Question

我在我的数据库中有一个表，我打电话来在页面上显示该表中的所有内容表中的一个项目是一个图像，当moused需要显示一个弹出窗口显示来自单独表格的数据时那个项目。我有一个JQuery根据第一个表中的行来调用数据但是当我尝试将两者放在一起时，我得到的第一个表显示正常，但现在我没有弹出鼠标悬停。

此页面目前包含在index.php页面中，该页面包含调用.js / .css文件的脚本。

以下是我要做的事情的代码：

     <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($('#hidden-table').html(), 400);' href='somewhere.html'><img src='/testarea/include/mobs/dead/$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>

Answer 1

你在这条线上有两个分号......可能是问题

$query = mysql_query("SELECT * FROM tbl_drop WHERE mob_name='$mob_name'") or die(mysql_error());;

onmouseover没有显示div

1 个答案: