下面的代码在通过网络(电子邮件)发送之前对 URL 进行编码:
private static String urlFor(HttpServletRequest request, String code, String email, boolean forgot) {
try {
URI url = forgot
? new URI(request.getScheme(), null, request.getServerName(), request.getServerPort(), createHtmlLink(),
"code="+code+"&email="+email+"&forgot=true", null)
: new URI(request.getScheme(), null, request.getServerName(), request.getServerPort(), createHtmlLink(),
"code="+code+"&email="+email, null);
String s = url.toString();
return s;
} catch (URISyntaxException e) {
throw new RuntimeException(e);
}
}
/**
* Create the part of the URL taking into consideration if
* its running on dev mode or production
*
* @return
*/
public static String createHtmlLink(){
if (GAEUtils.isGaeProd()){
return "/index.html#ConfirmRegisterPage;";
} else {
return "/index.html?gwt.codesvr=127.0.0.1:9997#ConfirmRegisterPage;";
}
}
问题在于生成的电子邮件如下所示:
http://127.0.0.1:8888/index.html%3Fgwt.codesvr=127.0.0.1:9997%23ConfirmRegisterPage;?code=fdc12e195d&email=demo@email.com
链接时,?
标记和 #
符号替换为 %3F
和 %23
从浏览器打开它不会打开,因为它不正确。
正确的做法是什么?
最佳答案
您需要组合网址的查询部分并将片段添加为正确的参数。
这样的事情应该有效:
private static String urlFor(HttpServletRequest request, String code, String email, boolean forgot) {
try {
URI htmlLink = new URI(createHtmlLink());
String query = htmlLink.getQuery();
String fragment = htmlLink.getFragment();
fragment += "code="+code+"&email="+email;
if(forgot){
fragment += "&forgot=true";
}
URI url = new URI(request.getScheme(), null, request.getServerName(), request.getServerPort(), htmlLink.getPath(),
query, fragment);
String s = url.toString();
return s;
} catch (URISyntaxException e) {
throw new RuntimeException(e);
}
}
关于java - 用 Java 对字符串 URL 进行编码,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17924653/