I'm using jQueryZoom that need two images to be displayed like this (one image for the 'a' tag and another for the 'img'):
<a href="images/BIGIMAGE.JPG">
<img src="images/SMALLIMAGE.JPG">
</a>
But I need bring those images from two diferents directories from MySQL (the default directory: 'zoom' and the directory: 'normal'. So I'm trying this on my HTML body:
<?php if ( $imagePath = $results02['adaptador']->getImagePathFrontend() && $imagePathNormal = $results02['adaptador']->getImagePathFrontend( IMG_TYPE_NORMAL ) ) { ?>
<a href="<?php echo $imagePath ?>" rel='gal1' id="demo1" >
<img src="<?php echo $imagePathNormal ?>" />
</a>
<?php } ?>
It's bringing me just the second one. In the first one it's returns a true value (1). This is the code after load the page on the browser:
<a href="1" rel='gal1' id="demo1" >
<img src="../../images/produtos/adaptadores/normal/Adaptador 2P.png" />
</a>
Why the 'if' is returning true on the first option instead of the image path like in the second one?
You have to group the assignments separately. Right now you're assigning to $imagePath the result of a boolean operation.
if ( ($imagePath = $results02['adaptador']->getImagePathFrontend()) &&
($imagePathNormal = $results02['adaptador']->getImagePathFrontend( IMG_TYPE_NORMAL )) )
{
// ...
}
Either set image path outside or put parenthesis around it. your order of operations is probably confusing you
<?php
$imagePath = $results02['adaptador']->getImagePathFrontend();
if ( $imagePath && $imagePathNormal = $results02['adaptador']->getImagePathFrontend( IMG_TYPE_NORMAL ) ) { ?>
<a href="<?php echo $imagePath ?>" rel='gal1' id="demo1" ><img src="<?php echo $imagePathNormal ?>" /></a>
<?php } ?>
Related
I'm trying to display a link only if it has a value.
How can i get the image if the_field imdb is not empty?
<a href="<?php the_field('imdb'); ?>" >
<img style="width:60px;"src="/img/link.png" /></a>
Use get_field(); instead of the_filed();.
if(!empty(get_field("fildname"))){
#your code hear
}
Try:-
if(!empty(the_field('imdb'))){
<a href="<?php the_field('imdb'); ?>" ><img style="width:60px;"src="/img/link.png" /></a>
}
Use isset
if(isset(the_field('imdb'))&&!empty(the_field('imdb'))){
<img src=""/>
}
Since i didnt want to show the img or link on older posts i solved it by:
<?php
// assign image
$imdbimg = "<img src='http://mydomain.com/IMDb.png' class='imdb' />";
// imdb link from custom field
$imdblink = get_field('imdb');
?>
// display img and link on post newer then id 16
<?php global $post;
if ( $post->ID >= 16 ){
echo ' ' . $imdbimg . '';
}
?>
I am new in this so I'll try to be clear as I can.
I want to display\output, using php, page an image link html tag if user filed not empty like this on client side:
<a href="[dynamic from user filed]" title="My facebook">
<img src="images/facebook.png" alt="facebook" /></a>
so I wrote this but it always displays on html (client) the filed data as text without the all the html code (no html on client):
<div>
<?php
$Usr_Url = bp_member_profile_data('field=facebook' );
if ( !empty( $Usr_Url )) { ?>
<a href="<?php $Usr_Url; ?>" title="My facebook">
<img src="images/facebook.png" alt="facebook" /></a>
<?php } ?>
</div>
I suppose it is security issue so I need to make the code a server side code or something, can you give an advice please?
the current output seems to be always the user filed on html: www.facebook.com/try
*(need to add the I am retrieving buddypress field bp_member_profile_data(filed='filedname') and that sections works)
You are doing it the wrong way. Do it like this:
<?php
$Usr_Url = bp_member_profile_data('field=facebook' );
if ( !empty( $Usr_Url ))
{
echo "<a href=".$Usr_Url."title='My facebook' <img src='images/facebook.png' alt='facebook' /></a>";
}
?>
Comment if there are errors.
If wanting to output text from inside a PHP code block you need to use echo:
<?php
$Usr_Url = 'http://example.com';
echo 'Example URL';
...
?>
Or alternatively, you can mix in PHP code blocks into your HTML:
<?php $Usr_Url = 'http://example.com'; ?>
Example URL
...
You can't just stick HTML elements into your PHP code blocks though, as this breaks the syntax as in the original post.
To rework your code:
<div>
<?php
$Usr_Url = bp_member_profile_data('field=facebook' );
if ( !empty( $Usr_Url )) {
echo '<a href="'.$Usr_Url.'" title="My facebook">';
echo '<img src="images/facebook.png" alt="facebook" /></a>';
} ?>
</div>
You can try this out. It displays the image when Usr_Url is NOT empty:
<?php
$Usr_Url = bp_member_profile_data('field=facebook' );
if ( $Usr_Url != "" ) {
echo '<img src="images/facebook.png" alt="facebook" />'; } else { echo '$Usr_Url is empty...'; }
?>
I hope it helps :)
this is my first post for help on here, and man do I really need it. This is the first time I've developed a client's site using multisite, and I'm having trouble applying the appropriate header image to it's site. There are six sites in all, and I'm using the same template for all six sites' front pages. Also, the front page is static and doesn't have a specific page selected.
The conditional below is my attempt at specifying specific images depending on which sub-site I'm on. It keeps throwing a syntax error, (sublime calls it a parse error). I would be so grateful for any help!
<?php
if( get_bloginfo('All in with Laila Ali')) {
<img src="<?php bloginfo('template_directory');?>/images/Banner-LailaAli.jpg" />
} elseif{
if( get_bloginfo('Jaimies 15 Minute Meals')) {
<img src="<?php bloginfo('template_directory');?>/images/Banner-JamieOliver.jpg" />
}
} elseif{
if( get_bloginfo('Lucky Dog')) {
<img src="<?php bloginfo('template_directory');?>/images/Banner-LuckyDog.jpg" />
}
} elseif{
if( get_bloginfo('Game Changers with Kevin Frazier')) {
<img src="<?php bloginfo('template_directory');?>/images/Banner-GameChangers.jpg" />
}
} elseif{
if( get_bloginfo('Recipe Rehab')) {
<img src="<?php bloginfo('template_directory');?>/images/Banner-RecipeRehab.jpg" />
}
} else {
<img src="<?php bloginfo('template_directory');?>/images/Banner-PetVet.jpg" />
}
?>
You are getting this error because you have consecutive <?php tags with HTML in between them. Things like <img src=" aren't valid php, but you are inside <?php tags, so PHP gives you an error.
One of the ways you can fix this is by ending the tags when you need to switch back to HTML. Like this:
<?php
if( get_bloginfo('All in with Laila Ali')) {
?>
<img src="<?php bloginfo('template_directory');?>/images/Banner-LailaAli.jpg" />
<?php
} elseif{
if( get_bloginfo('Jaimies 15 Minute Meals')) {
?>
<img src="<?php bloginfo('template_directory');?>/images/Banner-JamieOliver.jpg" />
<?php
}
}
?>
Another way, is to have your PHP echo the HTML you want. Which would look something like this:
<?php
if( get_bloginfo('All in with Laila Ali')) {
echo '<img src="' . bloginfo('template_directory') . '/images/Banner-LailaAli.jpg" />';
} elseif{
if( get_bloginfo('Jaimies 15 Minute Meals')) {
echo '<img src="'. bloginfo('template_directory') . '/images/Banner-JamieOliver.jpg" />';
}
}
?>
Basically, you have to remember that once you open a <?php tag, you can only put valid PHP until you close it with ?>.
I have a PHP and a jQuery script that I use to display a few images. A large image on the left side and 4 thumbnails on the right side. Each time the user clicks an image-thumbnail it will show up on the large image placeholder on the left side.
This is the PHP code I'm using to display the large image and thumbnails:
<div class="pic"><img title="<?php echo $current->alttext ?>" alt="<?php echo $current->alttext ?>" src="<?php echo $current->imageURL; ?>" />
</div>
<ul class="ngg-gallery-list-ir">
<!-- Thumbnail list -->
<?php foreach ( $images as $image ) : ?>
<?php if ( $image->hidden ) continue; ?>
<li id="ngg-image-<?php echo $image->pid ?>" class="ngg-thumbnail-list <?php if ($image->pid == $current->pid) echo 'selected' ?>" >
<a href="<?php echo $image->imageURL ?>" title="<?php echo $image->description ?>" >
<img title="<?php echo $image->alttext ?>" alt="<?php echo $image->alttext ?>" src="<?php echo $image->thumbnailURL ?>" <?php echo $image->size ?> />
</a>
</li>
<?php endforeach; ?>
This is the jQuery I'm using to update the large image when an user clicks on any thumbnail-image:
jQuery(document).ready(function($){
// handle the click of thumbnail images
// redirect it to change the main image
$(".ngg-thumbnail-list a").click(function(){
var src = $(this).attr("href");
$(".ngg-galleryoverview-ir .pic img").attr("src", src);
return false;
});
// preload the large images
function preload(arrayOfImages) {
$(arrayOfImages).each(function(){
$('<img/>')[0].src = this;
});
}
// populate the list of images to load
preload(images);
});
Everything works fine in this setup but I also need to display below the main large-image its title and description. This is the code I'm using:
<div class="container-title-description">
<div class="title"><?php echo $current->alttext ?></div>
<div class="descripton"><?php echo $current->caption ?></div>
</div>
The problem is this: if I add this code inside the foreach loop I get the title and description below each thumbnail-image. If I add this code outside the foreach loop when the main image changes the title and description will stay the same. How can I solve this?
You can view how this setup looks like on this website.
You already add the title and description as hidden title attributes inside the anchor element, so just extract them and update the HTML on demand:
$(".ngg-thumbnail-list a").click(function(){
var src = $(this).attr("href"),
desc = $(this).attr('title'),
title = $(this).find('img').attr('title');
$(".ngg-galleryoverview-ir .pic img").attr("src", src);
$('.container-title-description .title').text(title);
$('.container-title-description .description').text(desc);
return false;
});
Initial HTML (outside your foreach loop):
<div class="container-title-description">
<p class="title"></p>
<p class="description"></p>
</div>
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"/>