java - 在java/android中获取带有匹配器的字符串列表

Question

我有一个像

这样的字符串

number +919999999990  time at:07:42:45 on 10.04.2014, number
+919999999991 time at:08:42:45 on 11.05.2014,  number +919999999992 time at:075:42:45 on 05.03.2014 , number +9199999999913 ,time at:27:40:45 on 09.04.2015.

我必须检索所有电话号码、日期和时间。我打算使用模式和匹配器。

我的代码是

String extra1="number +919999999990  time at:07:42:45 on 10.04.2014, number +919999999991 time at:08:42:45 on 11.05.2014, number +919999999992 time at:075:42:45 on 05.03.2014 , number +9199999999913 ,time at:27:40:45 on 09.04.2015.";
Pattern pattern = Pattern.compile("\\+\\d{12}");
Matcher matcher = pattern.matcher(extra1);
System.out.println("matcher.groupCount() "+matcher.groupCount());
if (matcher.find()) {
    while(matcher.find()) {
        System.out.println(matcher.group());
    }   
} else {
     System.out.println("Not Found");
}

根据this和 this它应该打印所有电话号码。但我只得到第一个电话号码。

任何人都可以提出解决方案吗...

Answer 1

如果您删除将消耗第一个数字的“if (matcher...)”语句，它对我来说效果很好:

    String extra1 = "number +919999999990  time at:07:42:45 on 10.04.2014, number +919999999991 time at:08:42:45 on 11.05.2014, number +919999999992 time at:075:42:45 on 05.03.2014 , number +9199999999913 ,time at:27:40:45 on 09.04.2015.";
    Pattern pattern = Pattern.compile("\\+\\d{12}");
    Matcher matcher = pattern.matcher(extra1);
    System.out.println("matcher.groupCount() " + matcher.groupCount());
    while (matcher.find()) {
        System.out.println(matcher.group());
    }

输出:

matcher.groupCount() 0
+919999999990
+919999999991
+919999999992
+919999999991