Which link version is right - if any?
contact-us
contact-us
and image links:
<img alt="xxx" src="<?php echo get_home_url(null, 'fl-images/banner/teaser.png', null); ?>">
<img alt="xxx" src="<?php echo get_home_url(); ?>/fl-images/banner/svatebni-teaser.png">
For <a> tag use.
kontakty
For <img> tag use.
<img alt="xxx" src="<?php echo home_url('/fl-images/banner/svatebni-teaser.png'); ?>">
In your case both will work assuming fl-images is in the public root.
However, if you want to reference a blog id or add a scheme to the home URL context then you can use the arguments. eg: get_home_url( 2, 'contact-us/', 'https');.
See docs here
Related
The file path is:
theme
assets
src
images
grey-arrow.svg
Markup:
<?php $getIcon = get_template_directory().'/assets/src/images/grey-arrow.svg';?>
<div><img src="<?php echo $getIcon; ?>"/></div>
<?php echo $getIcon;?>
The image doesn't load and an echo of $getIcon returns:
/var/www/html/wp-content/themes/theme/assets/src/images/grey-arrow.svg
... Which is the correct path. Ideas on why the image doesn't load?
You should echo it and also you are closing your php tag properly.
<img src="<?php echo get_template_directory_uri(); ?>/assets/src/images/grey-arrow.svg"/>
or you can use bloginfo which is easier to remember and use (You need not echo)
<img src="<?php bloginfo('template_url'); ?>/assets/src/images/grey-arrow.svg"/>
How do you change the logo to link to the home URL for all pages except one? I want one page to link to another page when the logo is clicked.
Here is the PHP code for the logo:
<div class="section-boxed section-header">
<?php do_action('pexeto_before_header'); ?>
<div id="logo-container">
<?php
$logo_image = pexeto_option('retina_logo_image') ? pexeto_option('retina_logo_image') : pexeto_option('logo_image');
if(empty($logo_image)){
$logo_image=get_template_directory_uri().'/images/logo#2x.png';
}
?>
<img src="<?php echo $logo_image; ?>" alt="<?php esc_attr(bloginfo('name')); ?>" />
</div>
What about creating a second section of PHP - with a slightly different DIV ID that is called when that paticular page is loaded, rather than this one which is called on all the other pages,
Copy, paste, change DIV ID <div id="logo-container2">, change link address.
In HTML - on the single page that takes them elsewhere - call
<div id="logo-container2">
Would that work?
Try to use template_tag is_page as condition
<div class="section-boxed section-header">
<?php do_action('pexeto_before_header'); ?>
<div id="logo-container">
<?php
$logo_image = pexeto_option('retina_logo_image') ? pexeto_option('retina_logo_image') : pexeto_option('logo_image');
if(empty($logo_image)){
$logo_image=get_template_directory_uri().'/images/logo#2x.png';
}
// Default logo url to home
$logo_url = esc_url(home_url('/');
// if is page about or id 5 anything inside is_page()
if(is_page('about') $logo_url = esc_url(home_url('about');
?>
<img src="<?php echo $logo_image; ?>" alt="<?php esc_attr(bloginfo('name')); ?>" />
</div>
I believe you should be able to use the get_permalink method to check which page you are on, and use an if statement to tell it what the href should be.
<a href="<?= (get_permalink() == '/my-page') ? esc_url(home_url('/go-to-page')) : esc_url(home_url('/')); ?>">
Haven't tested this, but it should work.
I want to keep src attribute in a variable. I am working on php and tried the following method but it doesn't display the image on html page.
Code:
<?php $path="C:/horizontal.jpg"; ?>
<image src="<?php echo $path; ?>" style="width:304px;height:228px" />
I think you can use this technique:
<?php $path="http://" . $_SERVER["SERVER_NAME"];?>
<img src="<?php echo $path.'/projectName/image.jpg'; ?>" style="width:304px;height:228px" />
Where:
$_SERVER["SERVER_NAME"]: to get the server name, for example: www.example.com.
Also you can use $_SERVER["SERVER_ADDR"], in this case you can get the address IP of your server, for example: "127.0.0.1".
I hope this information helps you.
Good Luck.
<?php $path="localhost/projectName/";?>
<img src="<?php echo $path.'image.jpg'; ?>" style="width:304px;height:228px" />
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 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"/>