我现在已经跑到了墙上,这段代码在每条记录的末尾带出一个带有按钮的表格。一旦按下此按钮,则会执行一个功能,以将健康记录更新为-5。更新单个记录通过ID = ID
这对工作很好,但它影响所有行,我试图让它只通过ID触摸一个记录,但没有运气!如果你能帮到那就太好了!
php的
$sql="SELECT `id` , `FirstName` , `Health` FROM ajax_demo WHERE `id` = `id` LIMIT 0 , 30";
$result = mysql_query($sql);
if(isset($_REQUEST['submit']))
{
counterminus();
}
function counterminus()
{
$cmeter = $cmeter - 1;
$id = $_POST["id"];
$FirstName = $_POST["FirstName"];
mysql_query("UPDATE ajax_demo SET `Health` = `Health` - `Damage` WHERE id = {$id}");
Header("location:oo_test.php");
}
这是PHP /形式
<?php
echo
"<table border='1'>
<tr>
<th>id</th>
<th>Firstname</th>
<th>health</th>
</tr>";
while($row = mysql_fetch_row($result)) {
echo '<tr>';
foreach($row as $cell) {
echo "\n<td>$cell</td>";
}
echo '<td><form id="theForm" action="" method="POST" >
<input type="submit" name="submit" id="submit" value="Attack" />
<input type="hidden" name="'.$row[1].'" /></form></td></tr>';
echo "\n\n";
}?>
“这是表单”不是表单,而是PHP代码。真正的形式在浏览器中呈现 – 2013-02-21 14:05:36