Node.js url.parse 结果返回字符串
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Node.js url.parse result back to string
提问by Ben Humphreys
I am trying to do some simple pagination.
To that end, I'm trying to parse the current URL, then produce links to the same query, but with incremented and decremented pageparameters.
我正在尝试做一些简单的分页。为此，我试图解析当前的 URL，然后生成指向同一查询的链接，但使用递增和递减的page参数。
I've tried doing the following, but it produces the same link, without the new pageparameter.
我尝试执行以下操作，但它生成相同的链接，但没有新page参数。
var parts = url.parse(req.url, true);
parts.query['page'] = 25;
console.log("Link: ", url.format(parts));
The documentation for the URL moduleseems to suggest that formatis what I need but I'm doing something wrong.
URL 模块的文档似乎表明这format是我需要的，但我做错了什么。
I know I could iterate and build up the string manually, but I was hoping there's an existing method for this.
我知道我可以手动迭代和构建字符串，但我希望有一个现有的方法。
回答by Alex Turpin
If you look at the latest documentation, you can see that url.formatbehaves in the following way:
如果您查看最新的文档，您会发现它的url.format行为方式如下：
searchwill be used in place ofqueryquery(object; see querystring) will only be used ifsearchis absent.
search将用于代替queryquery(object; see querystring) 仅在search不存在时使用。
And when you modify query, searchremains unchanged and it uses it. So to force it to use query, simply remove searchfrom the object:
当您修改query, 时search保持不变并使用它。所以要强制它使用query，只需search从对象中删除：
var url = require("url");
var parts = url.parse("http://test.com?page=25&foo=bar", true);
parts.query.page++;
delete parts.search;
console.log(url.format(parts)); //http://test.com/?page=26&foo=bar
Make sure you're always reading the latest version of the documentation, this will save you a lot of trouble.
确保您始终阅读最新版本的文档，这将为您省去很多麻烦。
回答by Marcel M.
Seems to me like it's a bug in node. You might try
在我看来，这是 node.js 中的一个错误。你可以试试
// in requires
var url = require('url');
var qs = require('querystring');
// later
var parts = url.parse(req.url, true);
parts.query['page'] = 25;
parts.query = qs.stringify(parts.query);
console.log("Link: ", url.format(parts));
回答by ampersand
The other answer is good, but you could also do something like this. The querystringmodule is used to work with query strings.
另一个答案很好，但你也可以做这样的事情。该querystring模块用于处理查询字符串。
var querystring = require('querystring');
var qs = querystring.parse(parts.query);
qs.page = 25;
parts.search = '?' + querystring.stringify(qs);
var newUrl = url.format(parts);
回答by d1b1
To dry out code and get at URL variables without needing to require('url') I used:
为了在不需要 require('url') 的情况下干燥代码并获取 URL 变量，我使用了：
/*
Used the url module to parse and place the parameters into req.urlparams.
Follows the same pattern used for swagger API path variables that load
into the req.params scope.
*/
app.use(function(req, res, next) {
var url = require('url');
var queryURL = url.parse(req.url, true);
req.urlparams = queryURL.query;
next();
});
var myID = req.urlparams.myID;
This will parse and move the url variables into the req.urlparams variable. It runs early in the request workflow so is available for all expressjs paths.
这将解析 url 变量并将其移动到 req.urlparams 变量中。它在请求工作流的早期运行，因此可用于所有 expressjs 路径。
