I'm trying to display an image that is stored in my database. I know it is not best practice to store an image in the database, but just for this purpose its what I need. My problem is when I go to display the image, all I get is a load of scrambled code, which I guess is the image in code but not visually what I want. When I use the header tag to identify the image all I get is a thumbnail to mean it isn't reading it in right. Any help would be great.
Below is code I'm using. I've tried it in both ways using it to display in either the php or in html but can't display the image:
while($row = mysql_fetch_array($result))
{
$Long = $row['Long'];
$Lat = $row['Lat'];
$img = $row['file'];
echo "<b><center>Database Output</b><br><br>";
echo "<td>" .$row['file']."</td>";
echo "----";
echo "<td>" .$row['Lat']."</td></center>";
//echo "<td>" .$row['file']."</td>";
}
?>
<div id="map">
<img src="<? $img?>" alt="">
Thanks for any input
Either you embed the image into your html page as a data URI
<img src="data:image/jpeg;base64,<?php echo base64_encode($img) ?>" />
which is hideously NASTILY horrible for 'large' files - you make it impossible for the browser to cache the image, forcing the user download the base64-encoded data EVERY time they load the page.
or you have a sub-script to serve up the actual image, and have something more like:
<img src="getimage.jpg?id=XXX">
and
<?php
$data = get_image_from_db($_GET['id']);
header('Content-type: image/jpeg');
echo $data;
Related
I don't know if I'm doing it right, but I have a bunch of images I'm retrieving from the page and since I don't wan a page to have too many images of big sizes, I have displayed them with a much smaller size but I have attached each of them to a link so that when a user click on a picture it opens that image with its original size. The problem is that those images are really big and my client wants the ability to zoom in and out which I don't know how to do. The client thought about resizing the size of the window (in the browser) but sadly it resizes all other windows (for the application) and this is not ok because he needs to see the image and compare it with some information on the app. SO Below is the code of the images displayed and after the user have clicked on the image.
small images
$count = 0;
echo " <div class=\"row\">";
while($row = $result->fetch_assoc()) {
$ext = $row['Extension'];
$ImageID=$row['ImageID'];
if(($count%3) ==0){
echo "</div>";
echo " <div class=\"row\">";
echo " <div class=\"col-sm-2\">";
echo " <a href=\"viewimage.php?ImageID=$ImageID\" class=\"thumbnail\">";
echo '<img id=\"myImg\" src="data:image/$ext;base64,'.base64_encode( $row['Image'] ).'" style=\"width:130px;height:130px\"/>';
echo"</a></div>";
++$count;
}else{
echo " <div class=\"col-sm-2\">";
echo " <a href=\"viewimage.php?ImageID=$ImageID\" class=\"thumbnail\">";
echo '<img id=\"myImg\" src="data:image/$ext;base64,'.base64_encode( $row['Image'] ).'" style=\"width:130px;height:130px\"/>';
echo"</a></div>";
++$count;
}
}
echo "</div>" ;
Image after link is clicked
<?php
$ImageID = $_GET['ImageID'];
$query = "Select * from $dbname.Images where ImageID = $ImageID";
$result = mysqli_query($conn,$query);
$row = $result->fetch_assoc();
$ext = $row['Extension'];
echo '<img src="data:image/$ext;base64,'.base64_encode( $row['Image'] ).'"/>';
?>
I don't know what to do at this point, how can I provide that zoom in/out functionality?
First things first: Generally don't add base64 encoded images directly into your html. Link to them, and host them on your server. It is quite an expensive way of making images appear, both for the server, database, and for the client. It also makes it impossible for the client to cache the images, and it means that each repeated page visit causes the entire data to be sent.
Make two folders on your webservers:
images/
thumbnails/
Put your small images in "thumbnails" and large images in "images"
And if you need to, store the image-names in your database, so you can do something more like this:
echo '<img src="images/'+$imageName+'">'
If you want to, you can do an on-demand resizing of your images, using gd-lib.
The basic idea being, in pseudocode:
//Before the echo command, but after fetching the filename from database
if thumbnails/$imageName exists
then use gdlib to read images/$imageName and save a small version to thumbnails/$imageName
This approach is also applicable if you want to use client-side javascript to show larger versions on the same page. See my page finalkey.net for an example http://finalkey.net/gallery
I have a image stored in mysql as mediumblob and now I want to show it in html page, so I am doing this with it:
$c = base64_encode($resu[0]->image);
$image = '<img src="data:image/jpeg;base64,'.$c.'" />';
echo $image;
But I am getting only half of original image, so am I missing something here ?
