PHP - Echo'ing东西而MySQL数据库

Question

我想，我可能需要帮助，我的PHP代码... 我想呼应在MySQL数据库中的信息，它给我的错误：Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /usr/local/ampps/www/php/db.php on line 18

我的代码是：

<?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(); 

?> 

将是很好，如果有人可以帮助我:)

编辑：

我固定它仅使用mysqli的，这是我worki ng代码：

// 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(); 

感谢您的快速建议。

Answer 1

您已使用mysqli进行连接设置，然后使用简单的mysql函数获取结果集和fetch_array逻辑。在这种情况下你需要统一。

我已将mysql_query()调用更改为mysqli_query()调用，并且同样调用调用mysqli_fetch_array()调用。

最终代码变成：

$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>"; 
}