如何從數據庫中選擇特定的數據並將其放入HTML表單中並更新？

Question

這是我的代碼，我需要幫助

我需要從數據庫中獲取數據，然後將表單放入表單後，我需要能夠將表​​單中的信息更改爲表單，然後能夠更新它

<?php 
print_r($_REQUEST); 


$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname = "artiest"; 

$dbh = new PDO('mysql:host=localhost;dbname=artiest', $username,$password); 
$id = $dbh->lastInsertId(); 


$name = $dbh->SELECT name FROM artist WHERE ID = $id; 
$email = $dbh->SELECT email FROM artist WHERE ID = $id; 
$gender = $dbh->SELECT gender FROM artist WHERE ID = $id; 
$comment = $dbh->SELECT comment FROM artist WHERE ID = $id; 
$website = $dbh->SELECT website FROM artist WHERE ID = $id; 

$dbh = null; 
?> 


<h2>Artiest Wijzigen</h2> 

<form method="post" action="add_artist.php"> 
Naam: <input type="text" name="name" value="<?php echo $name;?>"> 

<br><br> 
E-mail: <input type="text" name="email" value="<?php echo $email;?>"> 

<br><br> 
Website, artiest: <input type="text" name="website" value="<?php echo $website;?>"> 

<br><br> 
Extra toevoegingen:<br> <textarea name="comment" rows="5" cols="40"><?php echo $comment;?></textarea> 
<br><br> 
Geslacht: 
<input type="radio" name="gender" <?php if (isset($gender) && $gender=="female") echo "checked";?> value="female">Vrouw 
<input type="radio" name="gender" <?php if (isset($gender) && $gender=="male") echo "checked";?> value="male">Man 

<br><br> 
<input type="submit" name="submit" value="Veranderen"> 
</form> 

Answer 1

首先請閱讀如何構建和執行查詢。

http://php.net/manual/en/pdo.prepare.php

然後，我會建議得到只有一個查詢中的所有數據：

「SELECT * FROM藝術家WHERE ID =：ID」