In this little snippet of code ,i show how i take the "foto1" column of my database and transfer the value of it to a variable in c# named $foto.
The $foto contains the path of the image corresponding to the product that is showing up. Ive tried to copy and paste the path and ditch out the php part and it works. But when i put it in img src it gives me like the broken image thing.And i cant figure out why it does that.
All help is aprecciated .
Have a nice day :)
<div class="row shop_box-top">
<?php
$ligaBD=odbc_connect('basededadospap','','');
$sql="SELECT * FROM produto WHERE nome_produto LIKE '%ADIDAS%'";
$resultado=odbc_exec($ligaBD,$sql);
?>
<div class="row shop_box-top">
<?php
while (odbc_fetch_row($resultado))
{
$nome = odbc_result($resultado,2);
$preco= odbc_result($resultado,4);
$foto = odbc_result($resultado,9);
?>
<div class="col-md-3 shop_box"><a href="stansmithflatwhite.html">
<img src="<?php echo $foto; ?>" class="img-responsive" alt=""/>
<span class="new-box">
<span class="new-label">Novo</span>
</span>
<div class="shop_desc">
<h3><?php echo $nome; ?></h3>
<span class="actual"><?php echo $preco; ?></span><br>
</div>
</a></div>
<?php }?>
depends of what path contains the $foto var. If is the absolute path, you have to retrive the relative path.
Try also to append an / or an http[s] in front of the path
<img src="/<?php echo $foto;?>">
So it would be : //path/to/photo
As I can see it in your comment, your image paths contain spaces, so a possible solution can be to use urlencode() before echoing them.
Try passing full path to img tag like http://localhost/xyz/images/Cal�ado/Adidas/Homem/Stan Smith/ADIDAS STAN SMITH - RED/ch-adidas-stan-smith-red-5.jpg.
Replace "localhost/xyz" with your website directory path.
Related
In my MySQL database, the image links are saved in the format like http://www.old.com/image1.jpg. But I had to change the domains of the images and the new links appear like images.new.com/image1.jpg. I have been changing the images links with jQuery with the following function:
$(document).ready(function() {
$('img').each(function() {
var src = $(this).attr('src');;
$(this).attr('src', src.replace('www.old', 'images.new'));
});
});
But I am wondering is there any way to change part of the URL strings with PHP. I am getting the image URLs with the following PHP function.
<?php
$imageQuery = $mysqli->query("SELECT imageURL FROM images WHERE album = 'UK' ORDER BY date ASC");
$images = $imageQuery->fetch_all(MYSQLI_ASSOC);
$images = array_chunk($images, 2);
?>
<div class="row">
<div class="col-4" id="box1">
<?php foreach (array_column($images, 0) as $image): ?>
<img class="img-fluid" src="<?= $image['imageURL']; ?>">
<?php endforeach; ?>
</div>
<div class="col-4" id="box2">
<?php foreach (array_column($images, 1) as $image): ?>
<img class="img-fluid" src="<?= $image['imageURL']; ?>">
<?php endforeach; ?>
</div>
</div>
With PHP how can I echo the new links for the images directly here in the img src? <img class="img-fluid" src="<?= $image['imageURL']; \\modified link here ?>">
PHP has it's own method for replacing strings, str_replace. The equivalent to your jQuery in PHP is:
str_replace('www.old', 'images.new', $image['imageURL'])
A better idea would be to update the values in your database.
An even better idea would be to not store the duplicate root URLs anywhere, and stitch them together in the application if you need absolute URLs, for some reason.
You could alternatively convert the text in the sql select statement:
SELECT REPLACE(imageURL, 'www.old', 'images.new') AS imageURL FROM ...
By using AS imageURL none of the rest of the code would need to change.
I'm echoing some images using <div> <img src="<?php... construction.
The images are shown, but after each image there are always some strings left from my code.
part of my code is here:
getting image by:
$image01= wp_get_attachment_image(get_post_meta($id, 'image_case_01', true),'full');
and echoing image in <div>
<div class="images-container">
<div>
<img src="<?php echo $image01; ?>" >
</div>
</div>
Result - the image is echoed properly but after the image I have that string left from my code
> " >
If i use
<div class="images-container">
<div>
<img src="/.../.../xxx.jpg" />
</div>
</div>
then all is ok.
What is wrong with my code in 2. ?
could you try this ?
<div class="images-container">
<div> <?php echo '<img src="'.$image01.'"/>'; ?> </div>
</div>
you probably have quote probleme , i'm surprised the image actually showed up as your php code is inside a '"'
EDIT: some mistake
EDIT2: my bad
so i did this and it worked
<html lang="en">
<?php $image01="erreur.png";?>
<head>
<body>
<div>
<?php echo "<img src=".$image01."></img>"; ?>
</div>
</head>
</body>
I found a solution for me.
