Hi , I have placed in my php to get the image file from the folder, but it is giving me a error in the source saying
Warning: Division by zero in
My code is :
<img src="<?php echo get_bloginfo('template_directory') /imgs/Logo.png;?>" alt="Logo" class="logo">
It was working for me last week , i forget what I have changed but now can not fix this , can anyone please help?
<img src="<?php echo get_bloginfo('template_directory').'/imgs/Logo.png';?>" alt="Logo" class="logo">
Try this
<img src="<?php echo get_bloginfo('template_directory');?>/imgs/Logo.png" alt="Logo" class="logo">
Related
$images['result'][0]['image_path'] - here I have path of image. It is in public folder.
<img width="150" height="150" alt="150x150" src="<?php BASEPATH."../public/" echo $images['result'][0]['image_path'];?>" />
Can anyone help me to display this image?
Try This,
<img width="150" height="150" alt="150x150" src="<?php echo base_url('your/Folder/Structure/').$images['result'][0]['image_path'];?>" />
First concate your file name with path which causing the issue.
base_url() will also help you to fetch/show your image
I create a block from configuration. Inside this block, I write this code...
<div class="provider-bg" id="provider1">
<img alt="" class="img-responsive" src="<?php echo base_path().path_to_theme() ?>/images/provider-1.jpg" />
</div>
Then, I save with PHP in text format.
My virtual host is ... http://localhost:8888/drupal
So, the image path will be like this ...
<img alt="" src="/drupal/sites/all/themes/myancast/images/ios.png">
This image appears in the last few days ago. Today, I run the site and that image disappear immediately and got 403 error.
Failed to load resource: the server responded with a status of 403 (Forbidden).
I'm trying to find the solution the whole day. But, I still cannot solve.
Can anyone help me please ?
Add global $base_url in your code and use like below
<img alt="" class="img-responsive" src="<?php echo $base_url.'/'.path_to_theme(); ?>/images/provider-1.jpg" />
did you try this ?
<img alt="" class="img-responsive" src="<?php echo '/'.path_to_theme(); ?>/images/provider-1.jpg" />
try this,
<?php
$imgurl = file_create_url(path_to_theme().'/images/provider-1.jpg');
?>
<img alt="" class="img-responsive" src="<?php echo $imgurl ?>"/>
I've encountered this issue a couple times but have always found a "hack" way around it. Is there a special way to link an image in a WordPress template to an outside url beyond the typical a href tag? Here's the images I'm trying to link to outside urls:
<div class="socialMedia">
Follow Us: <br />
<img src="<?php echo get_bloginfo('template_url') ?>/images/facebook.png" alt="facebook"/>
<img src="<?php echo get_bloginfo('template_url') ?>/images/twitter.png" alt="twitter"/>
<img src="<?php echo get_bloginfo('template_url') ?>/images/googleplus.png" alt="google plus"/>
<img src="<?php echo get_bloginfo('template_url') ?>/images/instagram.png" alt="instagram"/>
</div><!--.socialMedia-->
<div class="socialMedia">
Follow Us: <br />
<a href="<?php echo get_bloginfo('template_url') ?>/images/facebook.png" target="_blank" > <img src="<?php echo get_bloginfo('template_url') ?>/images/facebook.png" alt="facebook"/> </a>
</div><!--.socialMedia-->
did you mean it!!?
<li><a href='#'>Core Transformation</a>
<img alt='arrow' src=**'http://localhost/wordpress1/wp-content/themes/twentytwelve/images/header-triangle.png'** /></li>
<li><a href=**'#'**>LINKS</a></li>
<li><a href=**'#'**>Contact Us</a></li>
How can I replace the src and href using the site_url() function?
You can use the following ways :
<img src="<?php bloginfo('template_url'); ?>/images/image.jpg" />
<img src="<?php echo get_template_directory_uri();?>/images/image.jpg" />
<img src="<?php home_url();?>/images/image.jpg" />
<img src="<?php echo site_url();?>/images/image.jpg"/>
you can do it same for href
More details refer codex
<img src="<?php echo site_url();?>/images/image.jpg"/>
or a better option would be:
<img src="<?php bloginfo('template_url'); ?>/images/image.jpg" />
You can do similarly for href
site_url() is used to append text on url.. so if you want to navigate to your directory and fetch something you can use it like this..
site_url('/images/default.jpg');
you can replace your src with following way:
<img src="<?php bloginfo('template_url'); ?>/images/yourimagename.extension" />
for more information about wordpress go to http://www.wpbeginner.com/wp-themes/wordpress-theme-cheat-sheet-for-beginners
that will help you more in future.
Still you getting any issues with that,then comment it.
Not even sure if methods is the correct terminology...
Here is the original working code:
<a href="<?php bloginfo('url'); ?>">
<img src="<?php bloginfo('stylesheet_directory'); ?>/images/logo.png" alt="Polished Logo" id="logo"/></a>
<img src="<?php bloginfo('stylesheet_directory'); ?>/images/separator.png" width="2" height="59" alt="Line" class="logo_line"/>
<p id="logo_title"><?php bloginfo('description'); ?></p>
I wanted it to only execute on the homepage, so I wrote this:
<?
if ( $_SERVER["REQUEST_URI"] == '/' ){
echo '<a href="'.bloginfo('url').'">
<img src="'.bloginfo('stylesheet_directory').'/images/logo.png" alt="Polished Logo" id="logo"/></a>
<img src="'.bloginfo('stylesheet_directory').'/images/separator.png" width="2" height="59" alt="Line" class="logo_line"/>
<p id="logo_title">'.bloginfo('description').'</p>';
}
?>
But it outputs the bloginfo() and the other declarations completely outside the html tags I have created. For instance, with bloginfo('stylesheet_directory') it will display the directory outside the IMG tags I created.
Any ideas? Apparently my syntax isn't correct or something....
bloginfo function directly echoes the output. In this case you should use get_bloginfo to add the returned value to the string and echo the complete string. I believe this will work
<?php
if ( $_SERVER["REQUEST_URI"] == '/' ) {
echo '<a href="'.get_bloginfo('url').'">
<img src="'.get_bloginfo('stylesheet_directory').'/images/logo.png" alt="Polished Logo" id="logo"/></a>
<img src="'.get_bloginfo('stylesheet_directory').'/images/separator.png" width="2" height="59" alt="Line" class="logo_line"/>
<p id="logo_title">'.get_bloginfo('description').'</p>';
}
?>
Here is a better alternative:
<?php if ( $_SERVER["REQUEST_URI"] == '/' ) { ?>
<a href="<?php bloginfo('url') ?>">
<img src="<?php bloginfo('stylesheet_directory') ?>/images/logo.png" alt="Polished Logo" id="logo"/>
</a>
<img src="<?php bloginfo('stylesheet_directory') ?>/images/separator.png" width="2" height="59" alt="Line" class="logo_line"/>
<p id="logo_title"><?php bloginfo('description') ?></p>
<?php } ?>
I also suggest using the is_home() function provided by wordpress to check for the homepage instead of checking the $_SERVER['REQUEST_URI'] value.
bloginfo() outputs data with echo and returns nothing, so instead of trying to concatenate everything, just output in sequence, e.g.
echo '<a href="';
bloginfo('url');
echo '"><img src="';
bloginfo('stylesheet_directory');
//etc...
Ugly I know, but see answer by Nithesh for a possible alternative.
if you want to get the template path without auto echoing it by the bloginfo() function use:
get_bloginfo( 'stylesheet_directory', 'display' )