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)?
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
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} ${1}.";
Map<String, Object> values = new HashMap<String, Object>();
values.put("0", "James");
values.put("1", "Bond");
System.out.println(format(format, values)); // My name is Bond. James Bond.
其他回答
我的答案是:
a)尽可能使用StringBuilder
b)保持“占位符”的位置(以任何形式:整数是最好的,特殊字符如dollar宏等),然后使用StringBuilder.insert()(参数的几个版本)。
当StringBuilder内部转换为String时,使用外部库似乎有些过度,而且我认为会显著降低性能。
基于这个答案,我创建了MapBuilder类:
public class MapBuilder {
public static Map<String, Object> build(Object... data) {
Map<String, Object> result = new LinkedHashMap<>();
if (data.length % 2 != 0) {
throw new IllegalArgumentException("Odd number of arguments");
}
String key = null;
Integer step = -1;
for (Object value : data) {
step++;
switch (step % 2) {
case 0:
if (value == null) {
throw new IllegalArgumentException("Null key value");
}
key = (String) value;
continue;
case 1:
result.put(key, value);
break;
}
}
return result;
}
}
然后我创建类StringFormat用于字符串格式化:
public final class StringFormat {
public static String format(String format, Object... args) {
Map<String, Object> values = MapBuilder.build(args);
for (Map.Entry<String, Object> entry : values.entrySet()) {
String key = entry.getKey();
Object value = entry.getValue();
format = format.replace("$" + key, value.toString());
}
return format;
}
}
你可以这样用:
String bookingDate = StringFormat.format("From $startDate to $endDate"),
"$startDate", formattedStartDate,
"$endDate", formattedEndDate
);
有Java插件使用字符串插值在Java(像在Kotlin, JavaScript)。支持Java 8,9,10,11…https://github.com/antkorwin/better-strings
在字符串字面量中使用变量:
int a = 3;
int b = 4;
System.out.println("${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。
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.
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)
;