java 如何从 HTTP“Referer”标头获取参数值？
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How to get the parameter values from HTTP "Referer" header?
提问by Thomas_
I got the url value using request.getHeader("Referer")e.g.:
我使用request.getHeader("Referer")例如获取 url 值：
string rr=request.getHeader("Referer");
<%= rr %>
I got the url as http://www.sun.com/questions?uid=21&value=gg
我得到了网址 http://www.sun.com/questions?uid=21&value=gg
Now I stored that url in as string, how do I get the value parameter value as uid=21and value=gg
现在我将该 url 存储为字符串，如何将 value 参数值作为uid=21和value=gg
回答by Brian Agnew
You need to:
你需要：
- take the string after the '?'
- split that on '&' (you can do this and the above by using the URL object and calling getQuery())
- You'll then have strings of the form 'x=y'. Split on the first '='
- URLDecodethe result parameter values.
- 在 '?' 后面取字符串
- 将其拆分为“&”（您可以通过使用 URL 对象并调用 getQuery() 来执行此操作和以上操作）
- 然后，您将拥有“x=y”形式的字符串。在第一个 '=' 处拆分
- URLDecode结果参数值。
This is all a bit messy, unfortunately.
不幸的是，这一切都有些混乱。
Why the URLDecode step ? Because the URL will be encoded such that '=' and '?' in parameter values won't confuse a parser.
为什么是 URLDecode 步骤？因为 URL 将被编码为 '=' 和 '?' in 参数值不会混淆解析器。
回答by Brian Agnew
This tutorialmight help a bit.
本教程可能会有所帮助。
What you need to do is parse the URL, then get the 'query' property, then parse it into name/value pairs.
您需要做的是解析 URL，然后获取 'query' 属性，然后将其解析为名称/值对。
So something like this:
所以像这样：
URL url = new URL(referer);
String queryStr = url.getQuery();
String[] params = queryStr.split("&");
for (String param: params) {
String key = param.substring(0, param.indexOf('=');
String val = param.substring(param.indexOf('=') + 1);
}
Disclaimer: this has not been tested, and you will need to do more error checking!
免责声明：这还没有经过测试，你需要做更多的错误检查！
回答by Joe23
Below an example how Apache Solr does it (SolrRequestParsers).
下面是 Apache Solr 如何做到这一点的示例 (SolrRequestParsers)。
/**
* Given a standard query string map it into solr params
*/
public static MultiMapSolrParams parseQueryString(String queryString)
{
Map<String,String[]> map = new HashMap<String, String[]>();
if( queryString != null && queryString.length() > 0 ) {
try {
for( String kv : queryString.split( "&" ) ) {
int idx = kv.indexOf( '=' );
if( idx > 0 ) {
String name = URLDecoder.decode( kv.substring( 0, idx ), "UTF-8");
String value = URLDecoder.decode( kv.substring( idx+1 ), "UTF-8");
MultiMapSolrParams.addParam( name, value, map );
}
else {
String name = URLDecoder.decode( kv, "UTF-8" );
MultiMapSolrParams.addParam( name, "", map );
}
}
}
catch( UnsupportedEncodingException uex ) {
throw new SolrException( SolrException.ErrorCode.SERVER_ERROR, uex );
}
}
return new MultiMapSolrParams( map );
}
Usage:
用法：
MultiMapSolrParams params = SolrRequestParsers.parseQueryString( req.getQueryString() );
And an example how Jersey does it (UriComponent):
以及 Jersey 如何做到这一点的示例 (UriComponent)：
/**
* Decode the query component of a URI.
*
* @param q the query component in encoded form.
* @param decode true of the query parameters of the query component
* should be in decoded form.
* @return the multivalued map of query parameters.
*/
public static MultivaluedMap<String, String> decodeQuery(String q, boolean decode) {
MultivaluedMap<String, String> queryParameters = new MultivaluedMapImpl();
if (q == null || q.length() == 0) {
return queryParameters;
}
int s = 0, e = 0;
do {
e = q.indexOf('&', s);
if (e == -1) {
decodeQueryParam(queryParameters, q.substring(s), decode);
} else if (e > s) {
decodeQueryParam(queryParameters, q.substring(s, e), decode);
}
s = e + 1;
} while (s > 0 && s < q.length());
return queryParameters;
}
private static void decodeQueryParam(MultivaluedMap<String, String> params,
String param, boolean decode) {
try {
int equals = param.indexOf('=');
if (equals > 0) {
params.add(
URLDecoder.decode(param.substring(0, equals), "UTF-8"),
(decode) ? URLDecoder.decode(param.substring(equals + 1), "UTF-8") : param.substring(equals + 1));
} else if (equals == 0) {
// no key declared, ignore
} else if (param.length() > 0) {
params.add(
URLDecoder.decode(param, "UTF-8"),
"");
}
} catch (UnsupportedEncodingException ex) {
// This should never occur
throw new IllegalArgumentException(ex);
}
}
回答by Thomas_
I combine two codes and result is below:
我结合了两个代码，结果如下：
StringBuffer url = request.getRequestURL();
if(request.getQueryString()!= null){
url.append("?");
url.append(request.getQueryString());
}
String url_param = url.toString();
String[] params = url_param.split("&");
String[][] param_key = new String[1320][3];
String[][] param_val = new String[1320][3];
String param = "";
String key_tmp = "";
String val_tmp = "";
y = 1;
for(x=1;x<params.length;x++)
{
param = params[x];
key_tmp = param.substring(0, param.indexOf("="));
param_key[x][y] = key_tmp;
y++;
val_tmp = param.substring(param.indexOf("=") + 1);
param_val[x][y] = val_tmp;
y=1;
}
You get two arrays. One contains key(name), second contains values.
你得到两个数组。一个包含键（名称），第二个包含值。
