PHP在URL上显示搜索结果标题

Question

我正在创建PHP搜索表单，已将搜索变量名称添加到html表单，但是当我搜索这样的URL显示时，例如http：// localhost / zblog / results / 1 /时，URL上没有显示标题，我想获得这样的URL http： //本地主机/ zblog /结果/ 1 / xxxxxx这是我的代码 <form action="<?php echo $url; ?>results/1/<?php $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>

Answer 1

您可以通过JavaScript中的键侦听器来执行此操作

<script>

var myUrl = "<?php echo $url; ?>results/1/";

function updateAction(ele) {

    var txt = ele.value; //get the user input

    ele.parentElement.setAttribute("action",myUrl+txt);

}

</script>

<input onkeyup="updateAction(this);" id="search-box" name="search" size="40" type="text" placeholder="what are you looking for............"/>    

Answer 2

您的表单是通过post方法提交的。您需要从中获取搜索查询$_POST['search']，并将其分配给变量$search

<?php $search = $_POST['search'] ?>

<form action="<?php echo $url; ?>results/1/<?php $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>