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
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。
其他回答
我最终得到了下一个解决方案: 使用substitute()方法创建类templatessubstitute,并使用它格式化输出 然后创建一个字符串模板,并用值填充它
import java.util.*;
public class MyClass {
public static void main(String args[]) {
String template = "WRR = {WRR}, SRR = {SRR}\n" +
"char_F1 = {char_F1}, word_F1 = {word_F1}\n";
Map<String, Object> values = new HashMap<>();
values.put("WRR", 99.9);
values.put("SRR", 99.8);
values.put("char_F1", 80);
values.put("word_F1", 70);
String message = TemplateSubstitutor.substitute(values, template);
System.out.println(message);
}
}
class TemplateSubstitutor {
public static String substitute(Map<String, Object> map, String input_str) {
String output_str = input_str;
for (Map.Entry<String, Object> entry : map.entrySet()) {
String key = entry.getKey();
Object value = entry.getValue();
output_str = output_str.replace("{" + key + "}", String.valueOf(value));
}
return output_str;
}
}
jakarta commons lang的StrSubstitutor是一种轻量级的实现方法,前提是您的值已经被正确格式化。
http://commons.apache.org/proper/commons-lang/javadocs/api-3.1/org/apache/commons/lang3/text/StrSubstitutor.html
Map<String, String> values = new HashMap<String, String>();
values.put("value", x);
values.put("column", y);
StrSubstitutor sub = new StrSubstitutor(values, "%(", ")");
String result = sub.replace("There's an incorrect value '%(value)' in column # %(column)");
上述结果为:
“第2列中的“1”值不正确”
当使用Maven时,您可以将此依赖项添加到pom.xml:
<dependency>
<groupId>org.apache.commons</groupId>
<artifactId>commons-lang3</artifactId>
<version>3.4</version>
</dependency>
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;
}
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();
}
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