如何使用參數名稱而不是數字格式化消息？

Question

我有類似：

String text = "The user {0} has email address {1}." 
// params = { "Robert", "myemailaddr@gmail.com" } 
String msg = MessageFormat.format(text, params); 

這並不是非常適合我，因爲我有時譯者不知道什麼那張{0}和{1}，也將是不錯的能改寫消息而不用擔心參數的順序。

我想用可讀名稱而不是數字替換參數。類似這樣的：

String text = "The user {USERNAME} has email address {EMAILADDRESS}." 
// Map map = new HashMap(... [USERNAME="Robert", EMAILADDRESS="myemailaddr@gmail.com"] 
String msg = MessageFormat.format(text, map); 

有沒有簡單的方法來做到這一點？

謝謝！ rob

Answer 1

您可以使用MapFormat這一點。在這裏瞭解詳情：

http://www.java2s.com/Code/Java/I18N/AtextformatsimilartoMessageFormatbutusingstringratherthannumerickeys.htm

String text = "The user {name} has email address {email}."; 
      Object[] params = { "nameRobert", "rhume55@gmail.com" }; 
      Map map = new HashMap(); 
      map.put("name", "Robert"); 
      map.put("email", "rhume55@gmail.com"); 

System.out.println("1st : " + MapFormat.format(text, map)); 

輸出：1：用戶羅伯特擁有電子郵件地址rhume55@gmail.com。

Answer 2

自己做個簡單。這是我使用（該main()功能只是爲了測試代碼）：

import java.util.HashMap; 
import java.util.Map; 
import java.util.regex.Matcher; 
import java.util.regex.Pattern; 

public class StringTemplate { 
    final private String template; 
    final private Matcher m; 
    static final private Pattern keyPattern = 
     Pattern.compile("\\$\\{([a-zA-Z][a-zA-Z0-9_]*(\\.[a-zA-Z][a-zA-Z0-9_]*)*)\\}"); 
    private boolean blanknull=false; 

    public StringTemplate(String template) { 
     this.template=template; 
     this.m = keyPattern.matcher(template); 
    } 

    /** 
    * @param map substitution map 
    * @return substituted string 
    */ 
    public String substitute(Map<String, ? extends Object> map) 
    { 
     this.m.reset(); 
     StringBuffer sb = new StringBuffer(); 
     while (this.m.find()) 
     { 
      String k0 = this.m.group(); 
      String k = this.m.group(1); 
      Object vobj = map.get(k); 
      String v = (vobj == null) 
       ? (this.blanknull ? "" : k0) 
       : vobj.toString(); 
      this.m.appendReplacement(sb, Matcher.quoteReplacement(v)); 
     } 
     this.m.appendTail(sb); 
     return sb.toString();  
    } 

    public StringTemplate setBlankNull() 
    { 
     this.blanknull=true; 
     return this; 
    } 

    static public void main(String[] args) 
    { 
     StringTemplate t1 = new StringTemplate("${this} is a ${test} of the ${foo} bar=${bar} ${emergency.broadcasting.system}"); 
     t1.setBlankNull(); 
     Map<String, String> m = new HashMap<String, String>(); 
     m.put("this", "*This*"); 
     m.put("test", "*TEST*"); 
     m.put("foo", "$$$aaa\\\\111"); 
     m.put("emergency.broadcasting.system", "EBS"); 
     System.out.println(t1.substitute(m)); 
    } 
} 

Answer 3

你的問題是密切相關的：How to replace a set of tokens in a Java String 你可以使用velocity或其他模板庫。但是會有一些痛苦，因爲Java沒有任何一種Map文字。

Answer 4

static final Pattern REPLACE_PATTERN = Pattern.compile("\\x24\\x7B([a-zA-Z][\\w\\x2E].*?)\\x7D"); 

/** 
* Check for unresolved environment 
* 
* @param str 
* @return origin if all substitutions resolved 
*/ 
public static String checkReplacement(String str) { 
    Matcher matcher = REPLACE_PATTERN.matcher(str); 
    if (matcher.find()) { 
     throw LOG.getIllegalArgumentException("Environment variable '" + matcher.group(1) + "' is not defined"); 
    } 
    return str; 
} 

// replace in str ${key} to value 
public static String resolveReplacement(String str, Map<String, String> replacements) { 
    Matcher matcher = REPLACE_PATTERN.matcher(str); 
    while (matcher.find()) { 
     String value = replacements.get(matcher.group(1)); 
     if (value != null) { 
      str = matcher.replaceFirst(replaceWindowsSlash(value)); 
     } 
    } 
    return str; 
} 

但你失去所有的格式選項（如##＃）。

Answer 5

見StrSubstitutor從org.apache.commons.lang3：

Map valuesMap = HashMap(); 
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); 

// resolvedString: "The quick brown fox jumped over the lazy dog." 

Answer 6

我知道我的回答來得晚了一點，但如果你還需要這個功能，無需下載一個完整的模板引擎，你可以看看aleph-formatter（我是其中一位作者）：

Student student = new Student("Andrei", 30, "Male"); 

String studStr = template("#{id}\tName: #{st.getName}, Age: #{st.getAge}, Gender: #{st.getGender}") 
        .arg("id", 10) 
        .arg("st", student) 
        .format(); 
System.out.println(studStr); 

或者你可以鏈中的參數：

String result = template("#{x} + #{y} = #{z}") 
        .args("x", 5, "y", 10, "z", 15) 
        .format(); 
System.out.println(result); 

// Output: "5 + 10 = 15" 

內部，它通過「解析」的表達，沒有字符串連接，正則表達式/進行更換工程，並使用StringBuilder的創建結果。