从通配符到正则表达式 [英] From wildcards to regular expressions
问题描述
我想允许两个主要通配符？和 * 来过滤我的数据。 b
$ b
以下是我现在正在做的事情（正如我在许多网站上看到的那样）： b
$ b
public boolean contains（String data，String filter）{
if（data == null || data.isEmpty（））{
返回false;
字符串regex = filter.replace（。，[。]）
.replace（？，。）
.replace（* ，。*）;
返回Pattern.matches（regex，data）;
$ b 但是我们不应该转义所有其他的正则表达式特殊字符，比如 | 或者（等等？也可以保存？ \ ，例如， 和 * b
filter.replaceAll（（[$ | \\ [\ \] {}（），。+ ^ - ]），\\\ $ 1）// 1.转义正则表达式特殊字符，但？，*和\ $ b $ b .replaceAll （（[^ \\\] | ^）\\，$ 1。）// 2.替换之前没有的任何一个\ by。
.replaceAll（（[^ \\\] | ^）\\ *，$ 1. *）// 3.替换前面没有\的任何* 。*
.replaceAll（\\\\（[^？*] | $），\\\\\\\\\ $ 1）; //替换任何不是由一个或一个*（可能是d如果你觉得这样做，你会怎么想呢？如果你同意，我错过任何其他的正则表达式特殊字符？
编辑＃1 考虑到dan1111's和m.buettner的建议）：
$ b
//用一个*
替换任意偶数的反斜杠regex = regex.replaceAll（（？<！\\\\）（\\\\\\\\\\\ \\\）+（？！\\\\\），*）;
//减少不在前面的冗余通配符\
regex = regex.replaceAll（（？<！\\\\）[？] * [*] [*？] +，*）;
//转义正则表达式特殊字符，但是\，？和*
regex = regex.replaceAll（（[| \\ [\\] {}（），。^ $ + - ]），\\\\\ $ 1 ）;
//替换？不是由\之前。
regex = regex.replaceAll（（？<！\\\\）[？]，。）;
//替换*之前没有的一个\。*
regex = regex.replaceAll（（？<！\\\\）[*]， ） *;
这个怎么样？
编辑＃2 （考虑到dan1111的建议后）：
//用一个* $ b $替换任意偶数的反斜杠regex = regex.replaceAll（（？<！\\ （\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\'）（
//减少不在前面的冗余通配符\
regex = regex.replaceAll（（？<！\\\\）[？] * [*] [*？] +，*）;
//转义正则表达式特殊字符（如果用户还没有转义），但是\，？和*
regex = regex.replaceAll（（？<！\\\\）（[| \\ [\\] {}（），。^ $ + ]），\\\ $ 1）;
//替换？不是由\之前。
regex = regex.replaceAll（（？<！\\\\）[？]，。）;
//替换*之前没有的一个\。*
regex = regex.replaceAll（（？<！\\\\）[*]， ） *;
目标在望？
下面是我通过的解决方案（使用 Apache Commons Lang 库）：
$ b
public static boolean isFiltered（String data，String过滤器）{
//无过滤器：返回true
if（StringUtils.isBlank（filter））{
return true;
//过滤但是没有数据：返回false
else if（StringUtils.isBlank（data））{
return false;
//过滤器和数据：
else {
//不区分大小写
data = data.toLowerCase（）;
filter = filter.toLowerCase（）;
// .matches（）auto-anchors，所以加上[*]（即contains）
字符串regex =*+ filter +*;
//将任意一对反斜杠替换为[*]
regex = regex.replaceAll（（？<！\\\\）（\\\\\ \\\+++++++++++++++++++++++++++++++++++++++++++++++++++++++）
//最小化非冗余通配符
regex = regex.replaceAll（（？<！\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\' *）;
//转义正则表达式特殊字符，但[\]，[？]和[*]
regex = regex.replaceAll（（？<！\\\\） （[| \\ [\\] {}（），。^ $ + - ]），\\\\\ $ 1）;
//用[。]替换未转义的[？]
regex = regex.replaceAll（（？<！\\\\）[？]，。）;
//用[。*]替换未转义的[*]
regex = regex.replaceAll（（？<！\\\\）[*]，。* ）;
//返回数据是否匹配正则表达式
返回data.matches（regex）;
$ b 非常感谢@ dan1111和@ m.buettner for他们的宝贵帮助;）
I want to allow the two main wildcards ? and * to filter my data.
