i made a function for making a keywords from the post title and replace each word in the full post.
this is my function
function myseonew($title,$text){
$title = stripslashes($title);
$text = stripslashes($text);
$fburl = 'http://' . $_SERVER['SERVER_NAME'] . $_SERVER['REQUEST_URI'];
$keywords = explode(" ",$title);
$regex = '/('.implode('|', $keywords).')/i';
$output = preg_replace($regex, '<a id="smalltext" href="'.$fburl.'">\\1</a>', $text);
return $output;
}
but i faced a problem for the output the characters all of it comes ( �������� ).
so is there any way to solve this issue
by the way my encoding is UTF-8
regards
try using u modifier in your regexp:
$regex = '/('.implode('|', $keywords).')/iu';
Related
I'm trying to create a simple PHP find and replace system by looking at all of the images in the HTML and add a simple bit of code at the start and end of the image source. The image source has something like this:
<img src="img/image-file.jpg">
and it should become into this:
<img src="{{media url="wysiwyg/image-file.jpg"}}"
The Find
="img/image-file1.jpg"
="img/file-2.png"
="img/image3.jpg"
Replace With
="{{media url="wysiwyg/image-file.jpg"}}"
="{{media url="wysiwyg/file-2.png"}}"
="{{media url="wysiwyg/image3.jpg"}}"
The solution is most likely simple yet from all of the research that I have done. It only works with one string not a variety of unpredictable strings.
Current Progress
$oldMessage = "img/";
$deletedFormat = '{{media url="wysiwyg/';
$str = file_get_contents('Content Slots/Compilied Code.html');
$str = str_replace("$oldMessage", "$deletedFormat",$str);
The bit I'm stuck at is find the " at the end of the source to add the end of the required code "}}"
I don't like to build regular expressions to parse HTML, but it seems that in this case, a regular expression will help you:
$reg = '/=["\']img\/([^"\']*)["\']/';
$src = ['="img/image-file1.jpg"', '="img/file-2.png"', '="img/image3.jpg"'];
foreach ($src as $s) {
$str = preg_replace($reg, '={{media url="wysiwyg/$1"}}', $s);
echo "$str\n";
}
Here you have an example on Ideone.
To make it works with your content:
$content = file_get_contents('Content Slots/Compilied Code.html');
$reg = '/=["\']img\/([^"\']*)["\']/';
$final = preg_replace($reg, '={{media url="wysiwyg/$1"}}', $content);
Here you have an example on Ideone.
In my opinion what you are doing is not the best way this can be done. I would use abstract template for this.
<?php
$content = file_get_contents('Content Slots/Compilied Code.html');
preg_match_all('/=\"img\/(.*?)\"/', $content, $matches);
$finds = $matches[1];
$abstract = '="{{media url="wysiwyg/{filename}"}}"';
$concretes = [];
foreach ($finds as $find) {
$concretes[] = str_replace("{filename}", $find, $abstract);
}
// $conretes[] will now have all matches formed properly...
Edit:
To return full html use this:
<?php
$content = file_get_contents('Content Slots/Compilied Code.html');
preg_match_all('/=\"img\/(.*)\"/', $content, $matches);
$finds = $matches[1];
$abstract = '="{{media url="wysiwyg/{filename}"}}"';
foreach ($finds as $find) {
$content = preg_replace('/=\"img\/(.*)\"/', str_replace("{filename}", $find, $abstract), $content, 1);
}
echo $content;
I have a code for embedding a link for iframe.
$post_contetn = explode('htt',$content);
$content_with_link = $post_contetn[0];
$link = 'htt'.$post_contetn[1];
But the problem is that, if I write
http://www.espn.com was great
then it links "was great" is part of the $link.
How can I change (perhaps use regex) to only include the actual url?
======
If I incorporate siam's answer, should it be
$regex = '/https?:\/\/.*?(?=\s)/';
$post_contetn = preg_match($regex, $content, $linkarray);
$content_with_link = $post_contetn[0];
$link = $linkarray[0]
echo $content_with_link;
I then edited to
preg_match($regex, $content, $post_contetn);
$content_with_link = $post_contetn[0];
$link = $post_contetn[0]
echo $content_with_link;
But the error still occurs at echo line.
Try using the following regex :
(?:https?:\/\/\S+)?\S+\.\S+\.?\S+
see demo / explanation
PHP
<?php
$content = 'http://www.espn.com was great';
$regex = '/(?:https?:\/\/\S+)?\S+\.\S+\.?\S+/';
preg_match($regex, $content, $post_contetn);
$link = $post_contetn[0];
echo $link;
?>
I am using following code for convert url to hyperlink on text.But the problem is i want to use shorten title for hyperlink for example this is url http://stackoverflow.com/questions/ask?title=convert%20url%20to%20hyperlink%20on%20text%20as%20formatted and after convert like this :
http://stackoverflow.com/questions/ask?title=convert%20url%20to%20hyperlink%20on%20text%20as%20formatted
I want to this :
http://stackoverflow.com/...
This is my code :
$stringdata = preg_replace('|([\w\d]*)\s?(https?://([\d\w\.-]+\.[\w\.]{2,6})[^\s\]\[\<\>]*/?)|i', '$1 $2', $stringdata);
Title should be shorten but url should be same of original.
Thankyou.
You could always use parse_url found here
Here's an example:
$url = 'http://www.stackoverflow.com/questions/ask?title=convert%20url%20to%20hyperlink%20on%20text%20as%20formatted';
$splitUrl = parse_url($url);
echo '<a href="' . $url . '"/>' . $splitUrl['scheme'] . '://' . $splitUrl['host'] .'/... </a>';
parse_urlcreates an array from the URL provided.
