在java中构建URL
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Build URL in java
提问by rrr ppp
Trying to build http://IP:4567/foldername/1234?abc=xyz. I don't know much about it but I wrote below code from searching from google:
试图建立http://IP:4567/foldername/1234?abc=xyz. 我对此知之甚少，但我从谷歌搜索中编写了以下代码：
import java.net.MalformedURLException;
import java.net.URI;
import java.net.URL;
public class MyUrlConstruct {
public static void main(String a[]){
try {
String protocol = "http";
String host = "IP";
int port = 4567;
String path = "foldername/1234";
URL url = new URL (protocol, host, port, path);
System.out.println(url.toString()+"?");
} catch (MalformedURLException ex) {
ex.printStackTrace();
}
}
}
I am able to build URL http://IP:port/foldername/1234?. I am stuck at query part. Please help me to move forward.
我能够建立 URL http://IP:port/foldername/1234?。我被困在查询部分。请帮助我前进。
采纳答案by VGR
In general non-Java terms, a URL is a specialized type of URI. You can use the URIclass (which is more modern than the venerable URL class, which has been around since Java 1.0) to create a URI more reliably, and you can convert it to a URL with the toURLmethod of URI:
在一般的非 Java 术语中，URL 是一种特殊类型的 URI。您可以使用URI类（它比古老的 URL 类更现代，自 Java 1.0 以来就存在）来更可靠地创建 URI，并且您可以使用URI的toURL方法将其转换为 URL ：
String protocol = "http";
String host = "example.com";
int port = 4567;
String path = "/foldername/1234";
String auth = null;
String fragment = null;
URI uri = new URI(protocol, auth, host, port, path, query, fragment);
URL url = uri.toURL();
Note that the pathneeds to start with a slash.
注意path需要以斜线开头。
回答by vsminkov
You can just pass raw spec
你可以只传递原始规范
new URL("http://IP:4567/foldername/1234?abc=xyz");
Or you can take something like org.apache.http.client.utils.URIBuilderand build it in safe manner with proper url encoding
或者您可以采用类似的org.apache.http.client.utils.URIBuilder方法并使用正确的 url 编码以安全的方式构建它
URIBuilder builder = new URIBuilder()
builder.setScheme("http")
builder.setHost("IP")
builder.setPath("/foldername/1234")
builder.addParameter("abc", "xyz")
URL url = builder.build().toURL()
回答by Tyler Long
Use OkHttp
使用 OkHttp
There is a very popular library named OkHttpwhich has been starred 20Ktimes on GitHub. With this library, you can build the url like below:
有一个非常流行的库叫OkHttp，它在 GitHub 上已经被加星了20K次。使用此库，您可以构建如下所示的 url：
import okhttp3.HttpUrl;
URL url = new HttpUrl.Builder()
.scheme("http")
.host("example.com")
.port(4567)
.addPathSegments("foldername/1234")
.addQueryParameter("abc", "xyz")
.build().url();
Or you can simply parse an URL:
或者你可以简单地解析一个 URL：
URL url = HttpUrl.parse("http://example.com:4567/foldername/1234?abc=xyz").url();
回答by Philippe
If you happen to be using Spring already, I have found the org.springframework.web.util.UriComponentsBuilderto be quite nifty. Here is how you would use it in your case.
如果您碰巧已经在使用 Spring，我发现org.springframework.web.util.UriComponentsBuilder它非常漂亮。以下是您将如何在您的情况下使用它。
final URL myUrl = UriComponentsBuilder
.fromHttpUrl("http://IP:4567/foldername/1234?abc=xyz")
.build()
.toUri()
.toURL();
