从文本检索链接的java regex

Question

我有一个输入String，如：

String text = "Some content which contains link as <A HREF=\"/relative-path/fruit.cgi?param1=abc&param2=xyz\">URL Label</A> and some text after it";

我想将此文本转换为：

Some content which contains link as http://www.google.com/relative-path/fruit.cgi?param1=abc&param2=xyz&myParam=pqr (URL Label) and some text after it

所以在这里：

1)我想用普通链接替换链接标签。如果标签包含标签，那么它应该在URL之后进入大括号。

2)如果URL是相对的，我想在基本URL (http://www.google.com)前加上前缀。

3)我想在URL中附加一个参数。(&myParam=pqr)

我有问题，检索标签的URL和标签，并更换它。

我写了这样的东西：

public static void main(String[] args) {
    String text = "String text = "Some content which contains link as <A HREF=\"/relative-path/fruit.cgi?param1=abc&param2=xyz\">URL Label</A> and some text after it";";
    text = text.replaceAll("&lt;", "<");
    text = text.replaceAll("&gt;", ">");
    text = text.replaceAll("&amp;", "&");

    // this is not working
    Pattern p = Pattern.compile("href=\"(.*?)\"");
    Matcher m = p.matcher(text);
    String url = null;
    if (m.find()) {
        url = m.group(1);

    }
}

// helper method to append new query params once I have the url
public static URI appendQueryParams(String uriToUpdate, String queryParamsToAppend) throws URISyntaxException {
    URI oldUri = new URI(uriToUpdate);
    String newQueryParams = oldUri.getQuery();
    if (newQueryParams == null) {
        newQueryParams = queryParamsToAppend;
    } else {
        newQueryParams += "&" + queryParamsToAppend;  
    }
    URI newUri = new URI(oldUri.getScheme(), oldUri.getAuthority(),
            oldUri.getPath(), newQueryParams, oldUri.getFragment());
    return newUri;
}

Edit1:

Pattern p = Pattern.compile("HREF=\"(.*?)\"");

这个很管用。但我希望它是资本化不可知论者。Href、HRef、href、hrEF等都应该工作。

另外，如果我的文本有几个URL，我该如何处理。

Edit2:

一些进展。

Pattern p = Pattern.compile("href=\"(.*?)\"");
Matcher m = p.matcher(text);
String url = null;
while (m.find()) {
  url = m.group(1);
  System.out.println(url);
}

这将处理多个URL的情况。

最后一个悬而未决的问题是，我如何获得标签，并将原始文本中的href标记替换为URL和label。

Edit3：

在多个URL情况下，我的意思是在给定的文本中存在多个url。

String text = "Some content which contains link as <A HREF=\"/relative-path/fruit.cgi?param1=abc&param2=xyz\">URL Label</A> and some text after it and another link <A HREF=\"/relative-path/vegetables.cgi?param1=abc&param2=xyz\">URL2 Label</A> and some more text";

Pattern p = Pattern.compile("href=\"(.*?)\"", Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(text);
String url = null;
while (m.find()) {
 url = m.group(1); // this variable should contain the link URL
 url = appendBaseURI(url);
 url = appendQueryParams(url, "license=ABCXYZ");
 System.out.println(url);
}

Answer 1

public static void main(String args[]) {
    String text = "Some content which contains link as <A HREF=\"/relative-path/fruit.cgi?param1=abc&param2=xyz\">URL Label</A> and some text after it and another link <A HREF=\"/relative-path/vegetables.cgi?param1=abc&param2=xyz\">URL2 Label</A> and some more text";
    text = StringEscapeUtils.unescapeHtml4(text);
    Pattern p = Pattern.compile("<a href=\"(.*?)\">(.*?)</a>", Pattern.CASE_INSENSITIVE);
    Matcher m = p.matcher(text);
    while (m.find()) {
        text = text.replace(m.group(0), cleanUrlPart(m.group(1), m.group(2)));
    }
    System.out.println(text);
}

private static String cleanUrlPart(String url, String label) {
    if (!url.startsWith("http") && !url.startsWith("www")) {
        if (url.startsWith("/")) {
            url = "http://www.google.com" + url;
        } else {
            url = "http://www.google.com/" + url;
        }
    }
    url = appendQueryParams(url, "myParam=pqr").toString();
    if (label != null && !label.isEmpty()) url += " (" + label + ")";
    return url;
}

输出

Some content which contains link as http://www.google.com/relative-path/fruit.cgi?param1=abc&param2=xyz&myParam=pqr (URL Label) and some text after it and another link http://www.google.com/relative-path/vegetables.cgi?param1=abc&param2=xyz&myParam=pqr (URL2 Label) and some more text

Answer 2

您可以使用apache共用文本 StringEscapeUtils对html实体进行解码，然后使用replaceAll，即：

import org.apache.commons.text.StringEscapeUtils;

String text = "Some content which contains link as <A HREF=\"/relative-path/fruit.cgi?param1=abc&param2=xyz\">URL Label</A> and some text after it";
String output = StringEscapeUtils.unescapeHtml4(text).replaceAll("([^<]+).+\"(.*?)\">(.*?)<[^>]+>(.*)", "$1https://google.com$2&your_param ($3)$4");
System.out.print(output);
// Some content which contains link as https://google.com/relative-path/fruit.cgi?param1=abc&param2=xyz&your_param (URL Label) and some text after it

演示：

1. 胡桃
2. Regex解释

Answer 3

//这不管用

因为你的判断力是区分大小写的。

试着：-

Pattern p = Pattern.compile("href=\"(.*?)\"", Pattern.CASE_INSENSITIVE);

Edit1

要获得标签，请使用Pattern.compile("(?<=>).*?(?=</a>)", Pattern.CASE_INSENSITIVE)和m.group(0)。

Edit2

若要用最终字符串替换标记(包括标签)，请使用：-

text.replaceAll("(?i)<a href=\"(.*?)</a>", "new substring here")