Just an idea, may be you can seperate the logic to display the image.
What I mean is that you can create a file like image.php that accepts the id or filename
of the image and then display the image. Then you can simply refer the image in your HTML
by, for example, doing something like this:
<img src="image.php?imgId=12547"/>
in image.php file something like the following
$imgId=isset(GET['imgId'])?GET['imgId']:0;
$sql = "SELECT * FROM theBlogs WHERE ID = $imgId;";
$result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_array($result);
header("Content-type: image/jpeg");
echo $row['imageContent'];
Just perform a quick test with strlen on $resu[0]->image variable and check blob size in DB and check if they are different sizes for sure.
Display in li tag means use this
<li data-thumb='<?php echo "data:image/jpeg;base64,".base64_encode($img1 ); ?>'>
<div class="thumb-image">
<img <?php echo 'src="data:image/jpeg;base64,'.base64_encode( $img1 ).'"';?>
data-imagezoom="true" class="img-responsive" alt="">
</div>
</li>
I have a php variable which contain the whole source code of image.What I want is to echo the tag with its attributes src,height & width and send the source code of image through mail (i.e ).
$imagename = abc.jpg;
$concatpath = SITE_URL."/uploads/affiliatesAdv/".$imagename;
$imagesrc = '<img src="'.$concatpath.'" height="200px" width="200px">';
Now when I echo $imagesrc it displays the image. How can I show the source code instead?
Try this,
echo "<pre>";
echo $imagesrc;
echo "</pre>";
You need to use html encoding.
Some thing like this
<img src="abc.jpg"/>
to
<img src="abc.jpg"/>
See the reference.
You can try this site to do it online..
I've stored my Images into (Medium) BLOB fields and want to retrieve them embedded within my PHP-generated web pages.
When I test retrieving the stored images using
header('Content-type: ' . $image['mime_type']);
echo $image['file_data'];
everything looks just fine.
However, I have not yet found a way to retrieve the image(s) cleanly into the middle of my documents. For example, using
$image = $row['file_data'];
echo '<img src="data:image/jpeg;base64,'.$image['file_data'].'" alt="photo"><br>';
...or...
$im = imageCreateFromString($image);
I just wind up with a bunch of hexadecimal garbage on screen.
I intitially stored the Images using:
ob_start();
imagejpeg($resizedImage, null, 100);
$content = ob_get_contents();
ob_end_clean();
$sql = sprintf(
"insert into images (filename, mime_type, file_size, file_data, event_id)
values ('%s', '%s', %d, '%s',%d)",
mysql_real_escape_string($fileName),
mysql_real_escape_string($mimeType),
$imageSize,
mysql_real_escape_string($content),
$eventID
);
$result = $cn->query($sql);
Does anyone PLEASE have a working code snippet to successfully display the stored .jpg mid-file in the PHP output?
echo '<img src="data:image/jpeg;base64,'.base64_encode($image['file_data']).'" alt="photo"><br>';
However, remember that old IE versions do not support this kind of inline images! Besides that, the browser cannot cache such an image except together with its containing HTML page.
You should create some sort of "image server". You're already close to that.
For example, create something like image.php that will get a image name and will generate it on the fly.
So, for example, say you want to get somePic.jpg image. You can get it through:
image.php?name=somePic.jpg
<?php
header('Content-type: ' . $image['mime_type']);
echo $image['file_data'];
?>
Your tag:
<img src='image.php?name=somePic.jpg' />
Or more general:
echo "<img src='image.php?name={$image['filename']}' />"
Why not just call your test page image.php, then have it called from the browser on the rendered page:
<img src="image.php?imageid=123" alt="photo" />
I create a component in my joomla website. the component shows some photos (not big, only 8KB). the photos are stored in mysql blob. i can upload the photos to the joomla database but i cannot display it on the website. whatever i do it only show some encoding character or blank. I tried to create a separate page but but the result is same. Here is what i have done :
mycomp is my joomla component.
admin.mycomp.php
<?php
function showDetail($option)
{
$db = &JFactory::getDBO();
$id = mysql_real_escape_string(JRequest::getVar('id'));
$query = "select id,myphoto from jos_myphotos where id = ".$id;
$db->setQuery($query);
$rows = $db->loadObjectList();
HTML_myphoto::showPhoto($rows,$option);
}
?>
admin.mycomp.html.php
<?php
class HTML_myphoto
{
...
function showPhoto($row,$option)
{
...
header("Content-type: image/jpeg");
echo $row->myphoto; //this will show some encoding character
echo base64_decode($row->myphoto); //this will show blank page
//change echo with print get the same result.
