mysqli_fetch_array不工作

Question

$connection = mysqli_connect($server, $user_name, $password); 

if ($connection) { 
    echo "Database Found "; 
} 
else { 
die("Database connection failed"); 
} 
$query = "SELECT Topic, PostsMsg, DatePosted FROM posta"; 

$myData = mysqli_query($connection, $query); 


echo "<table border=1 bordercolor=#ffffff> 
<tr><th>Topic</th><th>Comments</th><th>Date</th> 
</tr>"; 
while ($posts = mysqli_fetch_array($myData, MYSQLI_BOTH)){ 
    echo "<tr>"; 
    echo "<td>" . $posts['Topic'] . "</td>"; 
    echo "<td>" . $posts['PostsMsg'] . "</td>"; 
    echo "<td>" . $posts['DatePosted'] . "</td>"; 
    echo "</tr>"; 
} 

echo "</table>"; 
mysqli_close($connection); 

這是我的上述代碼被創建此錯誤消息：

警告：mysqli_fetch_array（）預計參數1被mysqli_result，布爾在C中給出：\瓦帕\

爲了防止 sql injection。在上面的代碼

Answer 1

使用事先準備好的聲明，你忘記指定數據庫name.So嘗試像這樣...

<?php 
$connection = mysqli_connect($server, $user_name, $password,$database); 

if ($connection) { 
    echo "Database Found "; 
} 
else { 
die("Database connection failed"); 
} 
$query = "SELECT Topic, PostsMsg, DatePosted FROM posta"; 

$myData = mysqli_query($connection, $query); 


echo "<table border=1 bordercolor=#ffffff> 
<tr><th>Topic</th><th>Comments</th><th>Date</th> 
</tr>"; 
while ($posts = mysqli_fetch_assoc($myData)){ 
    echo "<tr>"; 
    echo "<td>" . $posts['Topic'] . "</td>"; 
    echo "<td>" . $posts['PostsMsg'] . "</td>"; 
    echo "<td>" . $posts['DatePosted'] . "</td>"; 
    echo "</tr>"; 
} 

echo "</table>"; 
mysqli_close($connection);