如何从数据库中选择特定数据并将其放入带有更新的HTML表单中?

时间:2016-09-22 15:30:10

标签: php html mysql forms pdo

这是我的代码,我需要帮助

我需要从数据库中获取数据,然后将其放入表单中,在表单之后我需要能够更改放入表单中的信息,然后才能更新它

<?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>

1 个答案:

答案 0 :(得分:0)

首先请阅读如何构建和执行查询。

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

然后,我建议只在一个查询中获取所有这些数据:

&#34; SELECT * FROM artist WHERE ID =:id&#34;