I checked php.net and read a few examples of how urlencode( ) works but somehow I just can't get it right. Can someone give me a hand?
it'll be a lot to example so hopefully my brief example would make sense.
I have a page called 2.php and it was called to show some contents of a .txt file choosen in 1.php.
I am told to make a link for 3.php and the link should look something like /3?filename=a.txt
with filename as GET parameter name and Ensure GET parameter value is urlencoded using the urlencode( ) function.
but I'm confused how and where I should put urlencode() to make it work.
I'll paste my 2.php code here...I simplified the codes a bit...
<?php
$fileContents = file("./aaa/" . $_GET["course"] . ".txt");
echo "<table border=\"1\">";
foreach($fileContents as $row)
{
echo "<tr>";
$contents = preg_split("/,/", $row);
foreach($contents as $eachline)
{
echo "<td>";
if(!(preg_match("/#/", $eachline)))
{
echo trim(ucfirst($eachline));
}
else
{
echo trim(strtolower($eachline));
}
echo "</td>";
}
echo "</tr>";
}
echo "</table>";
echo "<a href='./1.php'>Choose another txt file</a><br/>";
echo "or<br/>";
echo "<a href='.3.php?'>Work with this txt file</a>";
?>
BUT…the 3.php option must have a query string appended to it: the name of the text file that was selected in 1, so instead of ./3.php, the url should be something such as ./3?filename=asdf.txt
Use “filename” as the GET parameter name. Ensure the GET parameter value is urlencoded using the urlencode( ) function.
but I'm just not sure how to get it to work....
You can wrap the part that should be url encoded in the function within the string:
$url = 'http://www.google.com?q=' . urlencode($search);
OR in html
http://www.google.com?q=<?php echo urlencode($search); ?>
Where . is the concatenation of 2 outputs.
Related
PHP CODE:
$video1="welcome.mp4";
$video2="movie.mp4";
$video3="ends.mp4";
$num_id="1";
$get = '$' . "video" . $num_id;
$file = $get;
echo "<a href=Player.php?file=$file'>Play</a>";
html code results:
Play
HTML - expectancy:
Play
I would do something like this.
$videos = ['welcome.mp4','movie.mp4','ends.mp4'];
foreach($videos as $video){
echo "<a href=Player.php?file=$video'>Play</a><br>";
}
I would do this (If I wanted to be Kool, or confuse a junior developer .. lol )
$video1="welcome.mp4";
$video2="movie.mp4";
$video3="ends.mp4";
$num_id="1";
$get = ${"video".$num_id};
$file = $get;
echo "<a href=Player.php?file=$file'>Play</a>";
Output
<a href=Player.php?file=welcome.mp4'>Play</a>
But I am not sure you are ready to have access to that much power...
Sandbox
I have this issue with my jquery code:
<?php
//I oppen a text document, read the text inside of it and write it inside my html page.
//When someone clicks on a line, I want to take that very same line and send it via select(String) function.
$handle = fopen($_POST['lien'], 'r');
if ($handle)
{
while (!feof($handle))
{
$buffer = fgets($handle);
echo "<div onclick='select(\" ".$buffer." \");'>".$buffer."</div><br/>";
//It works when I put a simple string within the select param:
//echo "<div onclick='select(\" text \");'>".$buffer."</div><br/>";
}
fclose($handle);
}
?>
The jquery code :
function select(text){
alert(text);
//$("#selected").html();
}
where do you guys think is the problem ?
Thanks :)
Maybe Your $buffer contains some double-quotes, which then interfere with surrounding double-quotes. Look at produced HTML code if this is the case, maybe You'll see something like this:
<div onclick='select("He said: "Freeze!"")'>
... which Javascript can't parse correctly. If this is the case, consider using:
echo "<div onclick='select(".json_encode($buffer).")'>";
Try to change line:
echo "<div onclick='select(\" ".$buffer." \");'>".$buffer."</div><br/>";
to
echo "<div onclick='select(\" ".rtrim($buffer)." \");'>".$buffer."</div><br/>";
or to:
echo "<div onclick='select($(this).text());'>".$buffer."</div><br/>";
After looking around for something like octopress in php and not finding anything, I decided to create something myself in php that would do the trick.
I'd like to start with writing some code in php that reads php files and can extract meta-data from them, so I can build an archive page of blog posts, etc.
I thought I could create yaml files, and include php/html in these files for the main content of the blog posts, but it's not clear to me if this is possible at all? Googling around for "use php in yaml" didn't really get me much further.
So I thought I'd ask here what the best approach would be for doing something like this.
Can anyone help?
Thanks
B
I am not familiar with yaml - can you simply use PHP's get meta tags?
<?php
// Assuming the above tags are at www.example.com
$tags = get_meta_tags('http://www.example.com/');
// Notice how the keys are all lowercase now, and
// how . was replaced by _ in the key.
echo $tags['author']; // name
echo $tags['keywords']; // php documentation
echo $tags['description']; // a php manual
echo $tags['geo_position']; // 49.33;-86.59
var_dump($tags);// See any and all meta tags that have been picked up.
