Java正则表达式从日期较大的字符串中排除特定的权重

Question

I have some string

"Today 31.12.2014g we receive goods. These weight is 31.12g (23.03.2014)"

31.12.2014 g - its not mistake. Some text with date label have g letter (without space)

I need extract from string only weight value (without date value), but my regex:

[0-9]+\\.[0-9]+g

exctact date too :(

my results (two group):

12.2014g

31.12g <- i am need only this!!!

Answer 1

You can add negative look behind to make sure that before part you are interested in there is nothing you don't want which in your case seems to be

• lets say between 1 and 10 numbers with dot after it like in case

   31.12.2014g ^^^ 

• also to make sure that we will match entire value and not just part of it like in case

   31.12.2014g ^^^^^^^ 

  where 2.2014g fulfils condition of previous negative look behind we need to make sure that matched part should not have any digit before it

So try maybe something like

(?<!\\d{1,10}\\.)(?<!\\d)\\d+\\.\\d+g

BTW \\d (which in Java is written as "\\\\d" ) represents [0-9] . You can change it back if you want.

Demo:

String data = "Today 31.12.2014g we receive goods. These weight is 31.12g (23.03.2014)";
Pattern p = Pattern.compile("(?<!\\d{1,10}\\.)(?<!\\d)\\d+\\.\\d+g");
Matcher m = p.matcher(data);
while(m.find())
    System.out.println(m.group());

Output: 31.12g

Answer 2

You could search for white spaces:

\\s[0-9]+\\.[0-9]+g // -> " 31.12g"

… always assume there are exactly two decimal positions:

[^\\.][0-9]+\\.[0-9]{2}g // -> "31.12g" (though, this will fail if the date is spelled DD.MM.YYg)

… or work with the date:

[0-9]?[0-9]\\.[0-9]?[0-9]\\.[0-9]?[0-9]?[0-9][0-9]g?.+(\\b[0-9]+\\.[0-9]+g) // -> "31.12.2014g we receive goods. These weight is 31.12g", "31.12g"