Question

T有一个页面，动态创建结果。和我想要在点击名为view.jpg的图像上显示弹出窗口。该图像是超链接。所以请告诉我该怎么做。我正在展示我的代码如下。

代码：

echo "<table width='' height='' border='1' align='center'>
        <tr><td>Title</td>
            <td >Type</td>
            <td >Date</td>
            <td>Expiry Date</td>";
            if($typeofuser=='admin' || $typeofuser=='Accountant' || $typeofuser=='secretary')
            {
            echo "<td>View</td>
            <td>Edit</td>
            <td>Delete</td>";
            }
        echo "</tr>";

            $qry2 = mysqli_query($con,"SELECT nId,nTitle,nDescription,nDate,nExpiryDate FROM tblnoticemanager where Society_id = '$socId' and category='General'") or die(mysqli_error($con));
                while($NoticeData = mysqli_fetch_array($qry2))
                {
                echo "<tr>"; 
                    echo  "<td align='center' class='tdletter' style='text-transform:capitalize;'>" .$NoticeData['nTitle']. "</td>";
                    echo  "<td align='center' class='tdletter' ><div class='overflowDiv'>" . $NoticeData['nDescription']."</div></td>"; 
                    echo  "<td align='center' class='tdletter'>" . $NoticeData['nDate'] ."</td>"; 
                    echo  "<td align='center' class='tdletter'>" . $NoticeData['nExpiryDate'] ."</td>";
                    if($typeofuser=='admin' || $typeofuser=='Accountant' || $typeofuser=='secretary')
                    {
                    echo  "<td align='center' class='tdletter'><div><a href='?id=".urlencode(base64_encode($NoticeData['nId']))."'><img src='images/view.png' width='30' height='30' align='center' /></a></div></td>";
                    echo  "<td align='center' class='tdletter'><div><a href='?id=".urlencode(base64_encode($NoticeData['nId']))."' ><img src='images/edit.jpg' width='30' height='30' align='center'/></a></div></td>";  
                    echo  "<td align='center' class='tdletter'><span id='".$NoticeData['nId']."' class='trash'><img src='images/delete1.jpg' width='30' height='30' align='center' /></span></td>";
                    }
                echo "</tr>";              
                 }          
echo "</table>";

Answer 1

您可以使用attribute selector [attribute='value']

$("[src='images/view.png']").click(function(){
     alert("clicked");
});

如何在单击动态生成的链接时显示弹出窗口

1 个答案: