I want to display only the file name without displaying the url.
I want to replace the file name instead of 'document '.
while($fet=mysql_fetch_assoc($sql1))
{
$i=$i+1;
$next=$fet['f_name'];
echo '<h4><a class="astext" href="'.$next.'" title="'.$next.'" target="_blank" download>Document'.$i.'</a></h4>';
}
if I replace the word 'document' to $next it shows full url as http://www.sample/txt/1/sample.doc I need to display sample.doc.
echo '<h4><a class="astext" href="'.$next.'" title="'.$next.'" target="_blank" download>"'.$next.'"</a></h4>';
There are two options.
Assume you have $next=http://www.sample/txt/1/sample.doc
Option 1:
This works if you think http://www.sample/txt/1/ is same to all folders.
ltrim($next,"http://www.sample/txt/1/");
Option 2:
use basename($next)
This will extract the sample.doc
<?php
$link = "http://www.sample/txt/1/sample.doc";
$linkArray = explode('/', $link);
echo $linkArray[count($linkArray)-1];
?>
Magesh Kumaar also provide good one
<?php
$link = "http://www.sample/txt/1/sample.doc";
echo basename($link);
?>
I included $next1 = basename($next);
while($fet=mysql_fetch_assoc($sql1))
{
$i=$i+1;
$next=$fet['f_name'];
$next1 = basename($next);
echo '<h4><a class="astext" href="'.$next.'" title="'.$next.'" target="_blank" download>'.$next1.'</a></h4>';
}
Its working now.
<?php
$url = "http://www.sample/txt/1/sample.doc";
echo strstr($url,"sample.");
?>
For More Details
Related
I need to have a working remote IMG display using PHP, let's say like this:
<img src='$src' />
And I need to fetch the remote URL based on $id, where
<?php
$id = "brown_fox";
$url = "http://exampl.com/$id";
get_remote_img($url) {
// some code to get image which SRC is dynamic:
<img id="pic" src="sjf5d85v258d.jpg" />
$src="sjf5d85v258d.jpg";
return $src
}
?>
I hope I explained it understandably.
If I understand you correctly then you can do something like this:
<?php
...
get_remote_img($url) { ...
$src = get_remote_img($url);
// Concatenating the result to the elements src attribute:
echo '<img src='.$src.' />';
?>
What you're looking for is something like this:
<?php
$id = "brown_fox";
$url = "http://exampl.com/" . $id;
...
function get_remote_img($url) {
// some code to get image which SRC is dynamic:
$src="sjf5d85v258d.jpg";
echo "<img id=\"pic\" src=" . "\"" . $src . "\"" . "/>";
return $src;
}
?>
Also, if you want to send and receive query parameters in the URI dynamically through a form, you can take a look at GET Request in PHP.
I am trying to display an image on my webpage using a PHP script to determine which image is displayed.
The image link is as follows:
......
My PHP script is thus:
<?php
$result = $_GET['image'];
echo '<img src="images/gallery/'.$result.'.jpg">';
?>
So what I am trying to achieve in terms of HTML is:
<img src="images/gallery/image01.jpg">
The result I am getting is '"; ?>' displayed on the page.
Any help would be much appreciated!
You have to change your code like this
<?php
$result = $_GET['image'];
?>
<img src="images/gallery/<?php echo $result; ?>.jpg">
<?php
$result = filter_input ( INPUT_GET , 'image' );
if (isset($result) && !empty($result)) {
echo '<img src="images/gallery/'.$result.'.jpg">';
}
?>
You used echo wrong, here is how you should use it.
<?php
$result = $_GET['image'];
?>
<img src="images/gallery/<?php echo $result ?>.jpg">
I would change the gallery.php to this:
<?php $result = $_GET['image']; ?>
<img src="images/gallery/<?php echo $result; ?>.jpg">
That would simply it a little bit. You should echo out the result to see what you are getting when the variable is passed to the gallery page.
echo"<img src='{$image}'>";
$image = uploads/myImage.jpg
I think this is the simplest code. To use a php variable while echoing out html, use curly {} brackets to insert any php variable. For instance, a file upload...
<?php
if(isset($_POST['submit'])){
$filename=$_FILES['file']['name'];
$temp_dir=$_FILES['file']['tmp_name'];
$image = "img/".$filename;
}
?>
<?php if($row2['pack1']==1){ echo "<img src=".BASE_URL."images/1seo.png"; } ?>
I have stored images path in database and images in project folder. Now i am trying to view all IMAGES with their PRODUCTS in my HTML template. I have written the following code and its given me an empty images box in the output and when i open that image box in the new tab, it opens the following URL.
http://localhost/cms/1
Kindly tell me the way of referring images in 'img src ' tag.
include 'connect.php';
$image_query = "select * from product_images";
$image_query_run = mysql_query($image_query);
$image_query_fetch = mysql_fetch_array($image_query_run);
if (!$query=mysql_query ("select product_id,name from products")) {
echo mysql_error();
} else {
while ($query_array = mysql_fetch_array($query)) {
echo '</br><pre>';
$product_id = $query_array['product_id'];
echo "<a href='single_product.php?product_id=$product_id' >";
print_r ($query_array['name']);
echo "</a>";
echo "<img src=". $image_query_fetch
['images'] ." alt=\"\" />";
echo '</pre>';
}
}
} else {
echo "You need to Log in to visit this Page";
}
Add your local image path before source
example
echo "<img src='http://localhost/cms/1/". $image_query_fetch['images'] .'" alt=\"\" />";
*Use PHP *
<?php echo "http://" . $_SERVER['HTTP_HOST'] . $_SERVER['REQUEST_URI']; ?>
You can use the HTML base tag to set a base URL that relative URLs will be resolved from.
