How to Show Popup window on clicking dynamically generated link - php

Question

T have one page on which results are get created dynamically. And I
want to display popup window on clicking image named view.jpg. The
image is hyperlink. So please tell me how should i do it. I am showing
my code as follows.
code :
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

You can use attribute selector [attribute='value']
$("[src='images/view.png']").click(function(){
alert("clicked");
});