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
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));
其他回答
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) + "$";
int index = -1;
while ((index = convFormat.indexOf(formatKey, index)) != -1) {
convFormat.replace(index, index + formatKey.length(), formatPos);
index += formatPos.length();
}
valueList.add(values.get(key));
++currentPos;
}
return String.format(convFormat.toString(), valueList.toArray());
}
基于这个答案,我创建了MapBuilder类:
public class MapBuilder {
public static Map<String, Object> build(Object... data) {
Map<String, Object> result = new LinkedHashMap<>();
if (data.length % 2 != 0) {
throw new IllegalArgumentException("Odd number of arguments");
}
String key = null;
Integer step = -1;
for (Object value : data) {
step++;
switch (step % 2) {
case 0:
if (value == null) {
throw new IllegalArgumentException("Null key value");
}
key = (String) value;
continue;
case 1:
result.put(key, value);
break;
}
}
return result;
}
}
然后我创建类StringFormat用于字符串格式化:
public final class StringFormat {
public static String format(String format, Object... args) {
Map<String, Object> values = MapBuilder.build(args);
for (Map.Entry<String, Object> entry : values.entrySet()) {
String key = entry.getKey();
Object value = entry.getValue();
format = format.replace("$" + 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。
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.
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();
}