I am using codeigniter framework in my web development and passing the value of image id using php code such below
<a href="<?php echo base_url('account/gallery/view'); ?>?id=<?php echo $imageDetail['imageId']; ?>">
<img src="" />
</a>
and getting the value at controller with this code $imageId = $this->input->get('id');
With the get method I used, it will display the url like this :
http://my-domain-name/account/gallery/view?id=1
But I want to display the url like this :
http://my-domain-name/account/gallery/1
So what I did is setting in the router with this code but it's not working.
$route['account/gallery/(:num)'] = 'default/Gallery_controller/view/$1';
My web still able to show the template of the web if I type http://my-domain-name/account/gallery/1 but display error in the field that I fetch from database because unable to get the imageId
use Regex instead of :num in route.php
$route['account/gallery/([0-9]+)'] = 'default/Gallery_controller/view/$1';
And in your template file:
<a href="<?php echo base_url('account/gallery/'.$imageDetail['imageId']); ?>">
<img src="" />
</a>
Then you can access directly in your gallary controller in view function $id.
eg:
class Gallery extends MY_Controller {
public function view($id){
// Your Image ID = $id;
}
}
Set this url http://my-domain-name/account/gallery/1 proper in href
<a href="<?php echo site_url('account/gallery/view'.$imageDetail['imageId']); ?>">
<img src="" />
</a>
Replace
<a href="<?php echo base_url('account/gallery/view'); ?>?id=<?php echo $imageDetail['imageId']; ?>">
with
<a href="<?php echo base_url();?>account/gallery/view/<?php echo $imageDetail['imageId']; ?>">
Check your config.php
$config['enable_query_strings'] = true;
to
$config['enable_query_strings'] = false;
In this try this (:any)
$route['account/gallery/(:any)'] = 'default/gallery_controller/view/$1';
Or
$route['account/gallery/view?(:any)'] = 'default/gallery_controller/view/$1';
In the routes lower case
This code is make your url like http://domain-name/account/gallery/1
Try this:
<a href="<?php echo base_url();?>account/gallery/view/<?php echo $imageDetail['imageId'];?>">
<img src="" />
</a>
Related
I have a problem sending value to controller, there is a double on rsport23.000webhostapp.com
https://rsport23.000webhostapp.com/rsport23.000webhostapp.com/pesanan/tambah_form/2
what should I change?
<p><a href="<?php echo site_url('pesanan/tambah_form/'.$value->id_barang); ?>">PESAN SEKARANG
</a></p>
try to use with base_url. always work for me.
<p><a href="<?php echo base_url().'pesanan/tambah_form/'.$value->id_barang; ?>">PESAN SEKARANG
</a></p>
First of all set the base url in the application/config/config.php
$config['base_url'] = "https://rsport23.000webhostapp.com/";
then after that you can do like that:
<p>
<a href='<?php echo base_url("pesanan/tambah_form/$value->id_barang"); ?>'>PESAN SEKARANG</a>
</p>
i have a code which gets images from a database. I need to tell it to to display another image (which says no image available) if an image is not found. How would I do this????
Your suggestions would be very much appreciated
<a class="thumbimage" href="<?PHP mrd("$MyProductTitle", "$row[LID]", "$_GET[category]", "$rowxxx[MR]", "index.php?page=detail"); ?>"><img src="images/thumb/<?php echo "$row[IMAGENAME]"; ?>.jpg" border="1" /></a>
Try this:
<?php
$currentImage = "images/thumb/".$row[IMAGENAME].".jpg";
if(!file_exists($currentImage))
{
$currentImage = "PATH_TO_IMAGE_UNAVAILABLE";
}
?>
<a class="thumbimage" href="<?PHP mrd("$MyProductTitle", "$row[LID]", "$_GET[category]", "$rowxxx[MR]", "index.php?page=detail"); ?>"><img src="<?=$currentImage?>" border="1" /></a>
You can use onerror attribute.
Replace the error image (3331913_orig.gif) with yours error image on the code below:
<img src="{some_error_src}" onerror="this.onerror=null;this.src='http://availableservicesllc.weebly.com/uploads/2/2/3/9/22390468/3331913_orig.gif'">
Click here to see this example
I have built a site for someone else who will be updating it via a CMS (CushyCMS). They wanted a lightbox gallery included. I want to be able to allow them to upload a new image and for that image url to be copied into the a tag so lightbox works.
