Java Regex：替换为捕获组

Question

I have a string like this:

123456-1/1234/189928/2323 (102457921)

I want to get 102457921 . How can I achieve it with regex?

I have tried:

"123456-1/1234/189928/2323 (102457921)".replaceAll("(\\.*\()(\d+)(\))","$2");

But it does not work. Any hints?

Answer 1

怎么样

"123456-1/1234/189928/2323 (102457921)".replaceAll(".*?\\((.*?)\\).*", "$1");

Answer 2

好吧，你可以这样做：

"123456-1/1234/189928/2323 (102457921)".replaceAll(".*\((.+)\)","$1");

Answer 3

您可以这样做：

"123456-1/1234/189928/2323 (102457921)".replaceAll(".*?\(([^)]+)\)","$1");

Answer 4

怎么样“双”替换所有正则表达式简化了一个

"123456-1/1234/189928/2323 (102457921)".replaceAll(".*\\(", "").replaceAll("\\).*", "");

Answer 5

You can try something like that:

var str = '123456-1/1234/189928/2323 (102457921)';
    console.log(str.replace(/[-\d\/ ]*\((\d+)\)/, "$1"));
    console.log((str.split('('))[1].slice(0, -1));
    console.log((str.split(/\(/))[1].replace(/(\d+)\)/, "$1"));
    console.log((str.split(/\(/))[1].substr(-str.length - 1, 9));
    console.log(str.substring(str.indexOf('(') + 1, str.indexOf(')')));

In other case you must be familiar with the specifics of the input data to generate a suitable regexp.