Update: After looking at my HTML code in browser I figure out I need to only run the SQL code inside php just once in foreach loop and the images will be displayed horizontally. I was wondering how do I make that sql code run only once in that loop?
When I write the below code inside foreach ($ffs as $ff) {
Code:
echo "<h4 class='text-right'>{$Data[$increaseForText]["username_for_info"]}</h4>";
echo "<h2>{$Data[$increaseForText]["_name"]}</h2>";
echo "<p>{$Data[$increaseForText]["_desc"]}<p>";
So when this code is inside that foreach loop the images displays in vertical and I want my images to be displayed in horizontal (one next to other). But, if I remove that code from foreach and put it outside foreach code the image are displayed horizontally and works fine. I have tried CSS to display the image horizontally, but it only works if I remove that code from foreach. For some reason the above code (In foreach) somehow forcing the images to display in vertical, so no matter what I do it displays in vertical (the images).
I can't put my code outside foreach. I know I can use foreach to loop through my SQL code and it works fine, but the thing is I want it to work like first load images then first row only from sql, then 2nd image and 2nd row from sql and for that to make it work the only way is to put inside foreach my sql code, so it loads one at a time or else if I put it outside foreach It will load all the data of sql at once (1 row to 9 let's say) then all the images which makes no sense. I am storing my images in my hosting website files.
My question is how do I force my images to display horizontally one next to other?
My code:
<?php
session_start();
error_reporting(E_ALL);
ini_set('display_errors', '1');
require "navigationbar.php";
require "testing.php";
?>
<html>
<head>
<link rel="stylesheet" href="userprofilestyl.css">
</head>
<body>
<hr>
<?php
global $username;
//username to get data of specific user
$username = $_SESSION['name'];
//to get image by username
$image = "images/$username";
global $increaseForText;
$increaseForText = 0;
function listFolderFiles($dir, $username, $increaseForText)
{
//getting images
$ffs = scandir($dir);
unset($ffs[array_search('.', $ffs, true)]);
unset($ffs[array_search('..', $ffs, true)]);
// prevent empty ordered elements
if (count($ffs) < 1) {
return;
}
$column_count = 0;
$sql = "select username_for_info, _name, _desc
from info_desc where username_for_info = '$username'";
try {
require "testing.php";
$stmt = $conn->prepare($sql);
$stmt->execute();
$Data = $stmt->fetchAll(PDO::FETCH_ASSOC);
echo '<div class="image-container">';
foreach ($ffs as $ff) {
//select data from database
$s = "'<li>'.$ff";
$saving = "$dir/$ff";
$string = "$saving";
global $string_arr;
$string_arr = (explode("/", $string));
$sav;
$sav = '<li>' . $ff;
global $sa;
$sa = "$ff";
if (is_dir($dir . '/' . $ff)) {
listFolderFiles($dir . '/' . $ff, $username, $increaseForText);
}
//printing image
if (is_file($saving)) {
echo '<img src="' . $saving . ' " width="100" height="100" alt="Random image" />';
}
//printing text
echo "<h4 class='text-right'>{$Data[$increaseForText]["username_for_info"]}</h4>";
echo "<h2>{$Data[$increaseForText]["_name"]}</h2>";
echo "<p>{$Data[$increaseForText]["_desc"]}<p>";
$increaseForText++;
}
} catch (PDOException $e) {
echo '{error":{"text":' . $e->getMessage() . '}}';
}
echo '</div>';
}
listFolderFiles($image, $username, $increaseForText);
?>
</body>
</html>
Add this to your css file:
.image-container{
display:inline-flex;
flex-flow:row;
}
This will change the flow of every element inside the div with the class "image-container" from vertically to horizontally
Foreach has nothing to do with making the images display vertically, there should be something wrong with the HTML or CSS. Make sure your tags are closed, such as your p and li tags.
There are alot of ways to make the images display horizontally, it would easier if you can send a jsfiddle or a code snippet.
But try this on .image-container:
display: flex !important;
flex-flow: row;
Related
I have some images in a database which I would like to add to a html page
This is my current code
<div class = "gridrow">
<?php
foreach (range(1, 4) as $value) {
$result = $conn->query("select * from products where product_ID = '".$value."'");
$row = $result->fetch_array();
$name_p1 = $row['product'];
$price_p1 = $row['price'];
$image = "<img src='{$row['image']}'>";
echo "<div class = 'productwindow' >";
echo "<div class = 'productimage'><".$image."></div>";
echo "<div class = 'productvar'><p>".$name_p1."</p></div>";
echo "<div class = 'productvar'><p>$".$price_p1."</p></div>";
echo "</div>";
}
?>
</div>
This is what the page looks like
These are the errors I get
How can I make these images show correctly?
