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

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;
}

其他回答

试试Freemarker,模板库。

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.

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。

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。

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();
}