使用起始字符串和结束字符串从长字符串中提取子字符串？ [英] Extract substring from a long string using starting string and ending string?

问题描述

我有这个长字符串（它是一个长的连续字符串）：

I have this long string (its a one, long, continuous string):

家庭住址H.NO- 12 SECTOR- 12 GAUTAM BUDH NAGAR NOIDA- 121212，UTTAR PRADESH INDIA [phone]最近更新：12-JUN-12学期/学期 - 住宿类型：居住大厅（私人提供者）学期/学期时间地址A121A某些公寓部分LANE CITY COUNTY OX3 7FJ [phone]最近更新：12-SEP-12移动电话号码：[phone]

如果查看上面的字符串，可以使用以下模式生产：

If you look at the string above, the following pattern can be produced:

< home_address_text>< space>< the_address>< space>< last_updated_text>< last_updated_date><空间>< accomodation_t ype_text>< accomodation_type><空>< semester_time_address_text>< semester_time_address><空> last_updated_text><时间：LAST_UPDATED_DATELAST_UPDATED_TIME><空>< mobile_number_text>< MOBILE_NUMBER>

我想提取s这个字符串的特定部分，如：

1. H.NO- 12 SECTOR-12 GAUTAM BUDH NAGAR NOIDA- 121212，UTTAR PRADESH INDIA
2. Hall of住所（私人提供者）
3. A121A SOME PARARTMENT SOMELANE CITY COUNTY OX3 7FJ
4. 01212121212


I want to extract specific parts of this string, like: 1. H.NO- 12 SECTOR- 12 GAUTAM BUDH NAGAR NOIDA- 121212, UTTAR PRADESH INDIA 2. Hall of residence (private provider) 3. A121A SOME APARTMENT SOMELANE CITY COUNTY OX3 7FJ 4. 01212121212

这个信息是可变的，所以它因人而异，所以我不能只计算长度并使用子字符串来提取它，因为整个字符串的长度&我要提取的部分是可变的。

This information is variable, so it differs from person to person, so I can't just compute the length and use substring to extract it, because the length of the whole string & the part I want to extract is variable.

如何使用Java提取字符串的特定部分，如上所述？我很久以来一直在寻找方法，却找不到办法。非常感谢任何帮助

How can I extract specific parts of the string, as explained above, using Java? I've been looking for ways since a long time but couldn't find a way. Any help would be very much appreciated

推荐答案

这对我有用，基于您的（单个）示例。学习使用不情愿的修饰符来表达正则表达式。在这种情况下，他们会帮助你很多。

This worked for me, based on your (single) example. Learn to use the reluctant modifiers for regular expressions. They'll help you a lot in situations like this.

例如，要获得与第一部分相匹配的字符串：Home地址（。+？）\ + \d +最后更新：此正则表达式不会跳过我们不想要的上次更新字符串或+ dd（数字）。正则表达式表达式（。+？）是不情愿的（不是贪婪的），不会跳过+号或数字，让它们与表达式的其余部分匹配。

For example, to get a string of characters to match the first part: "Home address (.+?) \+\d+ Last Updated: this regex will not skip the "Last Updated" string or the "+dd" (digits) we don't want. The regex expression "(.+?)" is reluctant (not greedy) and won't skip over the + sign or the digits, leaving them to be matched by the rest of the expression.

你可以使用它来匹配静态文本包围的正则表达式中的子串。这里我使用捕获组来找到我想要的文本。（捕获组是括号中的部分。）

You can use this to match substrings in a regular expression that is surrounded by static text. Here I'm using capturing groups to locate the text I want. (Capturing groups are the parts in parenthesis.)

class Goofy
{

   public static void main( String[] args )
   {
      final String input
              = "Home address H.NO- 12 SECTOR- 12 GAUTAM BUDH NAGAR " +
              "NOIDA- 121212, UTTAR PRADESH INDIA [phone] " +
              "Last Updated: 12-JUN-12 Semester/Term-time " +
              "Accommodation Type: Hall of residence (private " +
              "provider) Semester/Term-time address A121A SOME " +
              "APPARTMENT SOME LANE CITY COUNTY OX3 7FJ +91 " +
              "1212121212 Last Updated: 12-SEP-12 Mobile Telephone " +
              "Number : 01212121212";

      final String regex = "Home address (.+?) \\+\\d+ Last Updated: " +
              "\\S+ Semester/Term-time Accommodation Type: (.+?) " +
              "Semester/Term-time address (.+?) \\+\\d\\d \\d+ " +
              "Last Updated.+ Number : (\\d+)";

      Pattern pattern = Pattern.compile( regex );
      Matcher matcher = pattern.matcher( input );
      if( matcher.find() ) {
         System.out.println("Found: "+matcher.group() );
         for( int i = 1; i <= matcher.groupCount(); i++ ) {
            System.out.println( "   Match " + i + ": " + matcher.group( i ));
         }
      }
   }
}

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