You're storing image data in the database, while the img src tag wants image URLs, that's why it's getting confused and you're getting errors.
The quick way around it is to convert the image data to base64 and pipe it in the src tag like so:
$image = '<img src="data:image/png;base64,'.base64_encode($row['image']).'">';
This is at best a hack, and not a great idea for a host of reasons, it also assumes all your images are PNG.
I'm using the below code to pull through user uploads into a portfolio. It's working great however one image (out of a list of many) is showing blank - yet displays on the next page, using the same path, but within a conventional <img /> tag.
<?php
$files= glob('./uploads/users/'. $item->user_id .'/'.$item->id.'/public_large/*');
$count = 0;
$user_avatar = '/themes/users/assets/img/noartwork.jpg';
foreach ($files as $file) {
$count++;
if ($count > 0 && is_file($file)) {
$user_avatar = $file;
}
}
?>
<div class="user-img" style="background-image:url('<?php echo site_url($user_avatar); ?>');"></div>
When I inspect element on the image that's not displaying, if I change the single quotes background-image:url('<?php echo site_url($user_avatar); ?>'); to double background-image:url("<?php echo site_url($user_avatar); ?>"); it pulls through, but this is not the case if I make this change in the actual code.
Any help would be great.
Thanks
have you tried removing single quotes? like this :
background-image:url(<?php echo site_url($user_avatar); ?>);
I found the fix for this. It was to do with the users file name containing a special character that I wasn't checking for e.g. my-user's-image.jpg breaks the code as the character needs to be escaped my-user\'s-image.jpg.
I amended the foreach to run a check as below and this is now working.
foreach ($files as $file) {
$count++;
if ($count > 0 && is_file($file)) {
$user_avatar = str_replace("'", "\'", $file);
}
}
Thanks
I'm working on a project and it's something new for me. I'll need to fetch rss content from websites, and display Descripion, Title and Images (Thumbnails). Right now i've noticed that some feeds show thumbnails as Enclosure tag and some others dont. right now i have the code for both, but i need to understand how i can create a conditional like:
If the rss returns enclosure image { Do something }
Else { get the common thumb }
Here follow the code that grab the images:
ENCLOSURE TAG IMAGE:
if ($enclosure = $block->get_enclosure())
{
echo "<img src=\"" . $enclosure->get_link() . "\">";
}
NOT ENCLOSURE:
if ($enclosure = $block->get_enclosure())
{
echo '<img src="'.$enclosure->get_thumbnail().'" title="'.$block->get_title().'" width="200" height="200">';
}
=================================================================================================
PS: If we look at both codes they're almost the same, the difference are get_thumbnail and get_link.
Is there a way i can create a conditional to use the correct code and always shows the thumbnail?
Thanks everyone in advance!
EDITED
Here is the full code i have right now:
include_once(ABSPATH . WPINC . '/feed.php');
if(function_exists('fetch_feed')) {
$feed = fetch_feed('http://feeds.bbci.co.uk/news/world/africa/rss.xml'); // this is the external website's RSS feed URL
if (!is_wp_error($feed)) : $feed->init();
$feed->set_output_encoding('UTF-8'); // this is the encoding parameter, and can be left unchanged in almost every case
$feed->handle_content_type(); // this double-checks the encoding type
$feed->set_cache_duration(21600); // 21,600 seconds is six hours
$feed->handle_content_type();
$limit = $feed->get_item_quantity(18); // fetches the 18 most recent RSS feed stories
$items = $feed->get_items(0, $limit); // this sets the limit and array for parsing the feed
endif;
}
$blocks = array_slice($items, 0, 3); // Items zero through six will be displayed here
foreach ($blocks as $block) {
//echo $block->get_date("m d Y");
echo '<div class="single">';
if ($enclosure = $block->get_enclosure())
{
echo '<img class="image_post" src="'.$enclosure->get_link().'" title="'.$block->get_title().'" width="150" height="100">';
}
echo '<div class="description">';
echo '<h3>'. $block->get_title() .'</h3>';
echo '<p>'.$block->get_description().'</p>';
echo '</div>';
echo '<div class="clear"></div>';
echo '</div>';
}
And here are the XML pieces with 2 different tags for images:
Using Thumbnails: view-source:http://feeds.bbci.co.uk/news/world/africa/rss.xml
Using Enclosure: http://feeds.news24.com/articles/news24/SouthAfrica/rss
Is there a way i can create a conditional to use the correct code and always shows the thumbnail?
Sure there is. You've not said in your question what blocks you so I have to assume the reason, but I can imagine multiple.
