我创建PHP搜索表单我要显示在搜索变量名的URL我的URL像这样的http://本地主机/ ZBLOG /结果/ 1 /我想这样的网址http://localhost/zblog/results/1/xxcxcxc这是我的代码<?php $search = $_POST["search"];?><form action="<?php echo $url; ?>results/1/<?php echo $search; ?>" method="post" name="search" id="searchthis" style="display:inline;"><input id="search-box" name="search" size="40" type="text" placeholder="what are you looking for............"/> <input id="search-btn" value="search" type="submit"/></form>
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UYOU
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您需要将form方法更改为method =“ get”并检查$ search获得正确的值。
<?php
$search = $_GET["search"];
?>
<form action="<?php echo $url; ?>results/1/<?php echo $search; ?>" method="get" name="search" id="searchthis" style="display:inline;">
<input id="search-box" name="search" size="40" type="text" placeholder="what are you looking for............"/>
<input id="search-btn" value="search" type="submit"/>
</form>
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