So far I am able to do that, but the client has to ensure the file names and format are exactly the same as what I have set them to in the php code. Is there a way to get the img url from the img tag (possibly identifying it using an id attribute) and put it directly into the tag?
Heres what I have so far:
<?php
$src1 = "images/test1.jpg";
$src2 = "images/test2.jpg";
?>
<a href="<?php echo $src1?>" data-lightbox="group-1">
<img class="cushycms profile" name="image1" id="slideshow_image" src="images/test1.jpg"/></a>
<a href="<?php echo $src2?>" data-lightbox="group-1">
<img class="cushycms profile" id="slideshow_image" src="images/test2.jpg"/></a>
Many thanks!
You can do it with jQuery:
<a data-lightbox="group-1">
<img name="image1" src="images/test1.jpg" />
</a>
<script>
var img = $('img[name="image1"]');
var imgSrc = img.attr('src');
img.parent().attr('href', imgSrc);
</script>
If you have multiple images and want to automate this you can use the map function:
<a data-lightbox="group-1">
<img name="image1" src="images/test1.jpg" class="slideshow" />
</a>
<a data-lightbox="group-1">
<img name="image2" src="images/test2.jpg" class="slideshow" />
</a>
<a data-lightbox="group-1">
<img name="image3" src="images/test3.jpg" class="slideshow" />
</a>
<script>
$(".slideshow").map(function() {
$(this).parent().attr('href', this.src);
});
</script>
Assuming you use jQuery somewhere there, you could/should use a lightbox plugin, like this one here:
http://www.jacklmoore.com/colorbox/
Using the xamples there, you would then use:
(http://www.jacklmoore.com/colorbox/example1/)
$('a.gallery').colorbox({rel:'group-1'});
which would result in a lighbox gallery showing on "click" on any anchor with class "gallery" while the gallery would show all the elements pointed by "href" from all anchors of that have class="group-1" (so each anchor would be
although you have to include some files for jquery and the lightbox plugin, I think it will make your life much easier in the end.
Also not sure what your php level is, but the example code you have there asks for:
<?php
$images = array(
'images/test1.jpg',
'images/test2.jpg'
);
?>
<?php foreach($images as $i => $url): ?>
<a href="<?php echo $url?>" data-lightbox="group-1">
<img class="cushycms profile" name="image<?php echo $i+1 ?>" id="slideshow_image" src="<?php echo $url ?>"/>
</a>
<?php endforeach; ?>
Just in case you got more images there.
i have a script in which output localhost/urlfromMysqlDatabase
i need output should be urlfromMysqlDatabase
please give me any suggestions to changes this script
$link = mysql_fetch_assoc(mysql_query("SELECT url FROM admins"));
and in between html body i have a code
<a href="<?php echo $link['url']; ?>" target="_self">
Try this:
<a href="<?php echo #end(explode("/", $link['url'])); ?>" target="_self">
end(explode("/", $link['url'])) will print out urlfromMysqlDatabase (everything after the last slash /).
$link = mysql_fetch_assoc(mysql_query("SELECT url FROM admins"));
$link=end(explode("/",$link['url']));
<a href="<?php echo $link; ?>" target="_self">
I have a PHP echo function inside of a HTML link, but it isn't working. I want to have an image location, defined in img src, be in part of the clickable link of the image. The page will have multiple images doing the same thing, so I am trying to use PHP to automate this.
<a href="http://statuspics.likeoverload.com/<?php echo $image; ?>">
<img src="<?php $image=troll/GrannyTroll.jpg?>" width="100" height="94" />
</a>
Turn
<?php $image=troll/GrannyTroll.jpg?>
into
<?php echo "troll/GrannyTroll.jpg"; ?>
?
Or provide more details on what you are trying to achieve.
Also, you might consider urlencode-ing some of those URL parameters.
Edit:
So you might try setting the variable beforehand:
<?php $image = "troll/GrannyTroll.jpg"; ?>
<img src="<?php echo $picture; ?>" width="100" height="94" />
So now i understand what you are trying to do.
One error is that you didn't enclose $image=troll/GrannyTroll.jpg with quotes like this:
$image = 'troll/GrannyTroll.jpg';
The second error is that you do it in the wrong order, you have to define $image first, before you use it.
That's what I believe you want to do:
<?php
$image = "troll/GrannyTroll.jpg";
?>
<img src="<?php echo $image; ?>" width="100" height="94"/>