div 'hidden-table'이 표시되지 않는 경우 'style : display : none'을 제거하면 표시됩니다. 올바른 데이터가 포함 된 표가 표시되어 작업을 시도하고 있습니다. 메아리에서 팝업을 꺼내서 별도로 표시하는 것입니다. div가 onmouseover와 함께 표시되지 않음
문제 만 발생했을 때의를 에코에서하고 'onmousover'를 중심 보인다 보인다전체 페이지 코드 :
<html>
<table border='0' cellpadding='0' cellspacing='0' class="center2">
<tr>
<td width='60'><img src="images/box_tl.png"></td>
<td style="background: url(images/box_tm.png)" align="center"><img src="images/news.png"></td>
<td width='25'><img src="images/box_tr.png"></td>
</tr>
<tr>
<td style="background: url(images/box_ml.png)"><h2>.</h2></td>
<td style="background: url(images/box_mm.png)">
<?php
include 'connect.php';
$query = mysql_query("SELECT * FROM tbl_img") or die(mysql_error());;
echo "<table border='0' cellpadding='1' cellspacing='1' width'90%' id='1' class='tablesorter'><thead>";
echo "<tr> <th> </th> <th>Mob Name</th> <th>Id</th> <th>Health</th> <th>Body</th> <th>Effects</th> <th>Spawn</th></tr></thead><tbody>";
// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array($query)) {
$mob_id = $row['mob_id'];
$mob = $row['mob'];
$body = $row['body'];
$mob_name = $row['mob_name'];
$health = $row['health'];
$level = $row['level'];
// Print out the contents of each row into a table
echo "<tr><td>";
echo "<img src='/testarea/include/mobs/$mob' />";
echo "</td><td>";
echo $mob_name;
echo "</td><td>";
echo $level;
echo "</td><td>";
echo $health;
echo "</td><td>";
echo
"
<a onmouseover='popup($('#hidden-table').html(), 400);' href=''><img src='/testarea/include/mobs/dead/$body' /></a>
";
echo "
<div id='hidden-table' style='display:none;'>
<table border='0' cellpadding='0' cellspacing='0' class='center3'>
<tr>
<td width='14'><img src='images/info_tl.png'></td>
<td style='background: url(images/info_tm.png)' align='center'></td>
<td width='14'><img src='images/info_tr.png'></td>
</tr>
<tr>
<td style='background: url(images/info_ml.png)'><h2>.</h2></td>
<td style='background: url(images/info_mm.png)'>
";
$query2 = mysql_query("SELECT * FROM tbl_drop WHERE mob_name='$mob_name'") or die(mysql_error());
echo "<table border='0' cellpadding='1' cellspacing='1' width='250' id='2' class='tablesorter'><thead>";
echo "<tr> <th> </th> <th>Item Name</th> <th>Qty</th></thead><tbody>";
// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array($query2)) {
$id = $row['id'];
$item_img = $row['item_img'];
$qty = $row['qty'];
$item_name = $row['item_name'];
// Print out the contents of each row into a table
echo "<tr><td width='50'>";
echo "<img src='/testarea/item/$item_img' />";
echo "</td><td width='150'>";
echo $item_name;
echo "</td><td width='50'>";
echo $qty;
echo "</td></tr>";
}
echo "</tbody></table>";
echo "
</td>
<td style='background: url(images/info_mr.png)'><h2>.</h2></td>
</tr>
<tr>
<td width='14'><img src='images/info_bl.png'></td>
<td style='background: url(images/info_bm.png)' align='center'><h2>.</h2></td>
<td width='14'><img src='images/info_br.png'></td>
</tr>
</table>
</div>"
;
echo "</td><td>";
echo "test";
echo "</td><td>";
echo "test";
echo "</td></tr>";
}
echo "</tbody></table>";
?>
</td>
<td style="background: url(images/box_mr.png)"><h2>.</h2></td>
</tr>
<tr>
<td width='60'><img src="images/box_bl.png"></td>
<td style="background: url(images/box_bm.png)" align="center"><h2>.</h2></td>
<td width='25'><img src="images/box_br.png"></td>
</tr>
</table>
</html>
javascript 함수'popup'가 정의됩니다. 여기서? – Hamish
코드가 너무 많아서이 공격을 알 수 없습니다. –
그래서 문제가 클라이언트에 있다면이 모든 서버 측 코드가 왜 포함되어 있습니까? –