上次提出了一個(gè)ajax下載中文編碼問(wèn)題，這次有遇到了ajax的上傳中文亂麻問(wèn)題
繼續(xù)bs各種編碼：）
遇到問(wèn)題：在使用xhr發(fā)出請(qǐng)求時(shí)，請(qǐng)求的url中參數(shù)帶中文，這時(shí)在后臺(tái)解析request得不到中文
如：
request = new ActiveXObject("Microsoft.XMLHTTP");
request.open("GET", url);中我的url需要帶中文參數(shù)
直接加參數(shù)是在后臺(tái)解析不出來(lái)的，所以我使用方法對(duì)url進(jìn)行編碼轉(zhuǎn)換
request?=?new?ActiveXObject("Microsoft.XMLHTTP");
request.open("GET",?uriFunc());
//省略了對(duì)象定義
//
..
function?uriFunc()?{
????var?separator?=?"?";
????if?(uri.indexOf("?")?>=?0)
????????separator?=?"&";????????
????return?uri?+?separator?+?paramName?+?"="?+?escape(inputField.value,'utf-8');
??}
然后就要在后臺(tái)進(jìn)行解碼了
以為沒(méi)法直接從request中取到編碼字符串，所以使用方法先取出url，然后再取出url的參數(shù)進(jìn)行解碼
取出url，以及取參數(shù)方法
String?exfirstword?=?getUrlParameter(unescape(request.getQueryString()),"excludeuser");?
?
?public?String?getUrlParameter(String?str,String?targetstr){
??????????
??????????String[]?stringarray?=?str.split("&")??;
????????????
????????????for(int?i=0;i<stringarray.length;i++){
???????????????if(stringarray[i].startsWith(targetstr)){
???????????????????stringarray=stringarray[i].split("=");????
???????????????????break;
???????????????}
????????????}
????????????return?stringarray[1];
??????}
?//解碼方法
?//以下是對(duì)js的escape進(jìn)行解碼
??????private?final?static?byte[]?val?=?{?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x00,?0x01,
??????????0x02,?0x03,?0x04,?0x05,?0x06,?0x07,?0x08,?0x09,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x0A,?0x0B,?0x0C,?0x0D,?0x0E,?0x0F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x0A,?0x0B,?0x0C,?0x0D,?0x0E,?0x0F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,
??????????0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F,?0x3F?};
?????
??????/**
???????*?解碼?說(shuō)明：本方法保證?不論參數(shù)s是否經(jīng)過(guò)escape()編碼，均能得到正確的“解碼”結(jié)果
???????*
???????*?@param?s
???????*?@return
???????*/
??????public?static?String?unescape(String?s)?{
??????StringBuffer?sbuf?=?new?StringBuffer();
??????int?i?=?0;
??????int?len?=?s.length();
??????while?(i?<?len)?{
??????int?ch?=?s.charAt(i);
??????if?('A'?<=?ch?&&?ch?<=?'Z')?{?//?'A'..'Z'?:?as?it?was
??????sbuf.append((char)?ch);
??????}?else?if?('a'?<=?ch?&&?ch?<=?'z')?{?//?'a'..'z'?:?as?it?was
??????sbuf.append((char)?ch);
??????}?else?if?('0'?<=?ch?&&?ch?<=?'9')?{?//?'0'..'9'?:?as?it?was
??????sbuf.append((char)?ch);
??????}?else?if?(ch?==?'-'?||?ch?==?'_'?//?unreserved?:?as?it?was
??????||?ch?==?'.'?||?ch?==?'!'?||?ch?==?'~'?||?ch?==?'*'
??????||?ch?==?'\''?||?ch?==?'('?||?ch?==?')')?{
??????sbuf.append((char)?ch);
??????}?else?if?(ch?==?'%')?{
??????int?cint?=?0;
??????if?('u'?!=?s.charAt(i?+?1))?{?//?%XX?:?map?to?ascii(XX)
??????cint?=?(cint?<<?4)?|?val[s.charAt(i?+?1)];
??????cint?=?(cint?<<?4)?|?val[s.charAt(i?+?2)];
??????i?+=?2;
??????}?else?{?//?%uXXXX?:?map?to?unicode(XXXX)
??????cint?=?(cint?<<?4)?|?val[s.charAt(i?+?2)];
??????cint?=?(cint?<<?4)?|?val[s.charAt(i?+?3)];
??????cint?=?(cint?<<?4)?|?val[s.charAt(i?+?4)];
??????cint?=?(cint?<<?4)?|?val[s.charAt(i?+?5)];
??????i?+=?5;
??????}
??????sbuf.append((char)?cint);
??????}?else?{?//?對(duì)應(yīng)的字符未經(jīng)過(guò)編碼
??????sbuf.append((char)?ch);
??????}
??????i++;
??????}
??????return?sbuf.toString();
??????}
這時(shí)候得到的
exfirstword 就是正確的中文了
問(wèn)題解決
posted on 2006-05-25 20:49
rocket 閱讀(1957)
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