使用PHP和MySQL更新帖子
[英]Updating post with PHP & MySQL
问题描述
我正在构建一种博客系统，但是在编辑系统中现有的“帖子”时遇到了麻烦。 它正在从数据库中收集数据，并显示它。 但是，当我单击更新按钮时，出现此错误。 我一直在寻找它，检查了几次我的代码，但是我什么都找不到。 这是错误。
SQLSTATE[HY093]: Invalid parameter number: number of bound variables does not match number of tokens
这是我的代码。
<?php //include config
require_once('../../includes/config.php');
//if not logged in redirect to login page
if(!$user->is_logged_in()){ header('Location: login.php'); }
?>
<!doctype html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Bewerk</title>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/2.2.3/jquery.js"></script>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/font-awesome/4.6.2/css/font-awesome.min.css">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" integrity="sha384-1q8mTJOASx8j1Au+a5WDVnPi2lkFfwwEAa8hDDdjZlpLegxhjVME1fgjWPGmkzs7" crossorigin="anonymous">
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js" integrity="sha384-0mSbJDEHialfmuBBQP6A4Qrprq5OVfW37PRR3j5ELqxss1yVqOtnepnHVP9aJ7xS" crossorigin="anonymous"></script>
<link rel="stylesheet" href="../style/normalize.css">
<link rel="stylesheet" href="../style/main.css">
<link rel="stylesheet" href="../style/login.css">
</head>
<body>
<?php
//check for any errors
if(isset($error)){
foreach($error as $error){
echo $error.'<br />';
}
}
try {
$stmt = $db->prepare('SELECT patient_id, voornaam, achternaam, leeftijd, lengte, gewicht, foto_url FROM patienten WHERE patient_id = :patient_id') ;
$stmt->execute(array(':patient_id' => $_GET['id']));
$row = $stmt->fetch();
} catch(PDOException $e) {
echo $e->getMessage();
}
?>
<?php
//if form has been submitted process it
if(isset($_POST['submit'])){
$_POST = array_map( 'stripslashes', $_POST );
//collect form data
extract($_POST);
if(!isset($error)){
try {
//insert into database
$stmt = $db->prepare('UPDATE patiënten SET patient_id, voornaam = :voornaam, achternaam = :achternaam, leeftijd = :leeftijd, lengte = :lengte, gewicht = :gewicht, foto_url = :foto_url WHERE patient_id = :patient_id') ;
$stmt->execute(array(
':patient_id' => $patient_id,
':voornaam' => $voornaam,
':achternaam' => $achternaam,
':leeftijd' => $leeftijd,
':lengte' => $lengte,
':gewicht' => $gewicht,
':foto_url' => $foto_url
));
//redirect to index page
header('Location: index.php?action=updated');
exit;
} catch(PDOException $e) {
echo $e->getMessage();
}
}
}
?>
<div class="row">
<div class="col-md-2 col-md-offset-5">
<img src="../style/images/logo.png">
</div>
</div>
<div class="keuze_link row">
<p><b><?php echo $row['voornaam'];?> <?php echo $row['achternaam'];?></b> bewerken</p>
</div>
<div class="container">
<form class="toevoegen" action="" method="post">
<input type="text" name="voornaam" value="<?php echo $row['voornaam'];?>">
<br>
<input type="text" name="achternaam" value="<?php echo $row['achternaam'];?>">
<br>
<input type="text" name="leeftijd" value="<?php echo $row['leeftijd'];?>">
<br>
<input type="text" name="lengte" value="<?php echo $row['lengte'];? >">
<br>
<input type="text" name="gewicht" value="<?php echo $row['gewicht'];?>">
<br>
<input type="text" name="foto_url" value="<?php echo $row['foto_url'];?>">
<br>
<input class="button" type="submit" name="submit" value="Updaten!">
</form>
</div>
</body>
</html>
如果你们想了解更多或看到更多代码，我想听听。 提前致谢。
编辑。 进行更新。
//insert into database
$stmt = $db->prepare('UPDATE patienten SET patient_id = :patient_id, voornaam = :voornaam, achternaam = :achternaam, leeftijd = :leeftijd, lengte = :lengte, gewicht = :gewicht, foto_url = :foto_url WHERE patient_id = :patient_id') ;
$stmt->execute(array(
'patient_id' => $patient_id,
':voornaam' => $voornaam,
':achternaam' => $achternaam,
':leeftijd' => $leeftijd,
':lengte' => $lengte,
':gewicht' => $gewicht,
':foto_url' => $foto_url
));
最新表格更新
<form class="toevoegen" action="" method="post">
<input type="hidden" name="voornaam" value="<?php echo $row['patient_id'];?>">
<input type="text" name="voornaam" value="<?php echo $row['voornaam'];?>">
<br>
<input type="text" name="achternaam" value="<?php echo $row['achternaam'];?>">
<br>
<input type="text" name="leeftijd" value="<?php echo $row['leeftijd'];?>">
<br>
<input type="text" name="lengte" value="<?php echo $row['lengte'];?>">
<br>
<input type="text" name="gewicht" value="<?php echo $row['gewicht'];?>">
<br>
<input type="text" name="foto_url" value="<?php echo $row['foto_url'];?>">
<br>
<input class="button" type="submit" name="submit" value="Updaten!">
</form>
2 个解决方案
解决方案1
0 2016-06-10 15:16:54
在更新查询中，您忘记了:patient_id进行绑定
是的，在列名前删除$。
解决方案2
0 已采纳 2016-06-10 15:19:24
$stmt->execute()函数中缺少：Patient_id。 请添加它，然后重试。
使用HTTP-POST方法将MySQL从xcode更新为PHP
声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.