Question

先决条件

Apache Tomcat 7
Spring 3.2.11.RELEASE
Apache Camel 2.14.1
Camel HTTP端点（<artifactId>camel-http</artifactId>）

问题

目前我使用以下代码将POST-Parameters设置为邮件正文。 camel HTTP-Component读取参数并发送它。

.setHeader(Exchange.HTTP_METHOD, constant(HttpMethods.POST.name()))
.setHeader(Exchange.CONTENT_TYPE, constant("application/x-www-form-urlencoded; charset: UTF-8"))
.setHeader(Exchange.CONTENT_ENCODING, constant("UTF-8"))
.setBody("parameter1=a&parameter2=b")

问题在于某些参数本身就是URL。所以像这样的东西应该作为POST-Request发送：

postparameter1=a&postparameter2=http://www.`...`.com?urlparam1=value1&urlparam2=value2&postparameter3=b

我的问题是如何发送“http://www。...。com？urlparam1 = value1＆amp; urlparam2 = value2”作为postparameter2的值。

提前致谢。

此致

最高

Answer 1

如上所述，以下内容适用于我。这个想法是解析一个给定的url拳头并在之后再次编码。这避免了双重编码。

import java.io.UnsupportedEncodingException;
import java.net.*;

public static String getEncodedURL(String urlString) {
    final String encodedURL;
    try {
        String decodedURL = URLDecoder.decode(urlString, "UTF-8");
        URL url = new URL(decodedURL);
        URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
        final URL urlFromDecoding = uri.toURL();
        encodedURL = URLEncoder.encode(urlFromDecoding.toString(), "UTF-8");
    } catch (UnsupportedEncodingException e) {
        ...
    } catch (MalformedURLException e) {
        ...
    } catch (URISyntaxException e) {
        ...
    }
    return encodedURL;
}

Answer 2

我在这里提到了如何通过camel HTTP发送字符串消息，我希望这对你有所帮助。另一个是我们需要添加基本的休息认证。用户名和密码是你的应用程序休息认证凭证。

from("seda:httpSender")  
    .log("Inside Http sender")
    .process(new Processor(){
        @Override
        public void process(Exchange exchange) throws Exception {
            // Camel will populate all request.parameter and request.headers, 
            // no need for placeholders in the "from" endpoint
            String content = exchange.getIn().getBody(String.class);      

            System.out.println("Outbound message string : "+content);

            // This URI will override http://dummyhost
            exchange.getIn().setHeader(Exchange.HTTP_URI, "http://localhost:9090/httpTest");

            // Add input path. This will override the original input path.
            // If you need to keep the original input path, then add the id to the 
            // URI above instead
         //   exchange.getIn().setHeader(Exchange.HTTP_PATH, id);

            // Add query parameter such as "?name=xxx"
            exchange.getIn().setHeader(Exchange.HTTP_QUERY, "outboundMessage="+content);  
            exchange.getIn().setHeader(Exchange.HTTP_METHOD, "POST");
        }
    })
    .doTry()
    .log("Message added as a parameter")
    .to("http4://localhost:9090/httpTest?authMethod=Basic&authPassword=admin&authUsername=admin")
    .log("HTTP message transfer success")
    .doCatch(Exception.class)
    .log("HTTP message transfer failed")
    .end();

Camel HTTP端点：如何将URL-String设置为POST参数

2 个答案: