Using a foreach loop in codeigniter, I am trying to display website links from database in href html tags, But it echos the siteurl instead of linking to the website link.
<?php foreach ($as as $a): {
<li>
<strong>Web</strong>
<a href="<?php echo $a['website']; ?>" target="_blank" rel="nowfollow"><?
php echo $a['website']; ?></a></div>
</li>
<?php }endforeach; ?>
for example www.instagram.com was in the database it will link similar to this
localhost/site/view/www.instagram.com/p/BgHH1TyDqvp/ instead of just linking too www.instagram.com
I've noticed that if the protocol (http or https) isn't at the front of an external link then it'll handle it as a relative path. If I know there's a chance of an external link in the database I'll check for the protocol and add it if needed.
Updated Code
<?php foreach($as as $a) :
// Check if we have the protocol
$pro = (substr($a['website'], 0, 4) != 'http' ? 'https://' : '');
?>
<li>
<strong>Web</strong>
<a href="<?= $pro . $a['website'] ?>" target="_blank" rel="nofollow">
<?= $a['website'] ?>
</a>
</li>
<?php endforeach; ?>
This is based off memory and haven't tested this specific code block.
I don't claim this to be the best way to handle this, but has worked for me in the past.
Related
Here in My View:
<div class="dropdown">
<button class="btn dropdown-toggle" type="button" data- toggle="dropdown">Dropdown Example
<span class="caret"></span></button>
<ul class="dropdown-menu">
<?php
foreach($site as $sites)
{
echo '<li >"'.$sites->site_title.'" </li>';
}
?>
</ul>
</div>
<?php
}
?>
i want to redirect when user click on $sites->site_title
but how it working is it automatically redirects to url
enter code here"<li><a href='shop/viewSiteId?id=".$sites->site_id."'>".$sites->site_title."</a></li>";
and get that id by using GET method
A redirect is the programmatic way to send a browser to a URL. In other words, a call to redirect is like clicking a link. They are not intended to work together in the way you have tried.
<?php
foreach($site as $sites) : ?>
<li>
<a href='<?= base_url("shop/viewSiteId?={$sites->site_id}"); ?>'><?=$sites->site_title; ?></a>
</li>
<?php endforeach;
If you are not familiar with the syntax, know that <?= is the shortcut way to write <?php echo
I've also used PHP Alternative Syntax for Control Structures and dropped in and out of the PHP processor a bunch of times. For me, that is the easier way to read and write this kind of thing. (Your mileage may vary.)
I am trying to see if a file in my database exists, and if it does, show that file as a hyperlinked button. The file is a URL. I am able to get the file to show accordingly, but the button shows even if there is no hyperlink. I have been researching everywhere to find out how I can echo the button only if there is a corresponding file but have had no luck. Just can get it right. Here is what I have and appreciate the help.
<?php if (file_exists($video))?><a class="videobutton" href="<?php echo $video; ?>"></a>
So based on Your Comment that you want to hide the whole button if the file is not found so this is how to do it :
<?php
if (file_exists($video)): ?>
<a class="videobutton" href="<?php echo $video; ?>"></a>
<?php endif; ?>
OR
if (file_exists($video)){ ?>
<a class="videobutton" href="<?php echo $video; ?>"></a>
<?php } ?>
<?php if (file_exists($video)){?>
<a class="videobutton" href="<?php echo $video; ?>"></a>
<?php }
else {
///
} ?>
In my opinion you missed the "{" brackets within the if statement.
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 am building a wordpress site with galleries in it and want to add a 'Back to Gallery' link in my single-attachment.php that will send the user back one directory.
For example, if the image is at www.site.com/galleries/gallery1/image1/ I would want the link to send it back to www.site.com/galleries/gallery1/.
Any suggestions would be appreciated.
You can either explode current URL,and remove last term from the URL and give it as link for Back to Gallery.
<?php $url = explode('/', 'www.site.com/galleries/gallery1/image1/');
array_pop($url);
$back=implode('/', $url);
?>
<a href="<?php echo $back; ?>" >Back to Gallery</a>
Or you can use $_SERVER['HTTP_REFERER'] for getting the previous URL.
<a href="<?php echo $_SERVER['HTTP_REFERER']; ?>" >Back to Gallery</a>
also by using javascript you can implement this in easy way :
<a href="javascript:history.go(-1)" >Back to Gallery</a>
I'm building a wordpress theme. In the backend, the user has the option to enter the url for their social networks (i.e. twitter, facebook, instagram etc). These URL's are then dynamically added to images in the theme front end linking to the respective networks.
The issue I have is that if a user doesn't enter a url for a network, I don't want the image to display. I am trying to write code that says, if the url is blank, echo 'class="hidden"' - this class has display:none in the css.
here is a snippet of my php:
<ul class="icons">
<li <?php if (get_theme_mod('footer_twitter')==0) {echo 'class="hidden"'; } ?>><span class="label">Twitter</span></li>
</ul>
and the css:
ul.icons li.hidden {
display:none;
}
The php code above currently outputs the echo statement for all cases, even when a url is entered in the backend. Can anyone help me with this code
Check the return of "get_theme_mod()" You can check this by using, cause i dont think it "== 0". http://codex.wordpress.org/Function_Reference/get_theme_mod
var_dump(get_theme_mod('footer_twitter'));
//string(0)
Here is your new code:
<ul class="icons">
<li class="<?php echo empty(get_theme_mod('footer_twitter')) ? 'hidden' : ''; ?>">
<a href="<?php echo get_theme_mod('footer_twitter'); ?> " class="icon circle fa-twitter">
<span class="label">Twitter</span>
</a>
</li>
</ul>
Please check this Syntax: http://php.net/manual/en/control-structures.alternative-syntax.php its the best way to code control structures in your "View" code.
<?php if (!empty (get_theme_mod( 'set_address2' ))) {
echo get_theme_mod( 'set_address2');
}?>
This seemed to work for me in Wordpress.
Let me know if this works.