java从字符串中提取数字

Question

String str = "POLYGON((39.4189453125 37.418708616699824,42.0556640625 37.418708616699824,43.4619140625 34.79181436843146,38.84765625 33.84817790215085,39.4189453125 37.418708616699824))";

我试图只获得39.4189453125 37.418708616699824,42.0556640625 37.418708616699824,43.4619140625 34.79181436843146,38.84765625 33.84817790215085,39.4189453125 37.418708616699824

同

final String coordsOnly = str.replaceAll("\\D\\(\\(\\w\\)\\)", "$2");

但我得到coordsOnly = "POLYGON((39.4189453125 37.418708616699824,42.0556640625 37.418708616699824,43.4619140625 34.79181436843146,38.84765625 33.84817790215085,39.4189453125 37.418708616699824))"

我错过了什么？

Answer 1

这里一个合理的方法是匹配以下模式，然后用第一个捕获组替换all：

POLYGON\(\((.*?)\)\)

示例代码：

final String coordsOnly = str.replaceAll("POLYGON\\(\\((.*?)\\)\\)", "$1");

Demo

编辑：

如果你还需要隔离数字对，你可以在逗号上使用String#split()。实际上，这看起来像数据库查询的输出，我怀疑数据库可能提供更好的方法来获取单个值。但是，如果您无法获得所需的确切输出，这里给出的答案是一个选项。

Answer 2

实际上，很多。

"\\D" // Matches a non-digit character, but it only matches one,
      // while you need to match a word "POLYGON";
"\\(\\(" // Good. Matches the double left parentheses ((
"\\w"    // One word character? Same issue, you need to match multiple chars. And what about '.'?
"\\)\\)" // Good. Matches the double right parentheses ))

逃脱的()没有创建匹配组;和\\w只匹配一个单词字符[a-zA-Z_0-9]，它甚至不匹配.。

我相信你应该尝试这样的事情：

String coords = str.replaceAll("POLYGON\\(\\(([^)]+)\\)\\)", "$1");

Answer 3

也许这就是你想要的？：

String rex = "[+-]?([0-9]*[.])?[0-9]+";
Pattern p = Pattern.compile(rex);
String input = "POLYGON((39.4189453125 37.418708616699824,42.0556640625 37.418708616699824,43.4619140625 34.79181436843146,38.84765625 33.84817790215085,39.4189453125 37.418708616699824))";
Matcher matcher = p.matcher(input);

while(matcher.find()) {
    System.out.println(matcher.group());
}

Answer 4

试试这个版本：

str = str.replaceAll("[^\\d\\.\\s,]", "");

Answer 5

String str = "POLYGON((39.4189453125 37.418708616699824,42.0556640625 37.418708616699824,43.4619140625 34.79181436843146,38.84765625 33.84817790215085,39.4189453125 37.418708616699824))";
String sp = "(([0-9]+[.])?[0-9]+[,]?\\s*)+";
Pattern p = Pattern.compile(sp);
Matcher matcher = p.matcher(str);
if (matcher.find()) {
    System.out.println(matcher.group());
}

输出：

39.4189453125 37.418708616699824,42.0556640625 37.418708616699824,43.4619140625 34.79181436843146,38.84765625 33.84817790215085,39.4189453125 37.418708616699824

Answer 6

另一种选择是替换所有非数字，空格，点或逗号：

str.replaceAll("[\\D&&\\S&&[^,\\.]]", "")

输出：

39.4189453125 37.418708616699824,42.0556640625 37.418708616699824,43.4619140625 34.79181436843146,38.84765625 33.84817790215085,39.4189453125 37.418708616699824