URL抛出URISyntaxException异常,请问怎么解决

如题所述

java.net.URISyntaxException的解决办法

近日在用HttpClient访问抓取汇率时,为了省力,直接采用
String url = "http://api.liqwei.com/currency/?exchange=usd|cny&count=1";
HttpClient client = new DefaultHttpClient();
HttpGet httpget = new HttpGet(url);
HttpResponse response = client.execute(httpget);

以前用这种方法都没有问题,但这次却报如下错误:
java.net.URISyntaxException: Illegal character in query at index 44

查找了一些网上资料,说地址中涉及了特殊字符,如‘|’‘&’等。所以不能直接用String代替URI来访问。必须采用%0xXX方式来替代特殊字符。但这种办法不直观。所以只能先把String转成URL,再能过URL生成URI的方法来解决问题。代码如下
URL url = new URL(strUrl);
URI uri = new URI(url.getProtocol(), url.getHost(), url.getPath(), url.getQuery(), null);
HttpClient client = new DefaultHttpClient();
HttpGet httpget = new HttpGet(uri);
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第1个回答  2016-09-02
近日在用HttpClient访问抓取汇率时,为了省力,直接采用
String url = "http://api.liqwei.com/currency/?exchange=usd|cny&count=1";
HttpClient client = new DefaultHttpClient();
HttpGet httpget = new HttpGet(url);
HttpResponse response = client.execute(httpget);

以前用这种方法都没有问题,但这次却报如下错误:
java.net.URISyntaxException: Illegal character in query at index 44

查找了一些网上资料,说地址中涉及了特殊字符,如‘|’‘&’等。所以不能直接用String代替URI来访问。必须采用%0xXX方式来替代特殊字符。但这种办法不直观。所以只能先把String转成URL,再能过URL生成URI的方法来解决问题。代码如下
URL url = new URL(strUrl);
URI uri = new URI(url.getProtocol(), url.getHost(), url.getPath(), url.getQuery(), null);
HttpClient client = new DefaultHttpClient();
HttpGet httpget = new HttpGet(uri);
第2个回答  2016-09-02
近日在用HttpClient访问抓取汇率时,为了省力,直接采用
String url = "http://api.liqwei.com/currency/?exchange=usd|cny&count=1";
HttpClient client = new DefaultHttpClient();
HttpGet httpget = new HttpGet(url);
HttpResponse response = client.execute(httpget);

以前用这种方法都没有问题,但这次却报如下错误:
java.net.URISyntaxException: Illegal character in query at index 44

查找了一些网上资料,说地址中涉及了特殊字符,如‘|’‘&’等。所以不能直接用String代替URI来访问。必须采用%0xXX方式来替代特殊字符。但这种办法不直观。所以只能先把String转成URL,再能过URL生成URI的方法来解决问题。代码如下
URL url = new URL(strUrl);
URI uri = new URI(url.getProtocol(), url.getHost(), url.getPath(), url.getQuery(), null);
HttpClient client = new DefaultHttpClient();
HttpGet httpget = new HttpGet(uri);
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