I have to scrape data from google.
For example, this URL:
https://www.google.com/searchbyimage?&image_url=https://traum-deutung.de/wp-content/uploads/2016/04/traumdeutung-offener-sarg-symbol-768x768.jpg
but when I scrape it and echo the data variable, it simply shows me the google page like this:
This is actual result from google image search but I didn't find its code in my php script:
the code i use for this
<?php
require 'simple_html_dom.php';
$link = $_GET['link'];
if (!isset($link) || empty($link)) {
echo "please provide a link". "<br> \n";
echo "EXAMPLE : <br>";
echo "example.com?link=HERE PUT IMAGE LINK";
die();
}
$html = file_get_html('https://www.google.com/searchbyimage?&image_url='. $link);
echo $html;
?>
Related
Im new to Php. I kinda need code to open Links in browser when script is loaded.
heres my code below.
<?php
$links = array_open("https://stackoverflow.com/",
"https://outlook.office.com/",
"https://www.protectedtext.com/",
"https://www.adobe.com/",
"https://www.linkedin.com");
echo $links[array_rand($links)];
?>
Hi change array_open to array
<?php
$links = array("https://stackoverflow.com/", "https://outlook.office.com/", "https://www.protectedtext.com/", "https://www.adobe.com/", "https://www.linkedin.com");
echo $links[array_rand($links)];
?>
And if you want to send the browser to the link you would send a location header like so...
<?php
$links = array("https://stackoverflow.com/", "https://outlook.office.com/", "https://www.protectedtext.com/", "https://www.adobe.com/", "https://www.linkedin.com");
header('Location: ' . $links[array_rand($links)]);
exit;
?>
I do PHP script, the script must copy the list of publications (from the homepage) and copy the information that is inside these publications.
I need to copy content from my previous site and add the content to the new site!
I have some success, my PHP script copies the list of publications on the home page. I need to make a script that pulled information inside each publication (title, photo, full text)!
For this, I wrote a function that extracts a link to each post.
Help me write a function that will copy information on a given link!
<?php
header('Content-type: text/html; charset=utf-8');
require 'phpQuery.php';
function print_arr($arr){
echo '<pre>' . print_r($arr, true) . '</pre>';
}
$url = 'http://goruzont.blogspot.com/';
$file = file_get_contents($url);
$doc = phpQuery::newDocument($file);
foreach($doc->find('.blog-posts .post-outer .post') as $article){
$article = pq($article);
$text = $article->find('.entry-title a')->html();
print_arr($text);
$texturl = $article->find('.entry-title a')->attr('href');
echo $texturl;
$text = $article->find('.date-header')->html();
print_arr($text);
$img = $article->find('.thumb a')->attr('style');
$img."<br>"; if (preg_match('!background:url.(.+). no!',$img,$match)) {
$imgurl = $match[1];
} else
{echo "<img src = http://goruzont.blogspot.com".$item.">";}
echo "<img src='$imgurl'>";
}
?>
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 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/"); ?>
I am trying to link a tumblr feed to a website. I found this code (As you can see, something must be broken with it as it doesnt even format correctly in this post):
<?php
$request_url = “http://thewalkingtree.tumblr.com/api/read?type=post&start=0&num=1”;
$xml = simplexml_load_file($request_url);
$title = $xml->posts->post->{‘regular-title’};
$post = $xml->posts->post->{‘regular-body’};
$link = $xml->posts->post[‘url’];
$small_post = substr($post,0,320);
echo ‘<h1>’.$title.’</h1>’;
echo ‘<p>’.$small_post.’</p>’;
echo “…”;
echo “</br><a target=frame2 href=’”.$link.”’>Read More</a>”;
?>
And i inserted the tumblr link that I will be using. When I try to preview my HTML, i get a bunch of messed up code that reads as follows:
posts->post->{'regular-title'}; $post = $xml->posts->post->{'regular-body'}; $link = $xml->posts->post['url']; $small_post = substr($post,0,320); echo '
'.$title.'
'; echo '
'.$small_post.'
'; echo "…"; echo "Read More"; ?>
Any help would be appreciated. Thank you!
That is PHP, not HTML. You need to process it with a PHP parser before delivering it to a web browser.
… it should also be rewritten so it can cache the remote data, and escape special characters before injecting the data into an HTML document.