Php 试图得到一个;找不到名称。”;我的while循环

Php 试图得到一个;找不到名称。”;我的while循环,php,html,while-loop,Php,Html,While Loop,我试图通过使用while循环为我的输出创建一个条件循环。这是我的输出: 警告:mysqli_num_rows()期望参数1是mysqli_result,资源在第26行的C:\xampp\htdocs\junhao\doLandingPage.php中给出 找不到名称 dbFunctions.php: $host = 'localhost'; $username = 'root'; //Change to your own one $password = ''; //Change to your

我试图通过使用while循环为我的输出创建一个条件循环。这是我的输出:

警告:mysqli_num_rows()期望参数1是mysqli_result,资源在第26行的C:\xampp\htdocs\junhao\doLandingPage.php中给出 找不到名称

dbFunctions.php:

$host = 'localhost';
$username = 'root'; //Change to your own one
$password = ''; //Change to your own one
$db = 'demo'; //Change to your own one

// Connect to the server
$connect = mysql_connect('localhost', 'root', '');

// Connect to the database
mysql_select_db('demo');
doLandingPage.php:

include 'dbFunctions.php';

$search = $_POST['search'];

// Query the database
$query = mysql_query("SELECT * FROM guests WHERE name = '$search'");
?>

<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
        <title>TITLE HERE.</title>
        <link rel="stylesheet" type="text/css" href="stylesheet.css">

        <!-- Foundation -->
        <link href="https://cdnjs.cloudflare.com/ajax/libs/foundation/6.1.2/foundation.min.css" rel="stylesheet" media="screen" />
        <link href="https://cdnjs.cloudflare.com/ajax/libs/foundation/6.1.2/foundation.min.js" media="screen">
    </head>    

    <body>

    <?php 
    if(mysqli_num_rows($query)>0){
        while($row = mysqli_fetch_array($query))
        {
            $id = $rows['id'];
            $name = $rows['name'];
            $table = $rows['table'];

            echo "<h1>$name</h1>"
            echo "<h2>$table</h2>"
        }
    }else{
        echo "<tr><td colspan='4'>Name not found.";
    }
    ?>

    </body>
</html>
包括“dbFunctions.php”;
$search=$_POST['search'];
//查询数据库
$query=mysql_query(“从来宾中选择*,其中name='$search'”);
?>
标题在这里。

您必须首先运行查询。您正试图在常规字符串上使用
mysqli\u num\u rows()

您混合了mysql和mysqli,因为两者不同。请注意,mysql不推荐使用mysqli

请尝试以下代码:-

<?php 

         if(mysql_num_rows($query)>0){

                while($row = mysql_fetch_array($query))
                {
                    $id = $row['id'];
                    $name = $row['name'];
                    $table = $row['table'];

                    echo "<h1>$name</h1>"
                    echo "<h2>$table</h2>"

                }
          }else{
                echo "<tr><td colspan='4'>Name not found.';
          }

    ?>

混合
mysql
&
mysqli