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php - 无法从 PHP 更新/更改 MySQL 中的数据

转载 作者:行者123 更新时间:2023-11-29 11:51:28 24 4
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我就快到了,但更新效果不佳,尤其是底部。

<?php

require('dbconnect.php');//Connects to the database
session_start();
$user_check=$_SESSION['login_user'];
$ses_sql=mysqli_query($link,"SELECT username FROM members WHERE username='$user_check'");
$row=mysqli_fetch_array($ses_sql,MYSQLI_ASSOC);
$loggedin_session=$row['username'];
if(!isset($loggedin_session))
{
header("Location: login.php");
}//To ensure that you must be logged in to access this page

?>
<html>
<head>
<title>Healing Food Form</title>
<meta charset="iso-8859-1"> <!--charset specifies characters available-->
<meta name="author" content="Klarenz Kristoffer M. Qui;ntildeones">
<meta name="description" content="form to update healing food">
<meta name="keywords" content="healing food,form">
</head>

<body>
<?php
$id=$_GET['hf_id'];
$query = "SELECT * FROM healingfood WHERE hf_id='$id'";

if(!mysqli_query($link,$query))
{
die("Sorry. There's a problem with the query.");
}

//stores the result of the query
$result = mysqli_query($link,$query);

while($record = mysqli_fetch_assoc($result))
{
$hf_id=$record['hf_id'];
$hf_title=$record['hf_title'];
$a_id=$record['a_id'];
$hf_image=$record['hf_image'];
$hf_description=$record['hf_description'];
$hf_benefits=$record['hf_benefits'];
$hf_source=$record['hf_source'];

?>
<form action="updatehealingfood.php?hf_id=<?php echo $record['hf_id']; ?>" method="POST">
<table id="container" align="center">
<caption>Update healing food</caption>
<tr>
<td>Title:</td>
<td><input name="hf_title" type="text" value="<?php echo $hf_title; ?>"><br></td>
</tr>
<tr>
<td>Author ID:</td>
<td><input name="a_id" type="text" value="<?php echo $a_id; ?>"><br></td>
</tr>
<tr>
<td>Image URL:</td>
<td><input name="hf_image" type="url" value="<?php echo $hf_image; ?>"><br></td>
</tr>
<tr>
<td>Description:</td>
<td><textarea name ="hf_description" rows="18" cols="60"><?php echo $hf_description; ?></textarea><br></td>
</tr>
<tr>
<td>Benefits:</td>
<td><input name="hf_benefits" type="text" value="<?php echo $hf_benefits; ?>"><br></td>
</tr>
<tr>
<td>Source:</td>
<td><input name="hf_source" type="text" value="<?php echo $hf_source; ?>"><br></td>
</tr>
<tr>
<td colspan="2" align="right"><input type="submit" name="update" value="Update Healing Food"></td>
</tr>
</table>
</form>

</body>
</html>
<?php
}

$id=$_GET['hf_id'];
if(isset($_POST['update']))
{
$hf_title=$_POST['hf_title'];
$a_id=$_POST['a_id'];
$hf_image=$_POST['hf_image'];
$hf_description=$_POST['hf_description'];
$hf_benefits=$_POST['hf_benefits'];
$hf_source=$_POST['hf_source'];
$query2="UPDATE healingfood SET hf_title='$hf_title', a_id='$a_id', hf_image='$hf_image', hf_description='$hf_description', hf_benefits='$hf_benefits', hf_source='$hf_source' WHERE hf_id='$id'";
$result2=mysql_query($query2) or die();
echo "Updated";
}

?>

当我应该更新数据时,数据保持不变。没有人改变。我不明白 $id=$_GET['hf_id']; .

我的错误是什么?

最佳答案

您正在混合 mysqli_*mysql_*

在第一部分中,您使用mysqli_query(),稍后您使用尚未连接到数据库的mysql_query()

坚持使用mysqli_*

更改:

$result2=mysql_query($query2) or die();

至:

$result2=mysqli_query($link, $query2) or die( "MySQL error: " . mysqli_error($link) );

关于php - 无法从 PHP 更新/更改 MySQL 中的数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33974386/

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