Here is how I'm doing now (as I saw on many websites):
public boolean contains(String data, String filter) {
if(data == null || data.isEmpty()) {
return false;
}
String regex = filter.replace(".", "[.]")
.replace("?", ".")
.replace("*", ".*");
return Pattern.matches(regex, data);
}
But shouldn't we escape all the other regex special chars, like | or (, etc.? And also, maybe we could preserve ? and * if they are preceded by a \? For example, something like:
filter.replaceAll("([$|\\[\\]{}(),.+^-])", "\\\\$1") // 1. escape regex special chars, but ?, * and \
.replaceAll("([^\\\\]|^)\\?", "$1.") // 2. replace any ? that isn't preceded by a \ by .
.replaceAll("([^\\\\]|^)\\*", "$1.*") // 3. replace any * that isn't preceded by a \ by .*
.replaceAll("\\\\([^?*]|$)", "\\\\\\\\$1"); // 4. replace any \ that isn't followed by a ? or a * (possibly due to step 2 and 3) by \\
What do you think about it? If you agree, am I missing any other regex special char?
Edit #1 (after having taken into account dan1111's and m.buettner's advices):
// replace any even number of backslashes by a *
regex = regex.replaceAll("(?<!\\\\)(\\\\\\\\)+(?!\\\\)", "*");
// reduce redundant wildcards that aren't preceded by a \
regex = regex.replaceAll("(?<!\\\\)[?]*[*][*?]+", "*");
// escape regexps special chars, but \, ? and *
regex = regex.replaceAll("([|\\[\\]{}(),.^$+-])", "\\\\$1");
// replace ? that aren't preceded by a \ by .
regex = regex.replaceAll("(?<!\\\\)[?]", ".");
// replace * that aren't preceded by a \ by .*
regex = regex.replaceAll("(?<!\\\\)[*]", ".*");
What about this one?
Edit #2 (after having taken into account dan1111's advices):
// replace any even number of backslashes by a *
regex = regex.replaceAll("(?<!\\\\)(\\\\\\\\)+(?!\\\\)", "*");
// reduce redundant wildcards that aren't preceded by a \
regex = regex.replaceAll("(?<!\\\\)[?]*[*][*?]+", "*");
// escape regexps special chars (if not already escaped by user), but \, ? and *
regex = regex.replaceAll("(?<!\\\\)([|\\[\\]{}(),.^$+-])", "\\\\$1");
// replace ? that aren't preceded by a \ by .
regex = regex.replaceAll("(?<!\\\\)[?]", ".");
// replace * that aren't preceded by a \ by .*
regex = regex.replaceAll("(?<!\\\\)[*]", ".*");
Goal in sight?
解决方案 Here is finally the solution I adopted (using the Apache Commons Lang library):
public static boolean isFiltered(String data, String filter) {
// no filter: return true
if (StringUtils.isBlank(filter)) {
return true;
}
// a filter but no data: return false
else if (StringUtils.isBlank(data)) {
return false;
}
// a filter and a data:
else {
// case insensitive
data = data.toLowerCase();
filter = filter.toLowerCase();
// .matches() auto-anchors, so add [*] (i.e. "containing")
String regex = "*" + filter + "*";
// replace any pair of backslashes by [*]
regex = regex.replaceAll("(?<!\\\\)(\\\\\\\\)+(?!\\\\)", "*");
// minimize unescaped redundant wildcards
regex = regex.replaceAll("(?<!\\\\)[?]*[*][*?]+", "*");
// escape unescaped regexps special chars, but [\], [?] and [*]
regex = regex.replaceAll("(?<!\\\\)([|\\[\\]{}(),.^$+-])", "\\\\$1");
// replace unescaped [?] by [.]
regex = regex.replaceAll("(?<!\\\\)[?]", ".");
// replace unescaped [*] by [.*]
regex = regex.replaceAll("(?<!\\\\)[*]", ".*");
// return whether data matches regex or not
return data.matches(regex);
}
}
Many thanks to @dan1111 and @m.buettner for their precious help ;)
这篇关于从通配符到正则表达式的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
