i am trying to getting image from the popup, my query is ok but the image cannot be displayed in the popup. i give the id of the image but the id is not getting it. my code is ok but my result is not showing.
my code
this is the code where i used to try to shoing my image in the popup
<div align="center" style="margin-top:20px; color:#0F0;font-size:24px;"></div>
<div align="center" style="margin-top:20px;">
<table border="2">
<?php
$query = mysql_query("SELECT * FROM image");
while($row = mysql_fetch_array($query)) { ?>
<!-- Large modal -->
<div class="modal fade bs-example-modal-lg" tabindex="-1" role="dialog" aria-labelledby="myLargeModalLabel" aria-hidden="true">
<div class="modal-dialog modal-lg">
<div id="form" class="modal-content">
<!-- <img src="<?php\ echo $row['img_name']?> "width=400px; height=400px;/> -->
</div>
</div>
</div>
<tr>
<td width="100">
<a href="image.php?q=<?php echo $row['img_id'] ?>">DEL</a>
||
<td>
<button class="btn btn-primary" data-toggle="modal" data-target=".bs-example-modal-lg" onclick='update(id)'>view</button>
</tr>
<?php } ?>
</table>
<script>
<script type="text/javascript">
function update(str) {
$.get('image.php?q='+str,
function (data){
$("#form").html(data);
});
}
function showUser(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest) { // code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
}
else { // code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("dialog").innerHTML = xmlhttp.responseText;
$("#dialog").dialog();
}
}
xmlhttp.open("GET", "image.php?q=" + str, true);
xmlhttp.send();
}
</script>
there is also jQuery in which i getting the image id and send to the other page and there i m seeing my image
image.php
<title>Untitled Document</title>
</head>
<body>
<?php
if (isset($_GET['q']))
$id = $_GET['q'];
$query = mysql_query("SELECT img_name FROM image WHERE img_id = $id");
while($row = mysql_fetch_array($query)){ ?>
<img src="<?php echo $row['img_name']?>"width=280px; height=300px;/>
<?php } ?>