I have the following link:
http://anydomainname.com/user/username/about
I know that $smarty.server.REQUEST_URI will return user/username/about.
But I can't find a way to return the latest part of my link which is about.
How can I return it? I prefer a solution that does not require me to alter or add new functions in .php files.
Here's an alternative to Marcin's answer using PHP's basename() function instead of substr/strrpos:
{$smarty.server.REQUEST_URI|basename}
You can use:
{$smarty.server.REQUEST_URI|substr:($smarty.server.REQUEST_URI|strrpos:'/'+1)}
It finds the last / in this string and return everything after it.
It is working in Smarty 3.1.19
Related
I am using a script to check links on a given page. I am using simple html DOM to parse the information into an array. I have to check the href of all the a tags to find if they contain a file or something like # or JS.
I tried the following without success.
if(preg_match("|^(.*)|iU", $href)){
save_link();
}
I dont know it my pattern is wrong or if there is a better method to complete this function.
I want to be able to detect if $href contains .com .php .file extensions. This way it will filter out items like # "function()" and other items used in the href attribute.
EDIT:
parse_url will not work stop posting it. The value # returns as a valid url like I stated above I am trying to look for any string followed by .* with no more than 4 chars following the .
I believe that the function you're looking for is parse_url().
This function will take a URL string, and return an array of components, which will allow you to work out what kind of URL it is.
However note that it has issues with incomplete URLs in PHP versions prior to 5.4.7, so you need to have the very latest PHP to get the best out of it.
Hope that helps.
See http://php.net/manual/en/function.parse-url.php
I'm assuming you don't want to match fragments (#) because you are not concerned with following internal anchors.
parse_url breaks up the different parts of the url into an array. You can see the path component of the URL in this array and run your check against that.
You can use parse_url() , like this :
$res = parse_url($href);
if ( $res['scheme'] == 'http' || $res['scheme'] == 'https'){
//valid url
save_link();
}
UPDATE:
I've added code to filter only http and https urls, thanks to Baba for spotting this.
Say I have a url like this in a php variable:
$url = "http://mywebsite.extension/names/level/etc/page/x";
how would I automatically remove everything after the .com (or other extension) and before /page/2?
Basically I would like every url that could be in $url to become http://mywebsite.extension/page/x
Is there a way to do this in php? :s
thanks for your help guys!
I think parse_url() is the function you're looking for. You can use it to break down an URL into it's component parts, and then put it back together however you want, adding in your own things as needed.
As PeeHaa noted, explode() will be useful for dividing up the path.
I'm looking for something that Is really hard for me to do.. I really tried to search all over the net for Solution, But I couldn't seem to find any. I also tried doing this for hours.
What I'm doing: Making a theme for PHPBB2, Installed a MOD that can include PHP in themes.
What is the problem: When I'm doing {} tags in php, It just can't echo those tags.
Let's say I have a function that creates a Table for me, like that:
CreateMyTable(Name,Size,Color);
I put in the function those strings:
CreateMyTable("{FORUM_NAME}",1000,red);
The title stays blank, I actually want it to echo {FORUM_NAME}.
How can I do this?
P.S: I can't do this
CreateMyTable(?>{FORUM_NAME}<?php , 1000, red);
It's not going to work becuase <? = <!-- PHP --> , ?> = <!-- ENDPHP -->.
Thanks for your help :)
If you look in the PHPbb2 template class, you'll find that the template is simply an evaluated set of PHP using the eval() function. You can either print the contents of the PHP before it is parsed using eval() and then use the variable name that the template gives, IE something like (which may not work depending how your template is setup):
CreateMyTable(((isset($this->_tpldata['.'][0]['FORUM_NAME'])) ? $this->_tpldata['.'][0]['FORUM_NAME'] : '' ),1000,randomcolor());
Please note, in order to do it similar to the way above you'd actually have to insert this into your template class.
An much better solution is to avoid using the mod that allows PHP in templates and use JavaScript in the templates to create the function, then print a call to that JavaScript function.
This will work:
CreateMyTable(FORUM_NAME,1000,red);
I also noticed that red is used without quotes - is this also a constant? If it's a variable it needs to have a $ in front of it. If it's a string it should be between quotes.
CreateMyTable(FORUM_NAME,1000,"red");
I want to create a PHP script that grabs the content of a website. So let's say it grabs all the source code for that website and I say which lines of code I need.
Is there a function in PHP that allows you too do this or is it impossible?
Disclaimer: I'm not going to use this for any illegal purposes at all and not asking you too write any code, just tell me if its possible and if you can how I'd go about doing it. Also I'm just asking in general, not for any specific reason. Thanks! :)
file('http://the.url.com') returns an array of lines from a url.
so for the 24th line do this:
$lines = file('http://www.whatever.com');
echo $lines[23];
This sounds like a horrible idea, but here we go:
Use file_get_contents() to get the file. You cannot get the source if the web server first processes it, so you may need to use an extension like .txt. Unless you password protect the file, obviously anybody can get it.
Use explode() with the \n delimiter to split the source code into lines.
Use array_slice() to get the lines you need.
eval() the code.
Note: if you just want the HTML output, then ignore the bit about the source in step 1 and obviously you can skip the whole eval() thing.
So i am using this wordpress function to get the users image
the_author_meta('author_image', the_author_ID()
and it will either return something.jpg or something.png or something.gif if it finds an image otherwise it will return an integer like 2330. How would i do a preg_match or some conditional to let me know if an image is present. I was thinking of doing a preg_match to see if there is a period in the name but if someone has a better idea that would be great..
Simpler:
if (is_numeric($author_image)){
// this is presumably not a file
}
If all you want to do is check the extension of the file to see if it ends with something (ex. '.jpg', '.png', etc.) you can use the solution presented here:
startsWith() and endsWith() functions in PHP
I do not have familiarity with the library that you are using, but there really should be a better way to detect if the file is actually an image (some sort of meta data). Maybe reading the documentation will help?
EDIT: I misread the part about the function returning integers if an image is not found. The is_numeric() solution is probably enough, but I'll leave my answer up to give you options (for example, if you want to distinguish between image types).