基于这个答案，我创建了MapBuilder类:

public class MapBuilder {

    public static Map<String, Object> build(Object... data) {
        Map<String, Object> result = new LinkedHashMap<>();

        if (data.length % 2 != 0) {
            throw new IllegalArgumentException("Odd number of arguments");
        }

        String key = null;
        Integer step = -1;

        for (Object value : data) {
            step++;
            switch (step % 2) {
                case 0:
                    if (value == null) {
                        throw new IllegalArgumentException("Null key value");
                    }
                    key = (String) value;
                    continue;
                case 1:
                    result.put(key, value);
                    break;
            }
        }

        return result;
    }

}

然后我创建类StringFormat用于字符串格式化:

public final class StringFormat {

    public static String format(String format, Object... args) {
        Map<String, Object> values = MapBuilder.build(args);

        for (Map.Entry<String, Object> entry : values.entrySet()) {
            String key = entry.getKey();
            Object value = entry.getValue();
            format = format.replace("$" + key, value.toString());
        }

        return format;
    }

}

你可以这样用:

String bookingDate = StringFormat.format("From $startDate to $endDate"), 
        "$startDate", formattedStartDate, 
        "$endDate", formattedEndDate
);

谢谢你的帮助!使用所有的线索，我写了一个例程来做我想要的——使用字典的类似python的字符串格式化。因为我是Java新手，任何提示都是感激的。

public static String dictFormat(String format, Hashtable<String, Object> values) {
    StringBuilder convFormat = new StringBuilder(format);
    Enumeration<String> keys = values.keys();
    ArrayList valueList = new ArrayList();
    int currentPos = 1;
    while (keys.hasMoreElements()) {
        String key = keys.nextElement(),
        formatKey = "%(" + key + ")",
        formatPos = "%" + Integer.toString(currentPos) + "$";
        int index = -1;
        while ((index = convFormat.indexOf(formatKey, index)) != -1) {
            convFormat.replace(index, index + formatKey.length(), formatPos);
            index += formatPos.length();
        }
        valueList.add(values.get(key));
        ++currentPos;
    }
    return String.format(convFormat.toString(), valueList.toArray());
}

另一个Apache Common StringSubstitutor的简单命名占位符的例子。

String template = "Welcome to {theWorld}. My name is {myName}.";

Map<String, String> values = new HashMap<>();
values.put("theWorld", "Stackoverflow");
values.put("myName", "Thanos");

String message = StringSubstitutor.replace(template, values, "{", "}");

System.out.println(message);

// Welcome to Stackoverflow. My name is Thanos.

public static String format(String format, Map<String, Object> values) {
    StringBuilder formatter = new StringBuilder(format);
    List<Object> valueList = new ArrayList<Object>();

    Matcher matcher = Pattern.compile("\\$\\{(\\w+)}").matcher(format);

    while (matcher.find()) {
        String key = matcher.group(1);

        String formatKey = String.format("${%s}", key);
        int index = formatter.indexOf(formatKey);

        if (index != -1) {
            formatter.replace(index, index + formatKey.length(), "%s");
            valueList.add(values.get(key));
        }
    }

    return String.format(formatter.toString(), valueList.toArray());
}

例子:

String format = "My name is ${1}. ${0} ${1}.";

Map<String, Object> values = new HashMap<String, Object>();
values.put("0", "James");
values.put("1", "Bond");

System.out.println(format(format, values)); // My name is Bond. James Bond.

我的答案是:

a)尽可能使用StringBuilder

b)保持“占位符”的位置(以任何形式:整数是最好的，特殊字符如dollar宏等)，然后使用StringBuilder.insert()(参数的几个版本)。

当StringBuilder内部转换为String时，使用外部库似乎有些过度，而且我认为会显著降低性能。

有Java插件使用字符串插值在Java(像在Kotlin, JavaScript)。支持Java 8,9,10,11…https://github.com/antkorwin/better-strings

在字符串字面量中使用变量:

int a = 3;
int b = 4;
System.out.println("${a} + ${b} = ${a+b}");

使用表达式:

int a = 3;
int b = 4;
System.out.println("pow = ${a * a}");
System.out.println("flag = ${a > b ? true : false}");

使用功能:

@Test
void functionCall() {
    System.out.println("fact(5) = ${factorial(5)}");
}

long factorial(int n) {
    long fact = 1;
    for (int i = 2; i <= n; i++) {
        fact = fact * i;
    }
    return fact;
}

更多信息，请阅读项目README。