警告：mysql_fetch_assoc（）預計參數1是資源，對象在C中給出：\ XAMPP \ htdocs中\ FontLibrary \的index.php上線14

Question

線14
而（$行= mysql_fetch_assoc（$結果0））

請幫忙。由於

下面是代碼

$HOST = 'localhost'; 
$USERNAME = 'root'; 
$PASSWORD = ''; 
$DB = 'Font Library'; 
$link = mysqli_connect($HOST, $USERNAME, $PASSWORD, $DB) or die(mysqli_connect_error()); 
if (!$link) { 
die(mysqli_error($link)); 
} 
$sql0="SELECT f.id, f.Font_Name, f.Font_Family 
     FROM font f 
     GROUP BY f.Font_Family 
     ASC"; 
$option = ''; 
$result0 = mysqli_query($link, $sql0) or die(mysqli_error($link)); 

while ($row = mysql_fetch_assoc($result0)) 
{ 
$option .= '<option value = " '.$row['f.id'].'">'.$row['f.Font_Family'].'</option>'; 
}  

Answer 1

你混合mysql和mysqli的函數調用

變化

while ($row = mysql_fetch_assoc($result0)) 

到

while ($row = mysqli_fetch_assoc($result0)) 

Answer 2

=>。你不能使用GROUP BY ASC

所以更改

$sql0="SELECT f.id, f.Font_Name, f.Font_Family 
    FROM font f 
    GROUP BY f.Font_Family 
    ASC"; 

到

$sql0="SELECT f.id, f.Font_Name, f.Font_Family 
    FROM font f 
    GROUP BY f.Font_Family 
    "; 

或

$sql0="SELECT f.id, f.Font_Name, f.Font_Family 
    FROM font f 
    ORDER BY f.Font_Family 
    ASC"; 

=>。更改mysql到mysqli

while ($row = mysql_fetch_assoc($result0)) 

到

while ($row = mysqli_fetch_assoc($result0))