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
不幸的是,答案是否定的。然而,你可以非常接近一个合理的语法:
"""
You are $compliment!
"""
.replace('$compliment', 'awesome');
它比String更具可读性和可预测性。至少是格式!
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) + "$";
int index = -1;
while ((index = convFormat.indexOf(formatKey, index)) != -1) {
convFormat.replace(index, index + formatKey.length(), formatPos);
index += formatPos.length();
}
valueList.add(values.get(key));
++currentPos;
}
return String.format(convFormat.toString(), valueList.toArray());
}
这是一个旧的线程,但只是为了记录,你也可以使用Java 8风格,像这样:
public static String replaceParams(Map<String, String> hashMap, String template) {
return hashMap.entrySet().stream().reduce(template, (s, e) -> s.replace("%(" + e.getKey() + ")", e.getValue()),
(s, s2) -> s);
}
用法:
public static void main(String[] args) {
final HashMap<String, String> hashMap = new HashMap<String, String>() {
{
put("foo", "foo1");
put("bar", "bar1");
put("car", "BMW");
put("truck", "MAN");
}
};
String res = replaceParams(hashMap, "This is '%(foo)' and '%(foo)', but also '%(bar)' '%(bar)' indeed.");
System.out.println(res);
System.out.println(replaceParams(hashMap, "This is '%(car)' and '%(foo)', but also '%(bar)' '%(bar)' indeed."));
System.out.println(replaceParams(hashMap, "This is '%(car)' and '%(truck)', but also '%(foo)' '%(bar)' + '%(truck)' indeed."));
}
输出将是:
This is 'foo1' and 'foo1', but also 'bar1' 'bar1' indeed.
This is 'BMW' and 'foo1', but also 'bar1' 'bar1' indeed.
This is 'BMW' and 'MAN', but also 'foo1' 'bar1' + 'MAN' indeed.
public static String format(String format, Map<String, Object> values) {
StringBuilder formatter = new StringBuilder(format);
List<Object> valueList = new ArrayList<Object>();
Matcher matcher = Pattern.compile("\\$\\{(\\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.
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();
}
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)
;