PHP - Echo'ing東西而MySQL數據庫

Question

我想，我可能需要幫助，我的PHP代碼... 我想呼應在MySQL數據庫中的信息，它給我的錯誤：Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /usr/local/ampps/www/php/db.php on line 18

我的代碼是：

<?php 
$servername = "blabla"; //changed, connecting works 
$username = "blabla"; 
$password = "blabla"; 
$database = "blabla"; 

// Create connection 
$conn = new mysqli($servername, $username, $password, $database); 
// Check connection 
if ($conn->connect_error) { 
die("Connection failed: " . $conn->connect_error); 
} else { 
echo "Connection success \n"; 
} 

$sql = mysql_query("SELECT * FROM Schueler"); 
while($data = mysql_fetch_array($sql)) //This is line 18... 
{ 
echo "ID: " . $data['ID'] . " Vorname: " . $data['Vorname'] . " Nachname: " . $data['Nachname'] . " Klasse: " . $data['Klasse'] . "<br>"; 
} 

$conn->close(); 

?> 

將是很好，如果有人可以幫助我:)

編輯：

我固定它僅使用mysqli的，這是我worki ng代碼：

// Create connection 
$conn = new mysqli($servername, $username, $password, $database); 
// Check connection 
if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} else { 
    echo "Connection success \n"; 
} 

$sql = "SELECT ID, Vorname, Nachname FROM Schueler"; 
$result = $conn->query($sql); 

if ($result->num_rows > 0) { 
    // output data of each row 
    while($row = $result->fetch_assoc()) { 
     echo "id: " . $row["ID"]. " - Name: " . $row["Vorname"]. " " . $row["Nachname"]. "<br>"; 
    } 
} else { 
    echo "0 results"; 
} 
$conn->close(); 

感謝您的快速建議。

Answer 1

您已使用mysqli進行連接設置，然後使用簡單的mysql函數獲取結果集和fetch_array邏輯。在這種情況下你需要統一。

我已將mysql_query()調用更改爲mysqli_query()調用，並且同樣調用調用mysqli_fetch_array()調用。

最終代碼變成：

$sql = mysqli_query("SELECT * FROM Schueler", $conn); 
while($data = mysqli_fetch_array($sql)) 
{ 
echo "ID: " . $data['ID'] . " Vorname: " . $data['Vorname'] . " Nachname: " . $data['Nachname'] . " Klasse: " . $data['Klasse'] . "<br>"; 
}