你可以使用StringTemplate库，它提供了你想要的和更多。

import org.antlr.stringtemplate.*;

final StringTemplate hello = new StringTemplate("Hello, $name$");
hello.setAttribute("name", "World");
System.out.println(hello.toString());

你可以在字符串助手类上有这样的东西

/**
 * An interpreter for strings with named placeholders.
 *
 * For example given the string "hello %(myName)" and the map <code>
 *      <p>Map<String, Object> map = new HashMap<String, Object>();</p>
 *      <p>map.put("myName", "world");</p>
 * </code>
 *
 * the call {@code format("hello %(myName)", map)} returns "hello world"
 *
 * It replaces every occurrence of a named placeholder with its given value
 * in the map. If there is a named place holder which is not found in the
 * map then the string will retain that placeholder. Likewise, if there is
 * an entry in the map that does not have its respective placeholder, it is
 * ignored.
 *
 * @param str
 *            string to format
 * @param values
 *            to replace
 * @return formatted string
 */
public static String format(String str, Map<String, Object> values) {

    StringBuilder builder = new StringBuilder(str);

    for (Entry<String, Object> entry : values.entrySet()) {

        int start;
        String pattern = "%(" + entry.getKey() + ")";
        String value = entry.getValue().toString();

        // Replace every occurence of %(key) with value
        while ((start = builder.indexOf(pattern)) != -1) {
            builder.replace(start, start + pattern.length(), value);
        }
    }

    return builder.toString();
}

https://dzone.com/articles/java-string-format-examples字符串。format(inputString， [listOfParams])将是最简单的方法。字符串中的占位符可以按顺序定义。欲了解更多详细信息，请查看提供的链接。

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());
}

你可以使用StringTemplate库，它提供了你想要的和更多。

import org.antlr.stringtemplate.*;

final StringTemplate hello = new StringTemplate("Hello, $name$");
hello.setAttribute("name", "World");
System.out.println(hello.toString());

不幸的是，答案是否定的。然而，你可以非常接近一个合理的语法:

"""
   You are $compliment!
"""
.replace('$compliment', 'awesome');

它比String更具可读性和可预测性。至少是格式!