How do I properly echo images combining HTML and PHP? $offer['picture'] has link saved as example.com/picture.png. I've tried many different options, but nothing works. Can anyone help me out?
foreach($json['offers'] as $offer) {
$image = $offer['picture'];
?>
<img src="<?php echo $image ?>">
<?php
}
If there is only example.com/picture.png in $offer['picture'], problem is that images linked incorrectly. You should add http:// before image link to make browser sure you are loading image by absolute path.
foreach($json['offers'] as $offer) {
$image = $offer['picture'];
?>
<img src="http://<?php echo $image ?>">
<?php
}
Well it depends on your directory structure.
suppose your images are in example.com/assets/images/photo-1.jpg
your code should be
<?php foreach($json['offers'] as $offer) {
$image = $offer['picture']; ?>
<img src="http://www.example.com/assets/images/<?php echo $image; ?>">
<?php } ?>
In fact, you need to concatenate or hard code the path and image name will be appended dynamically.
Try this syntax,
<?php
foreach($json['offers'] as $offer) {
$image = $offer['picture'];
echo "<img src='$image' />";
}
?>
Related
I have the following code
<figure>
<?php if ($data[$i]->thumbnail) { ?>
<img src="<?php echo $img ?>" width="465" height="300" alt="news-report-juli19" title="" />
<?php } ?>
</figure>
I want to display images from mysql database dynamically,
the image_field_name=thumbnail.
I have 10 records in db and all have different images.
You should have to given the image path as well when you want to display.
<?php if ($data[$i]->thumbnail) { ?>
<img src="<?php echo base_url('path_of_img/$img'); ?>" width="465" height="300" alt="news-report-juli19" title="" />
<?php } ?>
If you use the Codeigniter framework, please try this.
In controller
$data['images'] = $this->model->get_images();
$this->load->view('view_name', $data);
In model you should make get_images() method that get image urls.
In view
<?php
foreach ($images as $value) {
echo '<img src="'.$value['thumbnail'].'" width="465"
height="300" alt="news-report-juli19" title="" />';
}
?>
Good luck.
use query variale with db colmn where images are present
use this in source img src
echo $query_variable['db_coln_name']
Hello I'm trying to echo an image on a view page in CodeIgniter but nothing is displayed on the page.
Here I'm making the variable:
<?php $image = "<img src='../../images/stoel.jpg' alt='img' />"; ?>
And here I'm trying to echo the image:
<img src="<?php echo $image;?>">
Intro
This is the most basic php, so what's the addition of this question? Please read the basics of echo here: http://php.net/manual/en/function.echo.php (Example 1).
Learn the basics first!
Solution
1. Assign full html tag to variable and echo full html:
<?php
$image = '<img src="../../images/stoel.jpg" alt="Foo">';
echo $image;
?>
2. Or assign image path to variable and echo concat string:
<?php
$path = '../../images/stoel.jpg';
echo '<img src="' . $path . '" alt="Foo">';
?>
3. Or assign image path to variable and echo only this with php:
<?php
$path = '../../images/stoel.jpg';
?>
<img src="<?= $path; ?>" alt="Foo">
You are inserting full Image tag into src attribute.
<?php $image = "<img src='../../images/stoel.jpg' alt='img' />"; ?>
and
<?php echo $image; ?>
or
<?php $image = "../../images/stoel.jpg"; ?>
and
<img src="<?php echo $image;?>">
<?php $image = "<img src='../../images/stoel.jpg' alt='img' />"; ?>
you try like this
<?php echo $image; ?>
You are already assign full image code in variable so you just need to print your variable:
<img src="<?php echo $image;?>">
To
<?php echo $image;?>
Make sure your images out side of the application folder then you can do something like
application
images
images > stoel.jpg
system
index.php
Use Base url from the url helper
<img src="<?php echo base_url('images/stoel.jpg');?>" />
Make sure you have set the base_url in config.php
$config['base_url'] = 'http://localhost/yourproject/';
You can also use HTML Helper img();
I think this is what you mean. Since you are echoing inside the src attribute, you do not need to store the whole <image>, just the path will do.
<?php
$image = "../../images/stoel.jpg";
?>
<img src="<?php echo $image;?>">
I'am using codeigniter file upload class to upload images and save full_path in to the database.
But when displaying the images i can't see it.
Here is my code
<?php foreach($images as $img) {?>
<img src="<?php echo $img;?>" width="150">
<?php }?>
From view-source i will open the image link it will mixed with base-url and image full_path.
http://localhost/holiday/admin/profile/package_explained/C:/xampp/htdocs/holiday/uploads/package-images/img-holoday-sample.png
the value of $img is (ie fullpath of the image) C:/xampp/htdocs/holiday/uploads/package-images/img-holoday-sample.png.
How can i display image with full_path?
Only get image name form Controller. Do not track full path.
Bcz your host server doesn't have path like this C:/xampp/htdocs/
<?php foreach($images as $img)
{
?>
<img src="<?php echo base_url()?>uploads/package-images/<?php echo $img;?>" width="150">
<?php
}
?>
Assume that you are not saving the path like that you have mentioned in DB:
Try this (add http:// before print):
<?php foreach($images as $img) {?>
<img src="http://<?php echo $img;?>" width="150">
<?php }?>
I am trying to display an image that is in a folder "upload" by getting the image name from the database.
while($row = mysqli_fetch_array($result))
{
$pic = $row['image'];
echo $row['item'] ;
echo $row['location'];
echo $row['description'];
echo $row['forum'];
echo $row['datetime'];
echo $row['username'];
?>
</br>
<img src="upload/<?php echo $pic ?>"/>
<?php echo $row['image']; } ?>
"upload/<?php echo $pic ?>"
</body>
</html>
As you can see it display everything except the img src.
This is my database (ignore BLOB that was a test). I can't seem to figure out where I'm going wrong.
Thanks
My recommendation is to store the full relative path in the database like this:
uploads/folder/file.jpg
My preferred MySQL field type is 'varchar(255)' the your echo in te PHP code will look like:
echo '<img src="'. $row['image'].'" />';
You are trying to print the image source outside the while-loop. The while-loop will only exit when $row is empty, so $row['image'] is also empty.
1) view source (or use firebug) to see the image tag and see what is the src given there.
2) try the url: "http://{LOCAL}/projects/projectviewposted.php/upload/happyball(1).jpg" and see if you can open the image
Try this
<img src="upload/<?php echo $pic; ?>"/>
<?php echo $row['image']; } ?>
"upload/<?php echo $pic ?>"
Else note down the link of the folder path. It might be that the folder path is not correct.
Hope this helps
You can try :
while($row = mysqli_fetch_array($result))
{
$pic = $row['image'];
echo $row['item'] ;
echo $row['location'];
echo $row['description'];
echo $row['forum'];
echo $row['datetime'];
echo $row['username'];
echo '<br />
<img src="upload/'.$pic.'"/>
'.$row['image'];
}
Well after 3 days I figured it out... my header location was incorrect!
../project/projectviewposted.php/ A stupid extra slash at the end!!
Thanks for your help and suggestions!
Here is my problem.
If I do:
$imagePath = "images/spalt_images/body_images/525A.JPG";
?>
<img src="<?php $imagePath ?>" alt="front image" class="productImage"/>
<?php
Then my image does not show up.
However, if I do:
<img src="images/spalt_images/body_images/525A.JPG" alt="front image" class="productImage"/>
Then my image shows up just fine. Why would it not work with php?
Thanks
You need to echo it:
<?php echo $imagePath ?>
or use a short tag (not recommended, but that's another discussion):
<?=$imagePath ?>
You need to echo $imagepath otherwise it won't 'print' it out.
Change <?php $imagePath ?> in your code to <?=$imagePath?>, or add an echo if your server doesn't allow shorttags.