Question

我正在尝试创建一个regexp来忽略String中的Freemarker FTL标记。我有一个包含文字的模板和FTL tags，例如"Hi, [#if gender="Male"]Mr[#elseif]Mrs[/#if] ${name} I've a special offer for you."

我想只替换FTL标签，所以只有[#something]模式的标签：我使用方括号语法。

理想情况下，我应该致电“myString".replaceAll("regex","");

但是我无法找到合适的正则表达式。

我想在替换后获得的最终字符串结果为"Hi, Mr Mrs ${name} I've a special offer for you."

Answer 1

以下模式应该符合您的期望：

\[/?#(.+?)\]

测试用例：

public class Test {

    private static final String PATTERN = "\\[/?#(.+?)\\]";
    private static final String TEXT = "Hi, [#if gender=\"Male\"]Mr[#elseif]Mrs[/#if] ${name} I've a special offer for you.";

    public static void main(String[] args) {
        String out = TEXT.replaceAll(PATTERN, "");
        System.out.println(out);
    }

}

输出： Hi, MrMrs ${name} I've a special offer for you.

Answer 2

这应该这样做：

public static void main(String[] args) {
    String template = "<html>[BR]\n"
                      + "<head>[BR]\n"
                      + "  <title>Welcome!</title>[BR]\n"
                      + "</head>[BR]\n"
                      + "<body>[BR]\n"
                      + "  [#-- Greet the user with his/her name --][BR]\n"
                      + "  <h1>Welcome ${user}!</h1>[BR]\n"
                      + "  <p>We have these animals:[BR]\n"
                      + "  <ul>[BR]\n"
                      + "  [#list animals as animal][BR]\n"
                      + "    <li>${animal.name} for ${animal.price} Euros[BR]\n"
                      + "  [/#list][BR]\n"
                      + "  </ul>[BR]\n"
                      + "</body>[BR]\n"
                      + "</html>";

    System.out.println("************ORIGINAL TEMPLATE**************");
    System.out.println(template);
    System.out.println("************REPLACED TEMPLATE**************");
    System.out.println(template.replaceAll("\\[/?#(.*?)\\]", "\u001B[33mREPLACEMENT\u001B[0m"));
}

Java regexp删除Freemarker FTL标签

2 个答案: