jakarta commons lang的StrSubstitutor是一种轻量级的实现方法，前提是您的值已经被正确格式化。

http://commons.apache.org/proper/commons-lang/javadocs/api-3.1/org/apache/commons/lang3/text/StrSubstitutor.html

Map<String, String> values = new HashMap<String, String>();
values.put("value", x);
values.put("column", y);
StrSubstitutor sub = new StrSubstitutor(values, "%(", ")");
String result = sub.replace("There's an incorrect value '%(value)' in column # %(column)");

上述结果为:

“第2列中的“1”值不正确”

当使用Maven时，您可以将此依赖项添加到pom.xml:

<dependency>
    <groupId>org.apache.commons</groupId>
    <artifactId>commons-lang3</artifactId>
    <version>3.4</version>
</dependency>

这是一个旧的线程，但只是为了记录，你也可以使用Java 8风格，像这样:

public static String replaceParams(Map<String, String> hashMap, String template) {
    return hashMap.entrySet().stream().reduce(template, (s, e) -> s.replace("%(" + e.getKey() + ")", e.getValue()),
            (s, s2) -> s);
}

用法:

public static void main(String[] args) {
    final HashMap<String, String> hashMap = new HashMap<String, String>() {
        {
            put("foo", "foo1");
            put("bar", "bar1");
            put("car", "BMW");
            put("truck", "MAN");
        }
    };
    String res = replaceParams(hashMap, "This is '%(foo)' and '%(foo)', but also '%(bar)' '%(bar)' indeed.");
    System.out.println(res);
    System.out.println(replaceParams(hashMap, "This is '%(car)' and '%(foo)', but also '%(bar)' '%(bar)' indeed."));
    System.out.println(replaceParams(hashMap, "This is '%(car)' and '%(truck)', but also '%(foo)' '%(bar)' + '%(truck)' indeed."));
}

输出将是:

This is 'foo1' and 'foo1', but also 'bar1' 'bar1' indeed.
This is 'BMW' and 'foo1', but also 'bar1' 'bar1' indeed.
This is 'BMW' and 'MAN', but also 'foo1' 'bar1' + 'MAN' indeed.

不完全是，但你可以使用MessageFormat多次引用一个值:

MessageFormat.format("There's an incorrect value \"{0}\" in column # {1}", x, y);

上面的事情也可以用string .format()来完成，但是如果你需要构建复杂的表达式，我发现messageFormat语法更干净，而且你不需要关心你放入字符串中的对象的类型

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());
}

我试了一下

public static void main(String[] args) 
{
    String rowString = "replace the value ${var1} with ${var2}";
    
    Map<String,String> mappedValues = new HashMap<>();
    
    mappedValues.put("var1", "Value 1");
    mappedValues.put("var2", "Value 2");
    
    System.out.println(replaceOccurence(rowString, mappedValues));
}

private static  String replaceOccurence(String baseStr ,Map<String,String> mappedValues)
{
    for(String key :mappedValues.keySet())
    {
        baseStr = baseStr.replace("${"+key+"}", mappedValues.get(key));
    }
    
    return baseStr;
}

试试Freemarker，模板库。