在Python中,格式化字符串时,我可以按名称而不是按位置填充占位符,如下所示:

print "There's an incorrect value '%(value)s' in column # %(column)d" % \
  { 'value': x, 'column': y }

我想知道这在Java中是否可能(希望没有外部库)?


当前回答

在编写本文时,Java中还没有内置任何东西。我建议编写自己的实现。我的偏好是一个简单流畅的构建器接口,而不是创建一个映射并将其传递给函数——你最终会得到一个漂亮的连续代码块,例如:

String result = new TemplatedStringBuilder("My name is {{name}} and I from {{town}}")
   .replace("name", "John Doe")
   .replace("town", "Sydney")
   .finish();

下面是一个简单的实现:

class TemplatedStringBuilder {

    private final static String TEMPLATE_START_TOKEN = "{{";
    private final static String TEMPLATE_CLOSE_TOKEN = "}}";

    private final String template;
    private final Map<String, String> parameters = new HashMap<>();

    public TemplatedStringBuilder(String template) {
        if (template == null) throw new NullPointerException();
        this.template = template;
    }

    public TemplatedStringBuilder replace(String key, String value){
        parameters.put(key, value);
        return this;
    }

    public String finish(){

        StringBuilder result = new StringBuilder();

        int startIndex = 0;

        while (startIndex < template.length()){

            int openIndex  = template.indexOf(TEMPLATE_START_TOKEN, startIndex);

            if (openIndex < 0){
                result.append(template.substring(startIndex));
                break;
            }

            int closeIndex = template.indexOf(TEMPLATE_CLOSE_TOKEN, openIndex);

            if(closeIndex < 0){
                result.append(template.substring(startIndex));
                break;
            }

            String key = template.substring(openIndex + TEMPLATE_START_TOKEN.length(), closeIndex);

            if (!parameters.containsKey(key)) throw new RuntimeException("missing value for key: " + key);

            result.append(template.substring(startIndex, openIndex));
            result.append(parameters.get(key));

            startIndex = closeIndex + TEMPLATE_CLOSE_TOKEN.length();
        }

        return result.toString();
    }
}

其他回答

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.

您应该看看官方的ICU4J库。它提供了一个类似于JDK的MessageFormat类,但前者支持命名占位符。

与本页提供的其他解决方案不同。ICU4j是ICU项目的一部分,由IBM维护并定期更新。此外,它还支持高级用例,如多元化等。

下面是一个代码示例:

MessageFormat messageFormat =
        new MessageFormat("Publication written by {author}.");

Map<String, String> args = Map.of("author", "John Doe");

System.out.println(messageFormat.format(args));

截至2022年,最新的解决方案是Apache Commons Text StringSubstitutor

医生说:

// Build map
 Map<String, String> valuesMap = new HashMap<>();
 valuesMap.put("animal", "quick brown fox");
 valuesMap.put("target", "lazy dog");
 String templateString = "The ${animal} jumped over the ${target} ${undefined.number:-1234567890} times.";

 // Build StringSubstitutor
 StringSubstitutor sub = new StringSubstitutor(valuesMap);

 // Replace
 String resolvedString = sub.replace(templateString)

;

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

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;
}