如何編寫MYNAME

Question

正確的語法應該

$值=蘇瑞

<?php 

    $value = print($D->myname); 
    $year = 1980; 

?> 

什麼是正確的方法是什麼？

Answer 1

基於W3Schools的例子：

<?php 
$servername = "localhost"; 
$username = "username"; 
$password = "password"; 
$dbname = "myDB"; 

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

$sql = "SELECT name FROM YourTable"; 
$result = $conn->query($sql); 

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

Answer 2

讓我們假設你做了要求您數據庫與PDO。
執行查詢後，取它將其用作array。

$req = $bdd->prepare("YOUR QUERY STRING"); 
$req->execute(); 

$myarray = $req->fetch() 
// The result of the request is saved in this variable as an array 
// use $req->fetch(PDO::FETCH_ASSOC) if you want an associative array 

echo $myarray['my_value']; 
// my_value is your myname 

而且，你給函數print()到您的變量$value。你應該給你的變量一個值並打印：

$value = $myarray['my_value'] // I continue with the code I wrote above 
print($value); // will write your variable