How to display the image from ajax success function?

Question

I have created ajax function like this...In this I will get the value from run time and i need to return the photo according to that value..In success function i need to display that image in particulat div

var num=document.getElementById('number').value;
$.ajax({
    url:"image.php?val="+num,
    contentType: "image/png",
    success:function(img)
    {
        $('#image').html('<img src="data:image/png;base64,' + img + '" />');
    }
    });

image.php page

$sql_sub = select_query("select pic from  photo  where picnum=".$_GET['val']."");
$img = $sql_sub[0][0]->load();
header("Content-type: image/png");
ob_start();
imagepng($img);
echo "data:image/png;base64,", base64_encode(ob_get_clean());

Looks about right. You'll need to base64 encode the image data using: btoa() — rrowland, Jun 05 '15 at 06:24
`contentType: "image/png",` — you aren't POSTing an image **to** the server. Don't claim that you are. — Quentin, Jun 05 '15 at 06:47
**Danger**: You are **vulnerable to [SQL injection attacks](http://bobby-tables.com/)** that you need to [defend](http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php) yourself from. — Quentin, Jun 05 '15 at 06:47
Poster updated the original post 30 minutes ago. He was originally not base64 encoding the image. — rrowland, Jun 05 '15 at 07:15

score 3 · Answer 1 · answered Jun 05 '15 at 06:29

It looks perfect..You may have an issue in tag. Check first that tag. However .append works great.

Have you tried this:

$('body').append('<img src="https://chart.googleapis.com/chart?cht=qr&chs=200x200&chl=http%3a%2f%2fwww.facebook.com" />');

$('#div_where_you_will_sho_qr_code').append(data.toString());

or:

$('#container').html('<img src="https://chart.googleapis.com/chart?cht=qr&chs=200x200&chl=http%3a%2f%2fwww.facebook.com" />');

where #container is some DOM element to harbor your image.

or the way I prefer:

$('#container').html(
    $('<img/>', {
        src: 'https://chart.googleapis.com/chart?cht=qr&chs=200x200&chl=http%3a%2f%2fwww.facebook.com',
        alt: ''
    })
);

score 1 · Answer 2 · answered Jun 05 '15 at 06:36

var num=document.getElementById('number').value;
$.ajax({
    url:"image.php?val="+num,
     type: "POST",
     dataType: "html",
    success:function(data)
    {
        $('#image').html(data));
    }
    });

image.php

$sql_sub = select_query("select pic from  photo  where picnum=".$_POST'val']."");
$img    = $sql_sub[0][0]->load();
$image  = '<img src="data:image/png;base64,'.$img.'" />';
echo $img;

score 0 · Accepted Answer · answered Jun 05 '15 at 07:07

var num=document.getElementById('number').value;
$.ajax({
url:"image.php?val="+num,
 type: "POST",
 dataType: "html",
success:function(img)
{
    $('#image').html('<img src="data:image/png;base64,' + img  + '" />');
}
});

image.php

$sql_sub = select_query("select pic from photo where picnum=".$_GET['val']."");
$img = $sql_sub[0][0]->load();
header("Content-type: image/png");
ob_start();
echo $img;
echo "data:image/png;base64,", base64_encode(ob_get_clean());

How to display the image from ajax success function?

3 Answers3