Java正则表达式URL解析 [英] java regular expression url parsing

问题描述

可能重复:
如何从给定的URL中提取参数

Possible Duplicate:
How to extract parameters from a given url

我正在尝试仅从此url中的参数中检索数字:

I'm trying to retrieve just the numbers from the parameter in this url:

htt://tesing12/testds/fdsa?communityUuid=45352-32452-52

我没有运气尝试过这个

^.*communityUuid=

任何帮助都会很好.

推荐答案

我建议不要使用简单的字符串操作路线.它更冗长，更容易出错.您还可以从内置类中获得一些帮助，然后利用所使用的URL知识(用&"分隔的参数)来指导您的实现:

I recommend against the simple string-manipulation route. It's more verbose and more error prone. You may as well get a little help from the built-in classes and then use your knowledge that you're working with a URL (parameters delimited with "&") to guide your implementation:

String queryString = new URL("http://tesing12/testds/fdsa?communityUuid=45352-32452-52").getQuery();

String[] params = queryString.split("&");

String communityUuid = null;
for (String param : params) {
    if (param.startsWith("communityUuid=")) {
        communityUuid = param.substring(param.indexOf('=') + 1);
    }
}

if (communityUuid != null) {
    // do what you gotta do
}

这为您提供了检查URL格式正确性的好处，并避免了因类似名称的参数而引起的问题(字符串操作路径将报告"abc_communityUuid"和"communityUuid"的值)

This gives you the benefit of checking the well-formed-ness of the URL and avoids problems that can arise from similarly named parameters (the string-manipulation route will report the value of "abc_communityUuid" as well as "communityUuid").

此代码的一个有用扩展是在您遍历"params"时构建一个映射，然后在映射中查询所需的任何参数名称.

A useful extension of this code is to build a map as you iterate over "params" and then query the map for any parameter name you want.

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