如果页面路由，我如何从href获取id？

Question

我想从下面的链接中获取ID。它可以跳过页面路由，但我需要使用页面路由并打破$_GET['id']。有没有办法让我使用页面路由，或者我应该看看另一个解决方案？

on HOME.PHP 我写这个：

<h1 class="post-title"><a href="viewlink?id=<?php echo $row['postID'] ?>"><?php echo $row['postTitle'] ?></a></h1>

我从根目录路由所有页面，似乎没有给我

viewlink.php

echo $_GET['id'];

这就是我的页面路线

if (isset($_GET['page'])) {
   $page = $_GET['page'];
       if ($page == "index") {
            require_once "frontend/pages/index.php";
 } elseif ($page == "viewlink") {
            require_once "frontend/pages/viewlink.php";
        }
    }

Answer 1

您没有显示如何路由，因此无法告诉您如何获得基本page密钥。如果你做一些标准的重写方法，你可能会留下类似于此的$_SERVER数组。这些值中的任何一个都可能是您提取id，QUERY_STRING，REQUEST_URI或REDIRECT_QUERY_STRING等所需的内容：

Array
(
    [DOCUMENT_ROOT] => /var/http/htdocs/iwantcreative
    [DOMAIN_NAME] => www.example.com
    [DOMAIN_NAME_REAL] => example.com
    [DOMAIN_PATH] => /var/http/htdocs/iwantcreative
    [DOMAIN_USER] => 2948313.3268049
    [FCGI_ROLE] => RESPONDER
    [GATEWAY_INTERFACE] => CGI/1.1
    [HTTP_ACCEPT_LANGUAGE] => en-us
    [HTTP_CONNECTION] => close
    [HTTP_COOKIE] => PHPSESSID=123123123123123123aaaaabbbbccccc
    [HTTP_DNT] => 1
    [HTTP_HOST] => www.example.com
    [HTTP_UPGRADE_INSECURE_REQUESTS] => 1
    [HTTP_USER_AGENT] => Mozilla/5.0 (Macintosh; Intel Mac OS 7_6_2) AppleWebKit/604.4.7 (KHTML, like Gecko) Version/11.0.2 Safari/604.4.7
    [HTTP_X_FORWARDED_FOR] => 123.123.123.1
    [HTTP_X_FORWARDED_PROTO] => http
    [HTTP_X_REAL_IP] => 111.111.111.1
    [PHP_SELF] => /index.php
    [PHP_WRAPPER_VERSION] => 560
    [PKEY] => 123123123.321321321
    [QUERY_STRING] => viewlink/&id=100
    [RAILS_URI] => INVAL
    [REDIRECT_DOMAIN_NAME] => www.example.com
    [REDIRECT_DOMAIN_NAME_REAL] => example.com
    [REDIRECT_DOMAIN_PATH] => /var/http/htdocs/myrootfolder
    [REDIRECT_PKEY] => 123123123.321321321
    [REDIRECT_QUERY_STRING] => viewlink/&id=100
    [REDIRECT_RAILS_URI] => INVAL
    [REDIRECT_SCRIPT_URI] => http://www.example.com/viewlink/
    [REDIRECT_SCRIPT_URL] => /viewlink/
    [REDIRECT_STATUS] => 200
    [REDIRECT_URL] => /viewlink/
    [REMOTE_ADDR] => 111.111.111.1
    [REMOTE_PORT] => 57480
    [REQUEST_METHOD] => GET
    [REQUEST_TIME] => 1515127759
    [REQUEST_TIME_FLOAT] => 1515127759.7582
    [REQUEST_URI] => /viewlink/&id=100
    [SCRIPT_FILENAME] => /var/http/htdocs/myrootfolder/index.php
    [SCRIPT_NAME] => /index.php
    [SCRIPT_URI] => http://www.example.com/viewlink/
    [SCRIPT_URL] => /viewlink/
    [SERVER_ADDR] => 123.123.123.1
    [SERVER_ADMIN] => [no address given]
    [SERVER_NAME] => www.example.com
    [SERVER_PORT] => 80
    [SERVER_PROTOCOL] => HTTP/1.0
    [SERVER_SIGNATURE] => 
    [SERVER_SOFTWARE] => Apache
    [argc] => 1
    [argv] => Array
        (
            [0] => viewlink/&id=100
        )

Answer 2

看来你也有某种网址重写设置，你没有提到。通常这个链接：<a href="viewlink?...可能会导致404，因为它缺少.php扩展名。

你如何从<a href="viewlink?...到达路由部分？我猜你有一个错误 - 检查RewriteRule（apache），他们需要[QSA] - ＆＃34;查询字符串追加＆＃34;在行尾。查询字符串是＆＃34;？＆＃34;之后的任何内容在网址中，您的情况为?id=