javascript 为什么我不能在 node.js 中对这个 URL 进行 url 编码？
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How come I can't url encode this URL in node.js?
提问by TIMEX
$node
querystring = require('querystring')
var dict = { 'q': 'what\'s up' };
var url = 'http://google.com/?q=' + querystring.stringify(dict);
url = encodeURIComponent(url);
console.log(url);
The result is this:
结果是这样的：
"http://google.com/?q=q=what's%20up"
Notice how the single quote is not encoded correctly. Is there something wrong with the node.js module?
注意单引号是如何编码不正确的。node.js 模块有问题吗？
回答by Gumbo
The 'is allowed in plain in the URI query. Here are the corresponding production rules for the URI query as per RFC 3986:
的'是允许在普通的URI的查询。以下是根据 RFC 3986 的 URI 查询的相应生产规则：
query = *( pchar / "/" / "?" ) pchar = unreserved / pct-encoded / sub-delims / ":" / "@" unreserved = ALPHA / DIGIT / "-" / "." / "_" / "~" pct-encoded = "%" HEXDIG HEXDIG sub-delims = "!" / "$" / "&" / "'" / "(" / ")" / "*" / "+" / "," / ";" / "="
query = *( pchar / "/" / "?" ) pchar = unreserved / pct-encoded / sub-delims / ":" / "@" unreserved = ALPHA / DIGIT / "-" / "." / "_" / "~" pct-encoded = "%" HEXDIG HEXDIG sub-delims = "!" / "$" / "&" / "'" / "(" / ")" / "*" / "+" / "," / ";" / "="
As you can see, sub-delimscontains a plain '. So the result is valid.
如您所见，sub-delims包含一个普通的'. 所以结果是有效的。
回答by Marek Sebera
it is encoded correctly, if you type the same query into google sear field manually you will get this address:
它被正确编码，如果您手动在 google sear 字段中键入相同的查询，您将获得以下地址：
http://www.google.cz/#hl=cs&cp=8&gs_id=u&xhr=t&q=what's+up&pf=p&sclient=psy&site=&source=hp&pbx=1&oq=what's+u&aq=0&aqi=g5&aql=&gs_sm=&gs_upl=&bav=on.2,or.r_gc.r_pw.&fp=792ecf51920895b2&biw=1276&bih=683
note that &q=what's+up&part
注意那&q=what's+up&部分
and encodeURIComponentis not a Node.js module, but part of standard javascript library
并且encodeURIComponent不是 Node.js 模块，而是标准 javascript 库的一部分
https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/encodeURIComponent
https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/encodeURIComponent
manual workaround:
手动解决方法：
$node
querystring = require('querystring')
var dict = { 'q': 'what\'s up' };
var url = 'http://google.com/?q=' + querystring.stringify(dict);
url = encodeURIComponent(url);
url = url.replace(/'/g,"%27");
