Question

当试图在PHP中创建一个表的编辑页面时，我在edit.php页面遇到问题，当我点击编辑表格行按钮时，它显然会将我带到正确的页面（编辑.php），然后当我输入已编辑的详细信息并提交它时，它会创建一个全新的表行条目，而不是更新我选择要更新的那个。

我已经检查过以确保将id正确设置为数据库中正确的表行，并且确实如此。我不知道为什么会这样做。任何帮助将不胜感激。

<?php require("manage_post.php"); ?>
<?php 
session_start();
if(!isset($_SESSION['userName'])){ //if login in session is not set
    header("Location: login.php");
    exit();
}
?> 

<?php

$con = mysql_connect("localhost","root",""); 
if (!$con) {
    die('Could not connect: ' . mysql_error()); 
}

mysql_select_db("cad", $con) 
    or die('Could not select database'); 

$query="SELECT `town`, `location`, `incident_type`, `time_date`, `admin`, `id` 
FROM `cad` 
WHERE `id` = 
$_GET[id]"; 

$result=mysql_query($query) 
  or die(mysql_error()); 

while( false!=($row=mysql_fetch_array($result)) )
{
    echo '<h1>Town name: ', htmlspecialchars($row['town']), '</td>';
}




$town=$row['town'] ;
$location= $row['location'] ;                   
$incident_type=$row['incident_type'] ;


if(isset($_POST['save']))
{   
    $town = $_POST['town'];
    $location = $_POST['location'];
    $incident_type = $_POST['incident_type'];

    mysql_query("UPDATE cad SET town ='{$town}', location ='{$location}',
         incident_type ='{$incident_type}' WHERE `id` = $_GET[id]") or die(mysql_error()); 

    echo "Saved! Redirecting back to the home page.";

}
mysql_close($con);

    $id=$_GET['id']; 
?>




<!DOCTYPE html>
<html>
<head>
<title>Edit Incident</title>
</head>

<body>
    <?php
    echo "<h1>You are editing incident number # $id</h1>";
?>
<form method="post">
<table>
    <tr>
        <td>Town</td>
        <td><input type="text" name="town" value="<?php echo $town; ?>"/></td>
    </tr>
    <tr>
        <td>Location</td>
        <td><input type="text" name="location" value="<?php echo $location; ?>"/></td>
    </tr>
    <tr>
        <td>Incident Type</td>
        <td><input type="text" name="incident_type" value="<?php echo $incident_type; ?>"/></td>
    </tr>
    <tr>
        <td>&nbsp;</td>
        <td><input type="submit" name="save" value="Save" /></td>
    </tr>
</table>

</body>
</html>

Answer 1

你必须使用如下的查询

mysql_query("UPDATE cad SET town ='{$town}', location ='{$location}',
     incident_type ='{$incident_type}' WHERE `id` = {$_GET['id']}") or die(mysql_error()); ;

Answer 2

添加新的隐藏密钥以发布

<input type="hidden" name="savedid" value="<?php echo $_GET['id']?>" />

并在更新sql中

$town = $_POST['town'];
$location = $_POST['location'];
$incident_type = $_POST['incident_type'];
$savedid = $_POST['savedid'];
mysql_query("UPDATE cad SET town ='{$town}', location ='{$location}',
     incident_type ='{$incident_type}' WHERE `id` = $savedid") or die(mysql_error()); 

echo "Saved! Redirecting back to the home page.";

如果安全性问题，您应该使用intval for theid

来过滤您的变量

MySQL更新查询使新表行而不是更新表行

2 个答案: