I´m running a php code inside an HTML code so that depending of the value of a variable a specific CSS file is selected.
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta name = "viewport" content="width=device-width, initial-scale=1.0">
<title>Soluciones Elisar</title>
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/font-awesome/5.13.0/css/all.min.css">
<link href="https://fonts.googleapis.com/css2?family=Oswald:wght#300;500;600&display=swap" rel="stylesheet">
<?php
$monitoreo = $_POST["actividad"];
$file='test.html';
$data = $monitoreo;
if ($data == 0) {
echo "<style type=\"text\css\" media=\"screen\">#import \"/css/styles.css\";</style>";
}
if ($data == 1) {
echo "<style type=\"text\css\" media=\"screen\">#import \"/css/styles2.css\";</style>";
}
if ($data == 2) {
echo "<style type=\"text\css\" media=\"screen\">#import \"/css/styles3.css\";</style>";
}
if ($data == 3) {
echo "<style type=\"text\css\" media=\"screen\">#import \"/css/styles4.css\";</style>";
}
if ($data == 4) {
echo "<style type=\"text\css\" media=\"screen\">#import \"/css/styles5.css\";</style>";
}
if ($data == 5) {
echo "<style type=\"text\css\" media=\"screen\">#import \"/css/styles6.css\";</style>";
}
?>
</head>
<body>
However the HTML file on my server ignores the CSS file and just prints
#import \"/css/styles.css\";"; } if ($data == 1) { echo ""; } if ($data == 2) { echo ""; } if ($data == 3) { echo ""; } if ($data == 4) { echo ""; } if ($data == 5) { echo ""; } ?>
I dont know if it´s my code or if I´m misunderstanding how php and HTML work. Thank you
How are you viewing what the server is producing? If you're using the browser's inspector, bear in mind that this does not show the actual source code — it shows the browser's best interpretation of the source code, with invalid HTML tidied up. You can view the actual source code with the "View Source" option in the browser; the usual shortcut in most browsers is Ctrl+U.
If you do that, my guess (I could be wrong) is that you'll see that the PHP isn't running at all, and what's being served to the browser is the actual PHP code you've written. If I do not mistake my guess, your problem is not a coding issue, but a server configuration issue.
The PHP file should be run and parsed on the server, and the output of that PHP should be sent to the browser, to be parsed there. Exactly how you tell the server to run and parse PHP varies from server to server, but usually it starts with naming the file with the .php extension (if you have full control of the server config, you can tell it to treat .html or even .asp or .whatever files as PHP if you want, but this is unusual and confusing, so you probably shouldn't). You'll also need to ensure that PHP is installed on the server, and that the server's PHP module is enabled for the site. You know your server better than we do, and can search for appropriate tutorials.
Try writing your code with single and double quotes like:
echo '<div class="card">';
It worked for me earlier when I had issues with php.
Related
Im setting up a xampp php website with auto creating css for site (If site named xyz.php/html is created, then a css is created too). Unfortunatelly css doesn't want to include in website using php echo's and html tags. No error.
In style.php:
$arr = explode("/",$_SERVER['PHP_SELF']);
$style = "";
foreach ($arr as $key){
if(strpos($key, ".php")){
$style = str_replace(".php", "style.css", $key);
}
}
if($fp = fopen($_SERVER['DOCUMENT_ROOT']."/TestPHP/".$addr,"wb+")){
fwrite($fp,"body{background-color:#666;}");
fclose($fp);
}
echo $addr = "lib/require/styles/".$style;
echo '<link href="'.$addr.'" rel="stylesheet">';
In index.php:
require_once 'lib/require/php/styles.php';
That's because the only HTML (as i can see in your code) doesn't have anything else than the style, no head, no body... Why don't you directly paste the HTML inside the file instead of making PHP echo it?
I am trying to send the information as a mail to the admin
Below is the mail code and I am reading this using file_get_contents() in another file.
<!doctype html>
<?php
require_once("mysqlconnect.php");
session_start();
?>
<html>
<head>
<link rel="stylesheet" href="css/mail.css">
</head>
<body>
<table border="2">
<?php
$q="select * from profile where mobile=2147483647";
$result=mysqli_query($conn,$q);
while($row = mysqli_fetch_assoc($result))
{
foreach($row as $key=>$value)
{
if($key=="Email")
echo"<tr><th>$key</th><td><a href='mailto:$value; target='_blank'>$value</td></tr>";
else
echo"<tr><th>".str_replace('_',' ',$key)."</th> <td>$value</td></tr>";
}
}
?>
</table>
</body>
</html>
<?php
session_destroy();
?>
The result of this code in the mail is the PHP code itself..
$value) { if($key=="Email") echo"
$key $value
"; else echo"
".str_replace('_',' ',$key)." $value
"; } } ?>
file_get_contents returns the content of a file as string. It does not executes the code inside it.
To use the HTML from a php file you can use php's buffer.
Like this:
ob_start();
include('file_path_here');
$HTML = ob_get_clean();
This will store the output of the file in the $HTML variable.
file_get_contents literally reads in a file as a string. If you're reading in a PHP file using that method, it'll literally return the code, which is what you're getting.
If you want to send HTML as an e-mail, use something like a template system (sitecrafting.com/blog/top-5-php-template-engines) or build a string and pass it onto the mail() function.
PHP documentation: file_get_contents
I would like to import a css stylesheet in a page depending on a php condition (or other), this condition is based upon the domain URL.
