什么是Java 1.4.2等效的Pattern.quote()

AHu*_*ist 7 java regex java1.4

什么是Java 1.4.2等效的Pattern.quote?

我在URI上使用Pattern.quote()但现在需要使它与1.4.2兼容.

aio*_*obe 4

嗯,源代码Pattern.quote是可用的,如下所示:

public static String quote(String s) {
    int slashEIndex = s.indexOf("\\E");
    if (slashEIndex == -1)
        return "\\Q" + s + "\\E";

    StringBuilder sb = new StringBuilder(s.length() * 2);
    sb.append("\\Q");
    slashEIndex = 0;
    int current = 0;
    while ((slashEIndex = s.indexOf("\\E", current)) != -1) {
        sb.append(s.substring(current, slashEIndex));
        current = slashEIndex + 2;
        sb.append("\\E\\\\E\\Q");
    }
    sb.append(s.substring(current, s.length()));
    sb.append("\\E");
    return sb.toString();
}
Run Code Online (Sandbox Code Playgroud)

基本上它依赖于

\Q  Nothing, but quotes all characters until \E
\E  Nothing, but ends quoting started by \Q
Run Code Online (Sandbox Code Playgroud)

\E并且对字符串中出现的情况进行特殊处理。