参数/反向引用限制数量的mod_rewrite [英] mod_rewrite number of parameters/back-references limitation

问题描述

显然有多少反向引用您可以在htaccess的访问限制（9）的RewriteRules ..

Apparently there is a limitation (9) on how many backreferences you can access in htaccess RewriteRules..

但是，我们有需要超过9个参数，像这样一个重写规则：

But we have a RewriteRule that requires more than 9 parameters, something like this:

RewriteRule ^([^/]+)/b([0-9]+)(/a([0-9]+))?(/v([0-9]+))?(,([0-9]+))?(/(ajax|share))?(,complete)?$ /index.php?control=sites&site=brands&control_file=version_select&name=$1&campaign_id=$2&answer=$4&page=$8&option=$10&video_id=$6&page_type=research [L]

那么，当您尝试访问$ 10，恰好是它使用了$ 1，将一个0就在旁边。

So what happens when you try to access "$10", is that it uses $1 and puts a 0 right next to it.

有没有人能解决这个问题，任何的替代解决方案或什么？

Has anybody been able to solve this problem, any alternative solutions or anything?

谢谢！

推荐答案

其实，你并不需要捕捉的一切。写非捕获组（使用介绍：）为您不想重用的事情，这应该再给你一些喘息的空间。比较：

Actually, you don't need to capture everything. Write non-capturing groups (introduced with "?:") for the things you don't want to reuse, this should give you some breathing space again. Compare:

Yours:                       Mine:
-------------------------    ---------------------------
^([^/]+)/b          $1       ^([^/]+)/b            $1
([0-9]+)            $2       ([0-9]+)              $2
(/a([0-9]+))?       $4       (?:/a([0-9]+))?       $3
(/v([0-9]+))?       $6       (?:/v([0-9]+))?       $4
(,([0-9]+))?        $8       (?:,([0-9]+))?        $5
(/(ajax|share))?    $10!     (?:/(ajax|share))?    $6
(,complete)?$       $11!     (,complete)?$         $7

但与单纯的mod_rewrite，你不能去高于9反向引用。如果您需要更多，使用另一种 - 例如捕获只有在改写的重要组成部分，并与URL在您的应用程序的其余部分做一些字符串处理

But with mod_rewrite alone, you can't go higher than 9 back-references. If you need more, use an alternative - for example capturing only the most important parts in rewrite and do some string-processing with the rest of the URL in your app.

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