我不能得到任何結果與此代碼顯示

Question

我得到「警告：mysqli_num_rows（）預計參數1被mysqli_result，布爾給予」當我使用此代碼：

$sql = "SELECT name, time, age FROM friends WHERE age = " .$age. "limit 0, 10"; 

$secquery = mysqli_query($conn, $sql); 

if (mysqli_num_rows($secquery) > 0){ 

    while($row = $secquery->fetch_assoc()) { 
    echo "This person === ".$row['name']. " matched your age, which is " .$row['age']. '<br>'; 

    } 
} 

我想匹配記錄在我的數據庫與輸入的記錄。我已經嘗試了一切。一點幫助？

Answer 1

$sql = "SELECT name, time, age FROM friends WHERE age='$age' limit 10"; 

Answer 2

您是否已經有連接字符串$ conn了？

<?php 
$conn=mysqli_connect("localhost","my_user","my_password","my_db"); 
// Check connection 
if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    }