Instead of using wp_get_attachment_image i used wp_get_attachment_image_src
Now everything works as expected.
I still don't understand why i had such quotes aoround my images, which probably lead to my problems.
May be it is because I'm getting images from a metabox meatfield.
Thank You for your support.
I have Image path like http://sits.in/imageDemo/image.php?image=R171-1.jpg in data base.
I can not load image in php by simply putting image link http://site.in/imageDemo/image.php?image=R171-1.jpg as src.
My code look like this
<?php
$count = 1;
while ($row = mysqli_fetch_array($result)) {
?>
<br>
<br>
<div class="card " style="width: 800px; margin-bottom: 25px;">
<div class="card-body">
<!-- <h5 class="card-title"><b>Product Number : <?php echo $count; ?> </b></h5> -->
<p>Order Id : <?php echo $row['orders_id']; ?></p>
<p>Product Name : <?php echo $row['product_name']; ?></p>
<img src="<?php echo $row['image1']; ?>" class="card-img-bottom" alt="Product Image"
style="width: 100px; height:100px">
</div>
<?php } ?>
I think you need to store path like that in your database
http://sits.in/imageDemo/R171-1.jpg
Also, create folder imageDemo in your local and store image as R171-1.jpg
If you want to use php file as image you need to make that php file to be shown as image file. Use GD libary. More info about how to use it for jpg here: https://www.php.net/manual/en/function.imagecreatefromjpeg.php
I wrote a code which is retrive image paths and display on the website in bootstrap grid style.But it does not showing the image. Code is working fine, Please help me. here is my code
<div class="row">
<?while ($row = mysql_fetch_assoc($query)) {?>
<div class = "col-md-3">
<?php echo $row['keywords'];?>
<?php $imagePath = $row['video_url'];?>
<?php echo $imagePath;?>
<div id="video_thumbnail">
<a href="#" class="thumbnail">
<?php echo '<img src="' . $imagePath . '">'; ?>
<img src="<?php echo file_dir . '/' . $imagePath; ?>" height="100" width="100"/>
</a>
</div>
</div>
<?php } ?>
</div>
unless file_dir is a constant using DEFINE, I suspect it's because you didn't put a $ in front of it $file_dir
You are also displaying two files. One with the file path and one without.
Chances are, the mysql query is returning a path which is not linked ....
ie <image src="myImage.jpg" /> is not the same as <image src="images/myImage.jpg" />
As #pmahomme said, right click the element and check the pat and if need be add the additional requirements
I am trying to display image from a blob field of a MySQL table. Looks like I have some sort of error in the following line. As soon as I put "header("Content-type: image/jpeg")" things get messed up and instead of displaying webpage, all source code of the page is displayed.
Please let me know how to correct.
<div class="image" align="left">
<a href="<?php header("Content-type: image/jpeg"); echo $rec['image']; ?>">
<img src="<?php echo $rec['image']; ?>" width="150" border="0"/>
</a>
</div><!-- image -->
You normally don't put the actual image contents in the src= attribute of the image tag. Instead, you point to the URL of an image file.
(There are ways to include the image source directly in the HTML, but it doesn't work consistantly with all browsers, and you still won't have your <a> link working properly.
Instead, the best way to do this is to create a separate PHP file to serve the image.
Your HTML:
<div class="image" align="left">
<img src="myimage.php?key=<?php echo($key) ?>" width="150" border="0"/>
</div><!-- image -->
myimage.php:
<?php
header("Content-type: image/jpeg");
$key = $_GET['key'];
// todo: load image for $key from database
echo $rec['image'];
You're trying to put the image data inline inside the content. The only feasible way to do this is via a Data URI data URI. Something like:
<img src="data:image/jpeg;base64,<?= base64_encode($rec['image']) ?>" width="150" border="0" />
However, what you probably want to do is put it into a separate script. So your HTML would be:
<img src="showimage.php?id=XXX" width="150" border="0" />
And your showimage.php script would be:
<?php
// Get $rec from database based on the $_GET['id']
header('Content-Type: image/jpeg');
echo $rec['image'];
?>
I've done something like that retrieving blob from my database in another way that you may find useful, here is the code example.. see if it suits your needs and if you needed anymore help let me know.
while ($row = mysql_fetch_array($hc_query2)) {
$title = $row['title'];
$text = $row['text'];
$image = $row ['image'];
$output ='<div class="HCInstance"><img src="data:image/jpeg;base64,' . base64_encode($image) . '" alt="High Council" width="100px" height="100px"/>
<div class="HCHeader"><h2>'.$title.'</h2></div><br/><div class="HCDetails"><p>'.$text.'</p></div></div>';
echo $output;
}