Java模式正则表达式，用于查找不以特定子字符串结尾的数字

Question

I´m trying to capture a substring of a line in Java using the pattern and matcher logic. For the current application I need to remove a substring of a line. therefore i want to capture the whole substring ending with a number which should not end with a specified 3 digit number reserved for other purposes. I also can´t simply Strip of a hardcoded amount of characters from my line, therefore a regex is needed. To simplify the Problem, the regex should just capture numbers of 9 digits not ending with 770.

this should be the results:

233456770 --> NO HIT
454683870 --> HIT
459987579 --> HIT

i tried the solution with the following solution, Stripping all numbers ending with a 0

\d{6}[^9][^9][^0]

I also tried a lookbehind, which resulted either in the substring beeing too Long or to short since there are 3 digits missing or too much

\\d{6}(?!990) and

\d{9}(?!990)

So how to detect and capture a number not ending with a specified substring while keeping the whole detected number (withoud removing the last three digits checked) in the resulting substring using Java Pattern language?

Answer 1

You may use

(?<!\d)\d{9}(?!\d)(?<!770)

See this demo .

Details

• (?<!\\d) - no digit immediately to the left of the current location is allowed
• \\d{9} - nine digits
• (?!\\d) - no digit immediately to the right of the current location is allowed
• (?<!770) - no 770 allowed immediately to the left of the current location.

Java string literal:

String regex = "(?<!\\d)\\d{9}(?!\\d)(?<!770)";

Answer 2

I would avoid regexex if I can since they are hard to debug and have worse performance.

If you can use plain java:

num !=null && num.length() == 9 && num.length.endsWith("770")