如何替换Java String中的一组标记？ [英] How to replace a set of tokens in a Java String?

问题描述

我有以下模板字符串：你好[姓名]请找到[截止日期] 到期的附加[发票编号]。

I have the following template String: "Hello [Name] Please find attached [Invoice Number] which is due on [Due Date]".

我还有名称，发票号和截止日期的字符串变量 - 用变量替换模板中的标记的最佳方法是什么？

I also have String variables for name, invoice number and due date - what's the best way to replace the tokens in the template with the variables?

（请注意，如果变量恰好包含令牌，则不应替换它。）

(Note that if a variable happens to contain a token it should NOT be replaced).

编辑

感谢@laginimaineb和@ alan-moore，这是我的解决方案：

With thanks to @laginimaineb and @alan-moore, here's my solution:

public static String replaceTokens(String text, 
                                   Map<String, String> replacements) {
    Pattern pattern = Pattern.compile("\\[(.+?)\\]");
    Matcher matcher = pattern.matcher(text);
    StringBuffer buffer = new StringBuffer();

    while (matcher.find()) {
        String replacement = replacements.get(matcher.group(1));
        if (replacement != null) {
            // matcher.appendReplacement(buffer, replacement);
            // see comment 
            matcher.appendReplacement(buffer, "");
            buffer.append(replacement);
        }
    }
    matcher.appendTail(buffer);
    return buffer.toString();
}

推荐答案

最有效的方法是使用匹配器不断查找表达式并替换它们，然后将文本附加到字符串构建器：

The most efficient way would be using a matcher to continually find the expressions and replace them, then append the text to a string builder:

Pattern pattern = Pattern.compile("\\[(.+?)\\]");
Matcher matcher = pattern.matcher(text);
HashMap<String,String> replacements = new HashMap<String,String>();
//populate the replacements map ...
StringBuilder builder = new StringBuilder();
int i = 0;
while (matcher.find()) {
    String replacement = replacements.get(matcher.group(1));
    builder.append(text.substring(i, matcher.start()));
    if (replacement == null)
        builder.append(matcher.group(0));
    else
        builder.append(replacement);
    i = matcher.end();
}
builder.append(text.substring(i, text.length()));
return builder.toString();

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