Hello I have in my DB a column called capalivro that has my image paths.
I want to make the img src the content of that column, but I'm not succeding in doing this since I don't know how to put that PHP var in the HTML.
This way I tried doesn't work because the php doesn't work because of the quotation marks.
<?php while ($livro = mysql_fetch_assoc($livrotodos)) { ?>
<div class="large-2 columns">
<div class="livro">
<div class="livro-overlay">
<h3><?php echo $livro['nomelivro'] ?></h3>
</div>
<img src= "<?php $livro['capalivro']?>" />
</div>
</div>
<?php }
?>
</div>
<img src= "<?php $livro['capalivro']?>" />
</div>
should be:
</div>
<img src= <?php echo "\"". $livro['capalivro']."\""?> />
</div>
change your line containing the img src= bit to:
<img src="<?php echo $livro['capalivro']?>" />
without the echo part, it won't display the "string" from the database.
Related
How to give the content of $topic_image to a variable like this at the source part?
The problem is here: --> src="image/<?php echo $topi_image?>" .
$display_content=<<<END_OF_TEXT
<div>
<img src="image/<?php echo $topic_image ?>" style="width:200px;height:150px;">
<p style="float:left">$topic_pris</p>
<p style="float:left">$topic_title</p>
<p style="float:left">$topic_name</p>
</div>
END_OF_TEXT;
Heredoc expands variables itself. You don't need the whole <?php echo $topic_image ?>, just the variable:
$display_content=<<<END_OF_TEXT
<div>
<img src="image/${topic_image}" style="width:200px;height:150px;">
<p style="float:left">$topic_pris</p>
<p style="float:left">$topic_title</p>
<p style="float:left">$topic_name</p>
</div>
END_OF_TEXT;
Im little bit stuck on a logic.scenario is i have a slider where two images are shown at a time taking 50%-50% part of screen,here is the simple html of it,
<div class="zeus-slider zeus-default" >
<div class="zeus-block">
<div class="zeus-slide s-50"> <img src="slider/slider1.jpg" data-effect="slideRight" alt="" /> </div>
<div class="zeus-slide s-50"> <img src="slider/slider2.jpg" data-effect="slideRight" alt="" /> </div>
</div>
</div>
here every block contains two images i have shown only one.now i have trying to make it dynamic, but i didnt get any idea how it will work, so i created database with two tables, each for individual image in a single block.Both table have fields like simg_id(A_I),img_name,simg_path.
here its my php on slider page.
<div class="zeus-block">
<?php
$slider_slt=getResultSet('select * from slider_img_master1 ORDER BY simg_id');
if(mysql_num_rows($slider_slt))
{
while($slider_data=mysql_fetch_assoc($slider_slt))
{
?>
<div class="zeus-slide s-50"> <img src="slider/first/<?php echo $slider_data['simg_path']?>" data-effect="slideRight" alt="" /> </div>
<?php }
} ?>
<?php
$slider_slt2=getResultSet('select * from slider_img_master2 ORDER BY simg_id');
if(mysql_num_rows($slider_slt2))
{
while($slider_data2=mysql_fetch_assoc($slider_slt2))
{
?>
<div class="zeus-slide s-50"> <img src="slider/second/<?php echo $slider_data2['simg_path']?>" data-effect="slideRight" alt="" /> </div>
<?php }
} ?>
</div>
now the problem is when i try to fetch path of image in slider, images are not changing one by one on both half of screen.instead it showing like both images from from both table top, another two images from both tables below 1st both, and so on , so full page is covered with images.
I know this idea of creating two tables for getting two images at once is silly,but i could not think any better.if any one can suggest any better way, it would be so helpful.
Update:getResultSet is function for mysql_query.
if anybody interested i found an answer for above question.
<div id="slide1" >
<div class="zeus-slider zeus-default" >
<?php
$slider_str="select *from slider_img_master1 where simg_status='Active'";
$i=1;
$result=mysql_query($slider_str);
if(mysql_num_rows($result)>0)
{
echo '<div class="zeus-block">';
while($row=mysql_fetch_assoc($result))
{
if($i%2==1 && $i!=1)
{
echo '<div class="zeus-block">';
}
?>
<div class="zeus-slide s-50"> <img src="slider/<?php echo $row['simg_path'];?>" data-effect="slideRight" alt="" /> </div>
<?php
if($i%2==0)
{
echo '</div>';
}
$i++;
}
} ?>
</div>
<div class="clear"> </div>
<div class="next-block"> </div>
<div class="prev-block"> </div>
</div>
Hello i am updating a WordPress site that was riddled with errors (over 1000...-'-) now i have gotten it down to 15 or so however one page has 105 errors and they are all caused by a stray p tag that is being generated after every image here's what the code is being outputted as
<div id="ngg-image-40" class="ngg-gallery-thumbnail-box" >
<div class="ngg-gallery-thumbnail" >
<a href="a link" title="the title" class="shutterset_set_5" ><br />
<img title="01596-01_1" alt="01596-01_1" src="the src" width="100" height="75" /><br />
</a>
</div>
</p></div>
As you can see there is a p tag there for no reason, I've tried Google but got no one with a solution to this problem, I've tried looking through all the php files for the nextgen gallery and couldn't figure it out the actual code that outputs the gallery is below.
<div id="ngg-image-<?php echo $image->pid ?>" class="ngg-gallery-thumbnail-box" <?php echo $image->style ?> >
<div class="ngg-gallery-thumbnail" >
<a href="<?php echo $image->imageURL ?>" title="<?php echo $image->description ?>" <?php echo $image->thumbcode ?> >
<?php if ( !$image->hidden ) { ?>
<img title="<?php echo $image->alttext ?>" alt="<?php echo $image->alttext ?>" src="<?php echo $image->thumbnailURL ?>" <?php echo $image->size ?> />
<?php } ?>
</a>
</div>
</div>
Again as you can see there is no reference to the p tag in the above. Any and all help is appreciated.
