php - 重写搜索选项的 url

Question

我想重写搜索选项的网址..目前我的搜索选项工作正常..但我想自定义网址和这是我的 html 脚本

<form name="search2" method="post" action="my-city.php">      
<select name="countryname">
<option value="usa">usa</option>
<option value="aus">aus</option>
</select>

<select name="num">
<option value="1">1</option>
<option value="2">2</option>
</select>

<select name="day">
<option value="1">1</option>
<option value="2">2</option>
</select>
<input type="submit" value="search"/>

当前结果: https://mydomain.com/ezhil/mysite/code/my-city.php

预期结果(例如，如果我选择countryname=usa、num=1、day=2网址应该是这样的): https://mydomain.com/ezhil/mysite/code/my-city/usa/1/2

Answer 1

您可以在 my-city.php 顶部添加以下内容:

<?php

if(array_key_exists("countryname",$_POST) and array_key_exists("num",$_POST) and array_key_exists("day",$_POST))
{
    preg_match("/^([a-z]+)$/",$_POST["countryname"],$countryname);
    $num = intval($_POST["num"]);
    $day = intval($_POST["day"]);
    if($contryname[1] != "" and $num > 0 and day > 0)
        Header("Location: http://mydomain.com/ezhil/mysite/code/my-city/".$countryname[1]."/".$num."/".$day);
}
?>

另外，您需要按如下方式修复 html 代码:

<form name="search2" method="post" action="my-city.php">      
<select name="countryname">
<option value="usa">usa</option>
<option value="aus">aus</option>
</select>

<select name="num">
<option value="1">1</option>
<option value="2">2</option>
</select>

<select name="day">
<option value="1">1</option>
<option value="2">2</option>
</select>
<input type="submit" value="search"/>