?>
Edit: I added the var_dump in so you can see all the tags you get. Test it out on the page you want to hit.
<?php
header('Content-Type:text/html; charset=utf-8');
$tags = get_meta_tags('http://www.narenji.ir');
var_dump($tags);
?>
Output is
array
'keywords' => string 'اخبار, تکنولوژی, نارنجی, گجت, فناوری, موبایل, خبر, تبلت, لپ تاپ, کامپیوتر, ربات, مانیتور, سه بعدی, تلویزیون' (length=186)
'description' => string 'مکانی برای آشنایی با ابزارها و اخبار داغ دنیای فناوری' (length=97)
Or you can use following code
<?php
$url = 'http://www.example.com/';
if (!$fp = fopen($url, 'r')) {
trigger_error("Unable to open URL ($url)", E_USER_ERROR);
}
$meta = stream_get_meta_data($fp);
print_r($meta);
fclose($fp);
?>
If your source file is image then you can try with it
<?php
echo "test1.jpg:<br />\n";
$exif = exif_read_data('tests/test1.jpg', 'IFD0');
echo $exif===false ? "No header data found.<br />\n" : "Image contains headers<br />\n";
$exif = exif_read_data('tests/test2.jpg', 0, true);
echo "test2.jpg:<br />\n";
foreach ($exif as $key => $section) {
foreach ($section as $name => $val) {
echo "$key.$name: $val<br />\n";
}
}
?>
For example if I had the script:
<?php
$page = "My Page";
echo "<title>" . $page . "</title>";
require_once('header.php');
require_once('content.php');
require_once('footer.php');
?>
Is there something I can add to the bottom of that page to show the entire pre-compiled php?
I want to literally echo the php code, and not compile it.
So in my browser I would see the following in code form...
// stuff from main php
$page = "My Page";
echo "<title>" . $page . "</title>";
// stuff from require_once('header.php');
$hello = "Welcome to my site!";
$name = "Bob";
echo "<div>" . $hello . " " . $name . "</div>";
// stuff from require_once('content.php');
echo "<div>Some kool content!!!!!</div>";
// stuff from require_once('footer.php');
$footerbox = "<div>Footer</div>";
echo $footerbox;
Is this possible?
There's no way to do it native to PHP, but you could try to hack it if you just wanted something extremely simplistic and non-robust:
<?php
$php = file_get_contents($_GET['file']);
$php = preg_replace_callback('#^\s*(?:require|include)(?:_once)?\((["\'])(?P<file>[^\\1]+)\\1\);\s*$#m', function($matches) {
$contents = file_get_contents($matches['file']);
return preg_replace('#<\?php(.+?)(?:\?>)?#s', '\\1', $contents);
}, $php);
echo '<pre>', htmlentities($php), '</pre>';
Notes:
Warning: Allowing arbitrary file parsing like I've done with the fist line is a security hole. Do your own authentication, path restricting, etc.
This is not recursive (though it wouldn't take much more work to make it so), so it won't handle included files within other included files and so on.
The regex matching is not robust, and very simplistic.
The included files are assumed to be statically named, within strings. Things like include($foo); or include(__DIR__ . '/foo.php'); will not work.
Disclaimer: Essentially, to do this right, you need to actually parse the PHP code. I only offer the above because it was an interesting problem and I was bored.
echo '$page = "My Page";';
echo 'echo "<title>" . $page . "</title>";';
echo file_get_contents('header.php');
echo file_get_contents('content.php');
echo file_get_contents('footer.php');
For clarity I'd put the title generation in it's own file, then just use a series of echo file_get_contents()...
echo file_get_contents('title.php');
echo file_get_contents('header.php');
echo file_get_contents('content.php');
echo file_get_contents('footer.php');
I wrote this code, it gets an image from a link that varies according to where you are:
<img src='http://chusmix.com/Imagenes/grupos/<?php echo substr(get_search_query(), 1); ?>.jpg'>
I want to make that code run if a PHP condition proves true, but I cannot make it work. It seems that the function doesn't return a value instead it takes the link textually. I mean it goes to http://chusmix.com/Imagenes/grupos/.jpg literally. However the code works correctly by itself.
This is the PHP code:
<?php
$search=get_search_query();
$first=$search[0];
if ($first=="#"){
echo "<html>";
echo "<img src='http://chusmix.com/Imagenes/grupos/<?php echo substr(get_search_query(), 1); ?>.jpg'>";
}
?>
You are already inside the php tag. So there is no need for <?php and ?>.
Try:
echo "<img src='http://chusmix.com/Imagenes/grupos/".substr($search,1).".jpg'>";
Replace line
echo "<img src='http://chusmix.com/Imagenes/grupos/<?php echo substr(get_search_query(), 1); ?>.jpg'>";
with
echo "<img src='http://chusmix.com/Imagenes/grupos/" . substr(get_search_query(), 1) . ".jpg'>";