I also did not find a solution within the gallery but I have to admit that I didn't search that well. I was lazy and fixed it with javascript as I use a custom javascript for my gallery. Maybe that helps you, too. It's MooTools btw. and assumes that the gallery div has the id "gallery":
var p = document.id('gallery').getPrevious();
if (p.get('tag') == 'p') {
p.dispose();
}
So I am working on my first wordpress site...
I am having a go at converting http://www.benjaminpotter.org/fleet/
to wordpress > http://www.benjaminpotter.org/test/
So there are already a few issues, but one of the things is that I am wanting to call content into the "about us" section on the index page via wordpress to act as a CMS.
This is an image of the back end:
nevertheless it ends up with no content in that section:
this is most likely because I need to put in a bit of code into the page of the website.
This is the current code for that page, and I have commented where I want to call the content mannaged bit... thanks for the help!
<? /* Template Name: Home Page
*/ ?>
<?php get_header(); ?>
<div id="tagline"></div>
<div id="sliderbox">
<img src="<?php bloginfo('template_url'); ?>/images/index/slider/slider_1.jpg" width="1000" height="466" alt="Don't settle for less" />
<img src="<?php bloginfo('template_url'); ?>/images/index/slider/slider_2.jpg" width="1000" height="466" alt="Don't settle for less" />
<img src="<?php bloginfo('template_url'); ?>/images/index/slider/slider_3.jpg" width="1000" height="466" alt="Don't settle for less" />
<img src="<?php bloginfo('template_url'); ?>/images/index/slider/slider_4.jpg" width="1000" height="466" alt="Don't settle for less" />
</div>
<a id="arrow2" class="arrow"></a>
<a id="arrow1" class="arrow"></a>
<a href="order.php" id="car_logo_slider" onmouseover="tooltip.show('');" onmouseout="tooltip.hide();">
<img src="<?php bloginfo('template_url'); ?>/images/index/car_slider/cars1.jpg" width="1001" height="86" alt="cars" />
<img src="<?php bloginfo('template_url'); ?>/images/index/car_slider/cars2.jpg" width="1001" height="86" alt="cars" />
<img src="<?php bloginfo('template_url'); ?>/images/index/car_slider/cars3.jpg" width="1001" height="86" alt="cars" />
</a>
<div id="wrapper-contunue">
<div id="mission_vision_header"></div>
<? include('inc/quickcontact.html'); ?>
<div id="mission_vision_text">
<!-- THE CONTENT MANAGED BIT! -->
<!-- THE CONTENT MANAGED BIT! -->
<!-- THE CONTENT MANAGED BIT! -->
<!-- THE CONTENT MANAGED BIT! -->
<!-- THE CONTENT MANAGED BIT! -->
</div>
<div style="margin-top:450px;" class="seperator"></div>
<div id="our_process" class="highlighed_div">
<div id="our_process_header"></div>
<div id="initial_text_our_process">
Here at Fleet Avenue, we have a simple and straight forward process towards acquiring our cars for you. This four step process is fast and effective, consisting of the following steps:
</div>
<div id="our_process_stages_graphic"></div>
<div id="explanatiory_text_our_process">
<div id="process_inquire">
<a id="inquire_button" href="order.php"></a>
</div>
</div> <?php get_footer(); ?>
<?php echo get_content(); ?>
Or
<?php
global $post;
echo $post->post_content;
?>
i believe you need to wrap you call to the_content() in the Wordpress loop
<?php
if ( have_posts() ) {
while ( have_posts() ) {
the_content();
} // end while
} // end if
?>
this is a very basic example if you make other calls later you will need to reset the query call after this one.
more info here: https://codex.wordpress.org/The_Loop
So, I'm working on a site where on the top of certain pages I'd like to display a static graphic and on some pages I would like to display an scrolling banner.
So far I set up the condition as follows:
<?php
$regBanner = true;
$regBannerURL = get_bloginfo('stylesheet_directory'); //grabbing WP site URL
?>
and in my markup:
<div id="banner">
<?php
if ($regBanner) {
echo "<img src='" . $regBannerURL . "/style/images/main_site/home_page/mock_banner.jpg' />";
}
else {
echo 'Slider!';
}
?>
</div><!-- end banner -->
In my else statement, where I'm echoing 'Slider!' I would like to output the markup for my slider:
<div id="slider">
<img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/1.jpg" alt="" />
<img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/2.jpg" alt="" />
<img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/3.jpg" alt="" />
.............
</div>
My question is how can I throw the div and all those images into my else echo statement? I'm having trouble escaping the quotes and my slider markup isn't rendering.
<div id="banner">
<?php if($regbanner): ?>
<img src="<?php echo $regBannerURL; ?>/style/images/main_site/home_page/mock_banner.jpg" />
<?php else: ?>
<div id="slider">
<img src="<?php echo ($bannerDir = bloginfo('stylesheet_directory') . '/style/images/main_site/banners/services_banners'); ?>/1.jpg" alt="" />
<img src="<?php echo $bannerDir; ?>/2.jpg" alt="" />
<img src="<?php echo $bannerDir; ?>/3.jpg" alt="" />
.............
</div>
<?php endif; ?>
</div><!-- end banner -->
If you don't like the offered solution using the syntaxif(...):...else...endif; you also have the possibilty of using heredoc-style to include bigger html-parts into an echo-statement without the need of escaping it.
The code-formatting in here unfortunatly messed up my example, which I wanted to post. But if you know the heredoc-notation, it should not be a problem ;)