php小白,各种谷歌,也试了好多stackoverflow上的代码,都只能正常display,无法update。代码如下:
<?php
define('DB_NAME', 'form');
define('DB_USER', 'root');
define('DB_PASSWORD', '123456');
define('DB_HOST', 'localhost');
$link = mysql_connect(DB_HOST, DB_USER DB_PASSWORD);
if (!$link)
{
die('could not connect: ' . mysql_error());
}
$db_selected = mysql_select_db(DB_NAME, $link);
if (!$db_selected)
{
die('Can\'t use ' .DB_NAME .':' . mysql_error());
}
$query = "SELECT * FROM articles";
$result = mysql_query($query) or die(mysql_error());
?>
<div class="content-holder">
<form action="" method="post">
<table border='1' cellpadding='10' id='ViewTable'>
<tr>
<th>title</th>
<th>id</th>
</tr>
<?php
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
$title = $row['title'];
?>
<tr>
<td>
<input type="textbox" class="TextAreaTitle" name="title" value="<?=$title?>"></input>
<input type="hidden" name="id" value="<?=$title?>"></input>
</td>
<td><?=$id?></td>
</tr>
<?php
}
?>
</table>
<input type="submit" name="update" class="submitlink" Value="update">
</form>
</div>
<?php
if(isset($_POST['update']))
{
for ($i=count($_POST['id']); $i--;)
{
$id = $_POST['title'][$i];
$title = $_POST['id'][$i];
mysql_query("UPDATE articles SET title= $title WHERE id= $id ");
}
}
?>