See -> http://www.w3schools.com/tags/tag_base.asp
I have a coupon site that display store urls on my store pages. What I want is for only .com at end of each store without showing http:// variations in the beginning
here is my code that displays a store url and I just want domain.com to be displayed instead of http://www.domain.com, also may show as http://domain.com
<p class="store-url"><a href="<?php echo $url_out; ?>" target="_blank"><?php echo $stores_url; ?>
It displays like this because of this function
<div class="store">
<?php // grab the store meta data
$term = get_term_by('slug', get_query_var('term'), get_query_var('taxonomy'));
$stores_url = esc_url(get_metadata(APP_TAX_STORE, $term->term_id, 'clpr_store_url', true));
$dest_url = esc_url(get_metadata(APP_TAX_STORE, $term->term_id, 'clpr_store_aff_url', true));
// if there's a store aff link, then cloak it. else use store url
if ($dest_url)
$url_out = esc_url(home_url(CLPR_STORE_REDIRECT_BASE_URL . $term->slug));
else
$url_out = $stores_url;
?>
What can be done................
Quick and dirty - to demonstrate the possible functions...
<?php
function cleanInputString($inputString) {
// lower chars
$inputString = strtolower($inputString);
// remove whitespaces
$inputString = str_replace(' ', '', $inputString);
// check for .com at the end or add otherwise
if(substr($inputString, -4) == '.com') {
return $inputString;
} else {
return $inputString .'.com';
}
}
// example
$inputStrings = array(
'xyzexamp.com',
'xyzexamp',
'xyz examp'
);
foreach($inputStrings as $string) {
echo('input: '. $string .'; output: '. cleanInputString($string) .'<br />');
}
?>
OUTPUT:
input: xyzexamp.com; output: xyzexamp.com
input: xyzexamp; output: xyzexamp.com
input: xyz examp; output: xyzexamp.com
The "right way" is probably to use the PHP URL-processing:
Break the URL up using http://php.net/manual/en/function.parse-url.php
Remove the scheme element of the resulting array using unset
Build it again using http://www.php.net/manual/en/function.http-build-url.php
This is what preg_replace was made for:
<?php
$http_url = 'http://www.somestore.com/some/path/to/a/page.aspx';
$domain = preg_replace('#^https?://(?:www\.)?(.*?)(?:/.*)$#', '$1', $http_url);
print $domain;
?>
This code will print out
somestore.com
I am trying to create some text based share buttons for my Wordpress that also output the shared amount. So far I got it working with Facebook and Delicious but I am not sure how to get it going with Twitter.
This is what I did for Delicious.
<?php
$shareUrl = urlencode(get_permalink($post->ID));
$shareTitle = urlencode($post->post_title);
$deliciousStats = json_decode(file_get_contents('http://feeds.delicious.com/v2/json/urlinfo/data?url='.$shareUrl));
?>
<a onclick='window.open("http://delicious.com/save?v=5&noui&jump=close&url=<?php echo $shareUrl; ?>&title=<?php echo $shareTitle; ?>", "facebook", "toolbar=no, width=550, height=550"); return false;' href='http://delicious.com/save?v=5&noui&jump=close&url=<?php echo $shareUrl; ?>&title=<?php echo $shareTitle; ?>' class='delicious'>
<?php
if($deliciousStats[0]->total_posts == 0) {
echo 'Save';
} elseif($deliciousStats[0]->total_posts == 1) {
echo 'One save';
} else {
echo $deliciousStats[0]->total_posts.' saves';
}
?>
</a>
I also got the API Url which calls the tweeted numbers and URL.
http://urls.api.twitter.com/1/urls/count.json?url=SOMESITEURLHERE&callback=twttr.receiveCount
Basically it calls the JSON encoded file, and then gives you the option to share the link in <A></A> tags but instead of showing some text such as Share, it will show the count instead. I'm basically creating some CSS share buttons.
Just use Twitter's own tweet button.
It does that for you and you can style it with .twitter-share-button
(I would post this as a reply but I don't have the privilege.)
Probably you have figured out a solution yourself by now. I just had the same problem and solved it this way:
$handle = fopen('http://urls.api.twitter.com/1/urls/count.json?url=nu.nl', 'rb');
$twitCount = json_decode(stream_get_contents($handle));
fclose($handle);
print_r($twitCount->count);
function get_tweets($url) {
$json_string = file_get_contents('http://urls.api.twitter.com/1/urls/count.json?url=' . $url);
$json = json_decode($json_string, true);
return intval( $json['count'] );
}
function total($url){
return get_tweets($url); }
Then, use this to get the twitte share count in required place.
<?php echo total("http://website.com/"); ?>