php中UPDATE的问题 - Problem with UPDATE in php

Problem with UPDATE in php

本文关键字：问题 UPDATE php | 更新日期: 2023-09-27

我的代码有什么问题?我可以显示文本从我的数据库，但更新不工作..

index . php

$i=0;
    while ($i < $num) {
    $first=mysql_result($result,$i,"FirstName");
    $last=mysql_result($result,$i,"LastName");
    $age=mysql_result($result,$i,"Age");
    echo "<b>$first $last</b><br>Age: $age<br>";
    $i++;
    }
    $query="SELECT * FROM persons WHERE  personID='$id'";
    $resulta=mysql_query($query);
    $list=mysql_numrows($resulta); 
$i=0;
while ($i < $list) {
$first=mysql_result($resulta,$i,"FirstName");
$last=mysql_result($resulta,$i,"LastName");
$age=mysql_result($resulta,$i,"Age");
++$i;
} 
?>
<form action="update.php">
<input type="hidden" name="ud_id" value="<?php echo "$id"; ?>">
First Name: <input type="text" name="ud_first" value="<?php echo "$first"?>"><br>
Last Name: <input type="text" name="ud_last" value="<?php echo "$last"?>"><br>
Age: <input type="text" name="ud_age" value="<?php echo "$age"?>"><br>
<input type="Submit" value="Update">
</form>

update.php

<?php
include_once('config.php');
$query="UPDATE persons SET FirstName='$ud_first', LastName='$ud_last', Age='$ud_age' WHERE personID='$ud_id'";
mysql_query($query);
?>

我想问题出在http://lt.php.net/manual/en/security.globals.php

所以你不能用$ud_id必须用$_POST['ud_id']

HTML:

<form action="update.php" method="post">

PHP:

$ud_id = mysql_real_escape_string($_POST['ud_id']);
...

我认为你唯一忘记的是在你的while函数中获取id，以及在update.php中需要使用$_GET的id我希望这能给你一个小小的开端。

为index . php

$i=0;
while ($i < $list) 
{
$id =mysql_result($resulta,$i,"id");
$first=mysql_result($resulta,$i,"FirstName");
$last=mysql_result($resulta,$i,"LastName");
$age=mysql_result($resulta,$i,"Age");
++$i;
}

为update.php

include_once('config.php');
$ud_id = $_GET['id'];
$query="UPDATE persons SET FirstName='$ud_first', LastName='$ud_last', Age='$ud_age' WHERE personID='$ud_id'";
if(!$query)
{
 die('Could not delete data: ' . mysql_error());
}
mysql_close($link);