长度为 3

回答by Tomasz Modelski

I've wrote my simple method for it :

我已经为它写了我的简单方法：

public class SomeCommons {
    /** Message Format like 'Some String {0} / {1}' with arguments */
    public static String msgFormat(String s, Object... args) {
        return new MessageFormat(s).format(args);
    }
}

so you can use it as:

所以你可以将它用作：

SomeCommons.msfgFormat("Step {1} of {2}", 1 , "two");

回答by Felipe Bejarano

This solution worked for me. I needed to create urls for a REST client dynamically so I created this method, so you just have to pass the restURL like this

这个解决方案对我有用。我需要为 REST 客户端动态创建 url，所以我创建了这个方法，所以你只需要像这样传递 restURL

/customer/{0}/user/{1}/order

and add as many params as you need:

并根据需要添加尽可能多的参数：

public String createURL (String restURL, Object ... params) {       
    return new MessageFormat(restURL).format(params);
}

You just have to call this method like this:

你只需要像这样调用这个方法：

createURL("/customer/{0}/user/{1}/order", 123, 321);

The output

输出

"/customer/123/user/321/order"

“/customer/123/user/321/order”

回答by DefconT34

I wrote this function it does just the right thing. Interpolate a word starting with $with the value of the variable of the same name.

我写了这个函数，它做了正确的事情。插入一个$以同名变量的值开头的单词。

private static String interpol1(String x){
    Field[] ffield =  Main.class.getDeclaredFields();
    String[] test = x.split(" ") ;
    for (String v : test ) {
        for ( Field n: ffield ) {
            if(v.startsWith("$") && ( n.getName().equals(v.substring(1))  )){
                try {
                    x = x.replace("$" + v.substring(1), String.valueOf( n.get(null)));
                }catch (Exception e){
                    System.out.println("");
                }
            }
        }
    }
    return x;
}