...
}
...
}
I tried to create a separate page like this :
admin.mycomp.html.php
<?php
class HTML_myphoto
{
...
function showPhoto($row,$option)
{
...
?>
<img src="show_image.php?myphoto=<?php echo $row->myphoto;?>" width=200 height=300>
<?php
...
}
...
}
show_image.php
<?php
$myphoto = (isset($_GET['myphoto'])) $_GET['myphoto'] : false;
if($myphoto)
{
header("Content-type: image/jpeg");
echo $myphoto; //this will show some encoding character
echo base64_decode($myphoto); //this will show blank page
//change echo with print get the same result.
}
?>
the result is same.
I think you have 2 options:
Either you make an image tag with its source in a PHP file receiving only a ID parameter and retrieving the photo's string in the DB and echoing it.
Or you echo directly your photo's string in your tag:
<img src="<?php echo base64_decode($myphoto); ?>" />
EDIT
I just checked in an old app where I store the favicons in a DB. You don't need to base64_decode when you display your image inline (my option 2).
So FYI, this image works:
<img alt="" src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABAAAAAQCAYAAAAf8/9hAAAABmJLR0QA/wD/AP+gvaeTAAAACXBIWXMAAABIAAAASABGyWs+AAAACXZwQWcAAAAQAAAAEABcxq3DAAACkElEQVQ4y42TTWyUZRSFn/e+n6W00E4l9ScSCZGODXFLIIZENyZu1ITEGBdE3ejGVQksunDlwoUrE0ICCQuomZAYF4YNNvEPlFpjiJRgKq10HDKdmVJxEOh0vu+9h0UrMSyAk9zVOffczXPRI8hdciVNTZ/X6MjLOnrkK3XTLbk64lEK5FIhaf87HyqznYKy5qsLyuUPL/D1OXzkC8U4IhhViC9o7OAnSkkKksQDVG81GR//iAvTbS7OXIQQ1p0Oc3O/YlCAAAcSQIHIEc43Z8/z6r5xfrsiBp9+ArK4FgprNye//oEMDIJDMBDU6jc4d/YXzpz5kcuzC5SGBvBkoMT257ah5FTn/4Qgpqd+InM3LEDr+j8cO/45pyrnmLk0w0uv7KV/cIDCI4GIhUAgY3h4C9W5Kma9LF//l2xpqc33301xcuJLTp+ehLiJbSNlcpxoEanAolF4okB0lRAiuVOt1cje/+AAly8tcHWhDnEDeAfXKo/FTbg7ZLdwCrDI0PAAK7fvQATkYIbN/l4jWh8pAb4BrJ/abI32301MXULKoIhE9bKxt8RKJ4D1gXpIitj+d19j6PGN7CiXQavgHYir3G5fY+tT/Wx/cgulnh6su0JYvUPodiHvAGJz32YoJC222qo1lvTH1b906NBngq06cPBjFZKkXIuNmhZbN3RlvqUTlUnt2fumsGc0NvapkJLcpeRrzF+r31R5dLd2PP+i6s3if//gcndJUqPRUKVSUbPZFCnlkpJcSWmd+dffeE/BntWJk9/eW7q/RJJSSjIzQxKBhIUcgN17dhFCxsREheXlZSStZUIghEBRFACYGRlwz3Q5FuDtt/Yx/fMFSqVB8jwnrPP/Xy7LMtwdM+Mu+2gfA7SP0igAAAAASUVORK5CYII=" style="margin-right: 5px; vertical-align: middle;" class="bbns_itemDragger">
And it is stored in my DB like this (base64 encoded):
data:image/png;base64,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
I'm sorry, did you skip some lines from show_image.php?!
Cause $myphoto is just the id of the photo. You can't base64_decode an ID.