Is the reason a decisions with more than two alternations?
You handle the scenario of a feed item having no image or an image already:
if ($enclosure = $block->get_enclosure())
{
echo '<img class="image_post" src="'.$enclosure->get_link().'" title="'.$block->get_title().'" width="150" height="100">';
}
With your current scenario there is only one additional alternation which makes it three: if the enclosure is a thumbnail and not a link:
No image (no enclosure)
Image from link (enclosure with link)
Image from thumbnail (enclosure with thumbnail)
And you then don't know how to create a decision of that. This is what basically else-if is for:
if (!$enclosure = $block->get_enclosure())
{
echo "no enclosure: ", "-/-", "\n";
} elseif ($enclosure->get_link()) {
echo "enclosure link: ", $enclosure->get_link(), "\n";
} elseif ($enclosure->get_thumbnail()) {
echo "enclosure thumbnail: ", $enclosure->get_thumbnail(), "\n";
}
This is basically then doing the output based on that. However if you assign the image URL to a variable, you can decide on the output later on:
$image = NULL;
if (!$enclosure = $block->get_enclosure())
{
// nothing to do
} elseif ($enclosure->get_link()) {
$image = $enclosure->get_link();
} elseif ($enclosure->get_thumbnail()) {
$image = $enclosure->get_thumbnail();
}
if (isset($image)) {
// display image
}
And if you then move this more or less complex decision into a function of it's own, it will become even better to read:
$image = feed_item_get_image($block);
if (isset($image)) {
// display image
}
This works quite well until the decision becomes even more complex, but this would go out of scope for an answer on Stackoverflow.
I'm using the basic way to doing the hover image as the CSS method doesn't work for me. Current I'm using the if/else statement to do so. If the contain the URL like abc.com it will hover the image.
But now I only can hover the group url but if there is sub categories in groups I won't able to hover, how can I do it all the activity inside the group, the image will hover?
How to doing if the URL contain the words or path. For example abc.com/groups/* it will hover the groups. Similar like we doing searching in MySQL the words/variable as using "%".
<?php
$request_url = apache_getenv("HTTP_HOST") . apache_getenv("REQUEST_URI");
$e = 'abc.com/dev/';
$f = 'abc.com/dev/groups/';
$g = 'abc.com/dev/user/';
?>
<div class="submenu">
<?php
if ($request_url == $e) {
echo '<div class="icon-home active"></div>';
} else {
echo '<div class = "icon-home"></div>';
}
?>
<?php
if ($request_url == $f) {
echo '<div class="icon-groups active"></div>';
} else {
echo '<div class = "icon-groups"></div>';
}
?>
</div>
I propose a javascript way to do so, with jQuery
$("a[href*='THE_URL_PATTERN_YOU_WANT_TO_MATCH']").children(".icon-home").addClass("active");
BTW, it is NOT a good idea to wrap a div into a a tag.
Im looking to create a condition in wordpress loop. if no image then image box (.thumbHome{display:none})
this is in my function.php
function getThumbImages($postId) {
$iPostID = get_the_ID();
$arrImages =& get_children('post_type=attachment&post_mime_type=image&post_parent=' . $iPostID );
if($arrImages) {
$arrKeys = array_keys($arrImages);
$iNum = $arrKeys[0];
$sThumbUrl = wp_get_attachment_thumb_url($iNum, $something);
$sImgString = '<img src="' . $sThumbUrl . '" alt="thumb Image" title="thumb Image" />';
echo $sImgString;}
else {
echo '<script language="javascript">noImage()</script>';
}
}
And my javascript:
window.onload = noImage();
function noImage(){
document.getElementByClassName('.thumbHome').css.display = 'none';
}
I tried:
window.onload = noImage();
function noImage(){
$('.thumbHome').addClass('hide');
}
RESULT: class hide added to all loop
I cant figure it another way, since im still new in coding.
thx
Well first of all, you don't want to call these functions on window.onload. That's going to immediately set all class instances of .thumbHome to hidden without any conditions.
Here's a very easy way to fix this issue. There are probably more intricate ways, but this works well.
In your main loop, add an unique id to each .thumbHome div based on the image id. So like:
echo '<div class="thumbHome" id="thumb-' . $iNum . '"> ... </div>';
// or you could you use the post ID, doesn't matter, as long as you are consistent
Then your else conditional (for whether there's a thumbnail) could be changed to:
else {
echo '<script type="text/javascript">noImage("#thumb-' . $iNum . '")</script>';
}
and your js function could be:
function noImage(var){
$(var).hide();
}
This is not necessary the best way to do this, it's just the best way with the situtation you find yourself in now.