尝试使用PHP从HTML更新MySQL；连接有效，但没有任何更新

Question

I'm trying to build an HTML form to update a MySQL database using PHP. 我正在尝试构建HTML表单以使用PHP更新MySQL数据库。 When I load the page, the table populates exactly as I intend for it. 当我加载页面时，表将完全按照我的预期进行填充。 However, when I change something in one of the text boxes and click the update button, the change doesn't get posted to the database. 但是，当我在其中一个文本框中更改某些内容并单击“更新”按钮时，所做的更改不会发布到数据库中。 It just reloads the database into the form/table right where I began. 它只是将数据库重新加载到我刚开始的表单/表中。 For reference, the name field is just a concatenation of first and last I use as the primary key/id. 作为参考， name字段只是我用作主键/ id的first和last的串联。 Where did I go wrong? 我哪里做错了？

<html>
<body>

<?php

define( 'DB_SERVER', '123.234.345.456' );
define( 'DB_USERNAME', 'subad' );
define( 'DB_PASSWORD', 'password' );
define( 'DB_NAME', 'membership' );
$link = mysqli_connect( DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_NAME);
if ( $link === false ) {die( "ERROR: Could not connect. " . mysqli_connect_error() );}
$sql = 'SELECT * FROM members WHERE county = "abc"';
$result = mysqli_query( $link, $sql )or die( "Error With $sql" );

?>

<?php

if(isset($_POST['update'])){
try{$pdo = new PDO("mysql:host=123.234.345.456;dbname=membership", "subad", "password");
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);} 
catch(PDOException $e){die("ERROR: Could not connect. " . $e->getMessage());}
try{
  $first = $_POST['first'];
  $last = $_POST['last'];
  $phone = $_POST['phone'];
  $email = $_POST['email'];

$sql = "UPDATE members SET first=:first, last=:last, phone=:phone, email=:email WHERE name=:name";

$stmt = $pdo->prepare($sql);

$stmt->bindValue(":first", $first, PDO::PARAM_STR);
$stmt->bindValue(":last", $last, PDO::PARAM_STR);
$stmt->bindValue(":phone", $phone, PDO::PARAM_STR);
$stmt->bindValue(":email", $email, PDO::PARAM_STR);
$stmt->bindValue(":name", $name, PDO::PARAM_STR);

$stmt->execute();}
catch(PDOException $e){die("ERROR: Could not able to execute $sql. " . $e->getMessage());}

}

unset($pdo);?>

<form method="post" action="<?php $_PHP_SELF ?>">
<table width="400" border="0" cellspacing="1" cellpadding="2">
<tr>

<?php while($row = mysqli_fetch_array($result)) {$first = $row['first']; $last = $row['last']; $email = $row['email']; $phone = $row['phone']; $name = $row['name']; ?>

<td width="100"></td>
<td><?=$name?></td>
</tr>
<tr>
<td width="100">First</td>
<td><input name="first" type="text" value="<?=$first?>"></td>
</tr>
<tr>
<td width="100">Last</td>
<td><input name="last" type="text" value="<?=$last?>"></td>
</tr>
<tr>
<td width="100">Email</td>
<td><input name="email" type="text" value="<?=$email?>"></td>
</tr>
<tr>
<td width="100">Phone</td>
<td><input name="phone" type="text" value="<?=$phone?>"></td>
</tr>

<?php } ?>
<tr>
<td>
<input name="update" type="submit" id="update" value="Update">
</td>
</tr>
</table>
</form>

</body>
</html>

Answer 1

This doesn't look right to me 这对我来说不合适

<form method="post" action="<?php $_PHP_SELF ?>">

I don't see where $_PHP_SELF is set and you're not actually echoing it out. 我看不到设置$_PHP_SELF位置，您实际上并没有将其回显。 Also you are probably thinking of $_SERVER["PHP_SELF"] but that is likely not what you want. 另外，您可能正在考虑$_SERVER["PHP_SELF"]但这可能不是您想要的。 Try just leaving action empty and it will post to the current url: 尝试将动作保留为空，它将发布到当前网址：

<form method="post" action="">

尝试使用PHP从HTML更新MySQL；连接有效，但没有任何更新

问题描述

1 个解决方案

解决方案1
0 2019-01-09 04:48:55

尝试使用PHP从HTML更新MySQL； 连接有效，但没有任何更新

问题描述

1 个解决方案

解决方案1 0 2019-01-09 04:48:55

尝试使用PHP从HTML更新MySQL；连接有效，但没有任何更新

解决方案1
0 2019-01-09 04:48:55