致命錯誤：調用上的空成員函數FETCH_ASSOC（）第9行

Question

我不明白

Непонимаю，вчемзаключаетсяошибка

<?php 

$mysqli = new mysqli ('localhost', 'root', '', 'UTS'); 
$mysqli->query("SET NAMES 'utf-8'"); 
$result_set = $mysqli->query ("SELECT * FROM 'product'"); 
$mysqli -> close(); 

function printResult ($result_set) { 
    while (($row = $result_set->fetch_assoc()) != false) { 
     print_r ($row); 
     echo "<br/>"; 
    } 
} 

Answer 1

你可以嘗試這樣的... 。你必須獲取該數組，你可以打印....

<?php 
$mysqli = new mysqli ('localhost', 'root', '', 'db_name'); 
$mysqli->query("SET NAMES 'utf-8'"); 
$result_set= mysqli_query($mysqli,"SELECT * FROM products"); 
$row=mysqli_num_rows($result_set); 
echo $row; 
while($row=mysqli_fetch_array($result_set)) 
{    
print_r($row); 
} 
?> 

Answer 2

1. 的$result_set是功能範圍之外。你不能在函數中訪問它。
2. 你從一個字符串不是一個表名

選擇嘗試以下變化：

<?php 



function printResult() { 
    $mysqli = new mysqli ('localhost', 'root', '', 'UTS'); 
    $mysqli->set_charset("utf-8"); //Better to do it this way 
    $result_set = $mysqli->query ("SELECT * FROM `product`"); //Not the ` instead of ' 

    while (($row = $result_set->fetch_assoc()) != false) { 
     print_r ($row); 
     echo "<br/>"; 
    } 
    $mysqli -> close(); 
} 

您可以從任何地方運行該功能。