PHP使用POST传参失败

问题遇到的现象和发生背景

PHP使用POST传参失败
用PHP写了一个从数据库查询并显示结果的功能，但无法利用post得到搜索框输入的内容
这是搜索框↓

img


这是点击查询后出现的界面↓

img


就是$_post['serchsym']是空的

用代码块功能插入代码，请勿粘贴截图

查询界面的代码

html>
<html>
<head>
<title>主页title>
<link rel="stylesheet" href="main.css">
<meta charset="UTF-8">
head>

<body>
<div id="kuangjia" style="width:1200px">
<div id="header" style="background-color:#99d6ea80;text-align:center">⚠本站结果仅供参考，若有不适请您及时就医⚠div>
<div id="serchsym" style="background-color:#99d6ea25;height:300px;width:400px;float:left">
<form action="serchactone.php" method="post">
<div id="inputbox">

    <p>症状查询<input type="text" class="text" id ="serchsym" name="serchsym" style="height:30px;width:210px" placeholder="请输入症状，可以少但不可以多">p>
    <p1><input onclick="{location.href='serchactone.php'}" type="button" class="btn" value="查询">p1>
div>
div>

div>
body>
html>

点击搜索后的反应代码


    $conn = mysqli_connect("localhost", "root", "123456") or die("数据库链接错误");
    $db_selected=mysqli_select_db($conn,"SafeAskM");
    $serchsym = isset($_post['serchsym']) ? $_post['serchsym']:"";
    if(!empty($serchsym)) {
        $sql = "SELECT * FROM ask_m WHERE serch_key LIKE '%$serchsym%'";
        $query = mysqli_query($sql);
        print_r($sql);
        while($r=mysql_fetch_array($query)) {
            foreach($query as $valuea){
                foreach($valuea as $valuesb){
                    echo $valuesb;
                    echo " ";
                        };
                echo "
";    
                }
            }
        }
    else{
    echo"啊哦，页面不见了哦";}
    
?>
我的解答思路和尝试过的方法

尝试输出了一下$_post['serchsym']，是空的，不知道怎么改正

按下边的代码来写就可以了，您试一下。

 <?php
    $conn = mysqli_connect("localhost", "root", "123456") or die("数据库链接错误");
    $db_selected=mysqli_select_db($conn,"SafeAskM");
    $serchsym = isset($_POST['serchsym']) ? $_POST['serchsym'] : '';
    if(!empty($serchsym)) {
        $sql = "SELECT * FROM ask_m WHERE serch_key LIKE '%$serchsym%'";
        $query = mysqli_query($conn,$sql);
        print_r($sql);
        while($row = mysqli_fetch_array($query)) {
            $valuea = '';
            foreach($row as $valuea){
                /* foreach($valuea as $valuesb){
                    echo $valuesb;
                    echo " ";
                } */
                //echo "";  
                echo $valuea;                
            }
        }
    }else{
        echo"啊哦，页面不见了哦";
    }  
?>

<!DOCTYPE html>
<html>
<head>
<title>主页</title>
<link rel="stylesheet" href="main.css">
<meta charset="UTF-8">
</head>
<body>
    <div id="kuangjia" style="width:1200px">
        <div id="header" style="background-color:#99d6ea80;text-align:center">⚠本站结果仅供参考，若有不适请您及时就医⚠</div>
        <div id="serchsym" style="background-color:#99d6ea25;height:300px;width:400px;float:left">
            <form action="serchactone.php" method="post">
                <div id="inputbox">
                    <p>症状查询<input type="text" class="text" id ="serchsym" name="serchsym" style="height:30px;width:210px" placeholder="请输入症状，可以少但不可以多"></p>
                    <p1><input type="submit" class="btn" value="查询"></p1>
                </div>
            </form>
        </div>
    </div>
</body>
</html>

注意大小写，$_post['serchsym']改为$_POST['serchsym']

$_post改为$_POST看看

少了

 </form>

这个，应该如下：

<form action="serchactone.php" method="post">
<div id="inputbox">
    <p>症状查询<input type="text" class="text" id ="serchsym" name="serchsym" style="height:30px;width:210px" placeholder="请输入症状，可以少但不可以多"></p>
    <p1><input onclick="{location.href='serchactone.php'}" type="button" class="btn" value="查询"></p1>
</div>
</form>

body中的div和form标签对要一一配对呢
比如

<div>...</div>
<form>...</form>
<div>...</div>