ReplaceAll with java8 lambda functions

Question

Given the following variables

templateText = "Hi ${name}";
variables.put("name", "Joe");

I would like to replace the placeholder ${name} with the value "Joe" using the following code (that does not work)

 variables.keySet().forEach(k -> templateText.replaceAll("\\${\\{"+ k +"\\}"  variables.get(k)));

However, if I do the "old-style" way, everything works perfectly:

for (Entry<String, String> entry : variables.entrySet()){
    String  regex = "\\$\\{" + entry.getKey() + "\\}";          
    templateText =  templateText.replaceAll(regex, entry.getValue());           
   }

Surely I am missing something here :)

Answer 1

Java 8

The proper way to implement this has not changed in Java 8, it is based on appendReplacement() / appendTail() :

Pattern variablePattern = Pattern.compile("\\$\\{(.+?)\\}");
Matcher matcher = variablePattern.matcher(templateText);
StringBuffer result = new StringBuffer();
while (matcher.find()) {
    matcher.appendReplacement(result, variables.get(matcher.group(1)));
}
matcher.appendTail(result);
System.out.println(result);

Note that, as mentioned by drrob in the comments, the replacement String of appendReplacement() may contain group references using the $ sign, and escaping using \\ . If this is not desired, or if your replacement String can potentially contain those characters, you should escape them using Matcher.quoteReplacement() .

Being more functional in Java 8

If you want a more Java-8-style version, you can extract the search-and-replace boiler plate code into a generalized method that takes a replacement Function :

private static StringBuffer replaceAll(String templateText, Pattern pattern,
                                       Function<Matcher, String> replacer) {
    Matcher matcher = pattern.matcher(templateText);
    StringBuffer result = new StringBuffer();
    while (matcher.find()) {
        matcher.appendReplacement(result, replacer.apply(matcher));
    }
    matcher.appendTail(result);
    return result;
}

and use it as

Pattern variablePattern = Pattern.compile("\\$\\{(.+?)\\}");
StringBuffer result = replaceAll(templateText, variablePattern,
                                 m -> variables.get(m.group(1)));

Note that having a Pattern as parameter (instead of a String ) allows it to be stored as a constant instead of recompiling it every time.

Same remark applies as above concerning $ and \\ – you may want to enforce the quoteReplacement() inside the replaceAll() method if you don't want your replacer function to handle it.

Java 9 and above

Java 9 introduced Matcher.replaceAll(Function) which basically implements the same thing as the functional version above. See Jesse Glick's answer for more details.

Answer 2

you also can using Stream.reduce(identity,accumulator,combiner) .

identity

identity is the initial value for reducing function which is accumulator .

accumulator

accumulator reducing identity to result , which is the identity for the next reducing if the stream is sequentially .

combiner

this function never be called in sequentially stream. it calculate the next identity from identity & result in parallel stream.

BinaryOperator<String> combinerNeverBeCalledInSequentiallyStream=(identity,t) -> {
   throw new IllegalStateException("Can't be used in parallel stream");
};

String result = variables.entrySet().stream()
            .reduce(templateText
                   , (it, var) -> it.replaceAll(format("\\$\\{%s\\}", var.getKey())
                                               , var.getValue())
                   , combinerNeverBeCalledInSequentiallyStream);

Answer 3

import java.util.HashMap;
import java.util.Map;

public class Repl {

    public static void main(String[] args) {
        Map<String, String> variables = new HashMap<>();
        String templateText = "Hi, ${name} ${secondname}! My name is ${name} too :)";
        variables.put("name", "Joe");
        variables.put("secondname", "White");

        templateText = variables.keySet().stream().reduce(templateText, (acc, e) -> acc.replaceAll("\\$\\{" + e + "\\}", variables.get(e)));
        System.out.println(templateText);
    }

}

output:

Hi, Joe White! My name is Joe too :)

However , it's not the best idea to reinvent the wheel and the preferred way to achieve what you want would be to use apache commons lang as stated here .

 Map<String, String> valuesMap = new HashMap<String, String>();
 valuesMap.put("animal", "quick brown fox");
 valuesMap.put("target", "lazy dog");
 String templateString = "The ${animal} jumped over the ${target}.";
 StrSubstitutor sub = new StrSubstitutor(valuesMap);
 String resolvedString = sub.replace(templateString);

Answer 4

Your code should be changed like below,

String templateText = "Hi ${name}";
Map<String,String> variables = new HashMap<>();
variables.put("name", "Joe");
templateText = variables.keySet().stream().reduce(templateText, (originalText, key) -> originalText.replaceAll("\\$\\{" + key + "\\}", variables.get(key)));

Answer 5

Performing replaceAll repeatedly, ie for every replaceable variable, can become quiet expensive, especially as the number of variables might grow. This doesn't become more efficient when using the Stream API. The regex package contains the necessary building blocks to do this more efficiently:

public static String replaceAll(String template, Map<String,String> variables) {
    String pattern = variables.keySet().stream()
        .map(Pattern::quote)
        .collect(Collectors.joining("|", "\\$\\{(", ")\\}"));
    Matcher m = Pattern.compile(pattern).matcher(template);
    if(!m.find()) {
        return template;
    }
    StringBuffer sb = new StringBuffer();
    do {
        m.appendReplacement(sb, Matcher.quoteReplacement(variables.get(m.group(1))));
    } while(m.find());
    m.appendTail(sb);
    return sb.toString();
}

If you are performing the operation with the same Map very often, you may consider keeping the result of Pattern.compile(pattern) , as it is immutable and safely shareable.

On the other hand, if you are using this operation with different maps frequently, it might be an option to use a generic pattern instead, combined with handling the possibility that the particular variable is not in the map. The adds the option to report occurrences of the ${…} pattern with an unknown variable:

private static Pattern VARIABLE = Pattern.compile("\\$\\{([^}]*)\\}");
public static String replaceAll(String template, Map<String,String> variables) {
    Matcher m = VARIABLE.matcher(template);
    if(!m.find())
        return template;
    StringBuffer sb = new StringBuffer();
    do {
        m.appendReplacement(sb,
            Matcher.quoteReplacement(variables.getOrDefault(m.group(1), m.group(0))));
    } while(m.find());
    m.appendTail(sb);
    return sb.toString();
}

m.group(0) is the actual match, so using this as a fall-back for the replacement string establishes the original behavior of not replacing ${…} occurrences when the key is not in the map. As said, alternative behaviors, like reporting the absent key or using a different fall-back text, are possible.

Answer 6

要更新@ didier-l的答案，在Java 9中这是一个单行程！

Pattern.compile("[$][{](.+?)[}]").matcher(templateText).replaceAll(m -> variables.get(m.group(1)))