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
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
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();
}
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();
}
}
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
);
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