Java正则表达式返回完整字符串而不是捕获

Question

Java代码：

String imagesArrayResponse = xmlNode.getChildText("files");
Matcher m = Pattern.compile("path\":\"([^\"]*)").matcher(imagesArrayResponse);

while (m.find()) {
    String path = m.group(0);
}

串：

[{"path":"upload\/files\/56727570aaa08922_0.png","dir":"files","name":"56727570aaa08922_0","original_name":"56727570aaa08922_0.png"}{"path":"upload\/files\/56727570aaa08922_0.png","dir":"files","name":"56727570aaa08922_0","original_name":"56727570aaa08922_0.png"}{"path":"upload\/files\/56727570aaa08922_0.png","dir":"files","name":"56727570aaa08922_0","original_name":"56727570aaa08922_0.png"}{"path":"upload\/files\/56727570aaa08922_0.png","dir":"files","name":"56727570aaa08922_0","original_name":"56727570aaa08922_0.png"}]

m.group退货

path":"upload\/files\/56727570aaa08922_0.png"

而不是路径的捕获​​值。我哪里错了？

Answer 1

请参阅group( int index )方法文档

用0调用时，它将返回整个字符串。第一组是第一个。

为避免此类陷阱，应使用带语法的命名组： "path\":\"(?<mynamegroup>[^\"]*)"

javadoc：

捕获组从左到右从一个索引开始。组零表示整个模式，因此表达式m.group（0）等效于m.group（）。