UPDATE:
Using Mikael Roos answer at that link I came up with what you needed it to do.
function make_clickable($text) {
$regex = '#\bhttps?://[^\s()<>]+(?:\([\w\d]+\)|([^[:punct:]\s]|/))#';
return preg_replace_callback($regex, function ($matches) {
$splitUrl = parse_url($matches[0]);
return "<a href='{$matches[0]}'>{$splitUrl['scheme']}://{$splitUrl['host']}/..</a>";
}, $text);
}
echo make_clickable('Some odd text here that makes https://stackoverflow.com/questions/ask?title=convert%20url%20to%20hyperlink%20on%20text%20as%20formatted clickable');
try this
$var='http://stackoverflow.com/questions/ask?title=convert%20url%20to%20hyperlink%20on%20text%20as%20formatted';
$array=explode('/', $var);
$url=$array[0]."//".$array[2]."/...";
What I want
If the URL in the string contains a .jpg at the end of the URL (not the string) then it should make an image from it with preg_replace else make a normal link.
so for example:
If I have http://www.example.com/images/photo.jpg then it should replace with:
<img src="http://www.example.com/images/photo.jpg" alt="http://www.example.com/images/photo.jpg">
The problem:
The URL is replaced with a link in any way and my regex isn't working :( .
What I have tried:
$content = preg_replace("/(http:\/\/[^\s]+(?=\.jpg))/i","<img src=\"$1\" alt = \"$1\"></img>",$content);
$content = nl2br(preg_replace("/(http:\/\/[^\s]+(?!\.jpg))/m", "$1", $content));
Try this
function replace_links($content)
{
if (preg_match('#(http://[^\s]+(?=\.(jpe?g|png|gif)))#i', $content))
{
$content = preg_replace('#(http://[^\s]+(?=\.(jpe?g|png|gif)))(\.(jpe?g|png|gif))#i', '<img src="$1.$2" alt="$1.$2" />', $content);
}
else
{
$content = preg_replace('#(http://[^\s]+(?!\.(jpe?g|png|gif)))#i', '$1', $content);
}
return $content;
}
$content = preg_replace('#\b(http://\S+\.jpg)\b#i', '<img src="$1" alt="$1" />', $content);
You don't need lookaround. Just go with
$content = preg_replace("#(http://[^ ]+\\.jpg(?= |$)#i","<img src=\"$1\" alt=\"$1\"/>", $content);
I think you used the lookahead operator when you wanted lookbehind. You could change (?=\.jpg) to (?<=\.jpg) but there are other, cleaner regex's I'm sure others will post.
This worked for me.
$parse_img='Hello, http://orbitco-ccna-pastquestions.com/images/Q5.jpg
In the figure above, router R1 has two point-to-point . ';
$parse_img=preg_replace('/(https?:\/\/(.\*)?\\.jpg|png|gif)[\s+]*/i',"< img src=\"$1\" alt = \"$1\">< /img >",$parse_img);
echo $parse_img;
Suyash
See i have an url in a html code
play
Now i want to print this url as it is written in a php page
http://b48.ve.vc/b/data/48/3746/05 Dabangg Reloaded_-_www.DjPunjab.Com.mp3
You can see that between the url 05 Dabangg Reloaded their is space. I made this program to print url from this html code..
$str = "play";
$pattern = '`.*?((http|ftp)://[\w#$&+,\/:;=?#.-]+)[^\w#$&+,\/:;=?#.-]*?`i';
if (preg_match_all($pattern,$str,$matches))
foreach($matches[1] as $data)
{
$str=$data;
echo $str;
}
Then i am getting this
http://b48.ve.vc/b/data/48/3746/05
please do not mention on foreach($matches[1] as $data) line bcoz i am using it with so many urls.. I just want to know how to print the whole url in this format.
http://b48.ve.vc/b/data/48/3746/05 Dabangg Reloaded_-_www.DjPunjab.Com.mp3
Spaces are become a huge matter.. Do not know how to fix it..
What i need to add inside
$pattern = '`.*?((http|ftp)://[\w#$&+,\/:;=?#.-]+)[^\w#$&+,\/:;=?#.-]*?`i';
For making it completely workable.
Please suggest me any idea.
$str = 'play';
$arr = explode("\"", $str);
$pattern = '`.*?((http|ftp)://[\w#$&+,\/:;=?#.-]+)[^\w#$&+,\/:;=?#.-]*?`i';
$url = preg_grep($pattern,$arr);
$url = implode('',$url);
Output: $url = 'http://b48.ve.vc/b/data/48/3746/05 Dabangg Reloaded_-_www.DjPunjab.Com.mp3'
Update: 2nd Solution [Reference-DOMElement].
$str = 'play';
$DOM = new DOMDocument;
$DOM->loadHTML($str);
$search_item = $DOM->getElementsByTagName('a');
foreach($search_item as $search_item) {
$url = $search_item->getAttribute('href');
}
echo $url; //Output: http://b48.ve.vc/b/data/48/3746/05 Dabangg Reloaded_-_www.DjPunjab.Com.mp3
You can str_replace each one -space- with %20 for encoding your URL
<?php
$url_org = 'http://b48.ve.vc/b/data/48/3746/05 Dabangg Reloaded_-_www.DjPunjab.Com.mp3';
$url_edited = str_replace(" ", '%20', $url_org);
?>
HERE
This will work.