从java中的字符串中提取long / float数字

Question

我有这样的输入==＆gt;
2本书在12.99
4个薯片3.99左右

我想从每一行中提取数值并将它们存储在变量中 例如在行中.. 2本书在12.99我想提取Qauntity = 2和Price = 12.99 来自给定的字符串

Answer 1

您可以使用：

Pattern p = Pattern.compile("(\\d+)\\D+(\\d+(?:.\\d+)?)");
Matcher mr = p.matcher("4 potato chips at 3.99");
if (mr.find()) {
    System.out.println( mr.group(1) + " :: " + mr.group(2) );
}

<强>输出：

4 :: 3.99

Answer 2

正则表达式

(\d+)[^\d]+([+-]?[0-9]{1,3}(?:,?[0-9]{3})*(?:\.[0-9]{2})?)

<强> Debuggex Demo

---

描述(Example)

/^(\d+)[^\d]+([+-]?[0-9]{1,3}(?:,?[0-9]{3})*(?:\.[0-9]{2})?)$/gm
^ Start of line
1st Capturing group (\d+)
    \d 1 to infinite times [greedy] Digit [0-9]
Negated char class [^\d] 1 to infinite times [greedy] matches any character except:
    \d Digit [0-9]
2nd Capturing group ([+-]?[0-9]{1,3}(?:,?[0-9]{3})*(?:\.[0-9]{2})?)
    Char class [+-] 0 to 1 times [greedy] matches:
        +- One of the following characters +-
    Char class [0-9] 1 to 3 times [greedy] matches:
        0-9 A character range between Literal 0 and Literal 9
    (?:,?[0-9]{3}) Non-capturing Group 0 to infinite times [greedy]
        , 0 to 1 times [greedy] Literal ,
    Char class [0-9] 3 times [greedy] matches:
        0-9 A character range between Literal 0 and Literal 9
    (?:\.[0-9]{2}) Non-capturing Group 0 to 1 times [greedy]
        \. Literal .
    Char class [0-9] 2 times [greedy] matches:
        0-9 A character range between Literal 0 and Literal 9
$ End of line
g modifier: global. All matches (don't return on first match)
m modifier: multi-line. Causes ^ and $ to match the begin/end of each line (not only begin/end of string)

捕获组1：包含数量

捕获第2组：包含金额

---

爪哇

try {
    Pattern regex = Pattern.compile("(\\d+)[^\\d]+([+-]?[0-9]{1,3}(?:,?[0-9]{3})*(?:\\.[0-9]{2})?)", Pattern.CASE_INSENSITIVE | Pattern.UNICODE_CASE);
    Matcher regexMatcher = regex.matcher(subjectString);
    while (regexMatcher.find()) {
        for (int i = 1; i <= regexMatcher.groupCount(); i++) {
            // matched text: regexMatcher.group(i)
            // match start: regexMatcher.start(i)
            // match end: regexMatcher.end(i)
        }
    } 
} catch (PatternSyntaxException ex) {
    // Syntax error in the regular expression
}

注意：这个java只是一个例子，我不用Java编码

Answer 3

您可以使用MessageFormat课程。以下是工作示例：

MessageFormat f = new MessageFormat("{0,number,#.##} {2} at {1,number,#.##}");
try {
    Object[] result = f.parse("4 potato chips at 3.99");
    System.out.print(result[0]+ ":" + (result[1]));
} catch (ParseException ex) {
    // handle parse error
}