正则表达式帮助-从字符串中提取值，而忽略先前的内容

Question

can someone help me to write a java regex to retrieve a value from the following string please?

XX0001  15NOV XXX SELECTED RAX                                       AXXXXX DXXXXXXXXX           REF NBR  002            SSSS 

I wanted to extract the value 002. All the strings / characters before 002 are fixed length and properly padded with trailing space (if req.). could have any string/numeric/special displayable characters.

I am looking for something like ... get 002 from that position ignoring whatever before. ?

Answer 1

Assuming that you want the last set of digits before the end of the string, you might want to do something like this:

^.*\\d{3}\\s+.{4}$

This should instruct the Regex Engine to start matching from the beginning of the string, match any characters and then, from the end, match 3 numbers, a space and any 4 characters.

Also, if you have fixed sizes and lengths, you can most likely get away with a .substring method, it is less complex.

Answer 2

What language?

If javascript:

myCode=myString.substring(56,58);

or

myCode=myString.substr(56,3);

If PHP:

$myCode=substr($myString,56,3);

This simpler option is preferable to regex because it is faster. You can use this because you're working with fixed length strings.

EDIT: Just saw your edit referencing Java. So in Java:

String myCode = myString.substring(56,59);

Answer 3

You don't need regex to do that. Just use the String method substring :

String myString = originalString.substring(106,109); // myString = "002"

106 is the begin index, and 109 the end index - 1. To simply get the first, just take the length of the original string just before the number you want to get, for instance:

System.out.println("XX0001  15NOV XXX SELECTED RAX                                       AXXXXX DXXXXXXXXX           REF NBR  ".length());

Answer 4

在Java中

string.substring(56,59);