For example, if the page loaded is "mydomain.com/about-us" import a "css/about-us.css" file.
I have tried with this code, but it does not work.
<?php
$url = $_SERVER['REQUEST_URI'];
if (strstr($url, "mydomain.com/about-us/")) {
include '/css/about-us.css';
}
?>
How can I import, or use a <style> tag conditionally?
solution correct:
the correct solution is use only the page name, so if you page is mydomain.com/about-us/
use " /about-us/" only.
now have other question, with the code posted you can import css for specific page , but I noticed that if the domain is mydomain.com/about-us/team.html example in the page team.html load also the css of "about-us" how load the css for about-us only in the page mydomain/about-us/ ??
How you can read here, strstr will return a string or FALSE. You can change it like this:
<!DOCTYPE html>
<head>
<?php
if (strstr($_SERVER['REQUEST_URI'], "mydomain.com/about-us/")!=false) {
echo '<link rel="stylesheet" type="text/css" href="/css/about-us.css">';
} ?>
</head>
...
</body>
</html>
Or:
<!DOCTYPE html>
<head>
<style type="text/css">
<?php
if (strstr($_SERVER['REQUEST_URI'], "mydomain.com/about-us/")!=false) {
echo file_get_contents('/css/about-us.css');
} ?>
</style>
</head>
...
</body>
</html>
In the first example your CSS is included through the <link> tag, in the second, the PHP-script loads your CSS file into the script-tags. You can not use include because it will load another php file and execute where it was included. You should use my first example, because it is more server-friendly because the CSS file doesn't need to be read. Your page will be faster.
You can add a stylesheet to the page with PHP by including this in the <head> of your html document:
<?php
echo '<link rel="stylesheet" type="text/css" href="' . $file . '">';
?>
Where $file is the name of the css file. You're going to have to provide some more information as to what you're trying to do for a better answer.
Update
The variable $_SERVER[REQUEST_URI] only gives the requested page, not the whole domain. From the PHP manual,
'REQUEST_URI'
The URI which was given in order to access this page; for instance, '/index.html'.
So the code should look as follows:
<?php
$requested_page = $_SERVER['REQUEST_URI'];
if ($requested_page === "/about-us" || $requested_page === "/about-us/") {
echo '<link rel="stylesheet" type="text/css" href="/css/about-us.css">';
}
?>
This will test if the requested page is "/about-us" (the client is requesting the "about-us" page) and if it does, the link to the stylesheet will be echoed.
use this:
<?php
$url = $_SERVER['REQUEST_URI'];
if (strstr($url, "mydomain.com/about-us/"))
{
// output an HTML css link in the page
echo '<link rel="stylesheet" type="text/css" href="/css/about-us.css" />';
}
else
{
// output an HTML css link in the page
echo '<link rel="stylesheet" type="text/css" href="/css/another.css" />';
}
?>
you can also do this to import the css contents directly, but probably some media/images links can break:
<?php
$url = $_SERVER['REQUEST_URI'];
if (strstr($url, "mydomain.com/about-us/"))
{
// output css directly in the page
echo '<style type="text/css">' .file_get_contents('./css/about-us.css').'</style>';
}
else
{
// output css directly in the page
echo '<style type="text/css">' .file_get_contents('./css/another.css').'</style>';
}
?>
So I downloaded and edited a script off the internet to pull an image and find out the hex values it contains and their percentages:
The script is here:
<?php
$delta = 5;
$reduce_brightness = true;
$reduce_gradients = true;
$num_results = 5;
include_once("colors.inc.php");
$ex=new GetMostCommonColors();
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en">
<head>
<title>Colour Verification</title>
</head>
<body>
<div id="wrap">
<img src="http://www.image.come/image.png" alt="test image" />
<?php
$colors=$ex->Get_Color("http://www.image.come/image.png", $num_results, $reduce_brightness, $reduce_gradients, $delta);
$success = true;
foreach ( $colors as $hex => $count ) {
if ($hex !== 'e6af23') {$success = false; }
if ($hex == 'e6af23' && $count > 0.05) {$success = true; break;}
}
if ($success == true) { echo "This is the correct colour. Success!"; } else { echo "This is NOT the correct colour. Failure!"; }
?>
</div>
</body>
</html>
Here is a pastebin link to the file colors.inc.php
http://pastebin.com/phUe5Pad
Now the script works absolutely fine if I use an image that is on the server, eg use /image.png in the Get_Color function. However, if I try and use an image from another website including a http://www.site.com/image.png then the script no longer works and this error appears:
Warning: Invalid argument supplied for foreach() in ... on line 22
Is anyone able to see a way that I would be able to hotlink to images because this was the whole point of using the script!
You must download a file to the server and pass its full filename to the method Get_Color($img) as $img parameter.
So, you need to investigate another SO question: Download File to server from URL
The error indicates that the value returned by Get_Color is not a valid object that can be iterated on, probably not a collection. You need to know how the Get_Color works internally and what is returned when it doesn't get what it wants.
In the mean-time, you can download [with PHP] the image from the external url into your site, and into the required folder and read the image from there.
$image = "http://www.image.come/image.png";
download($image, 'folderName'); //your custom function
dnld_image_name = getNameOfImage();
$colors=$ex->Get_Color("/foldername/dnld_image_name.png");
By the way, did you confirm that the image url was correct?
Why does file_get_contents not work for me? In the test file code below, it seems that everyone's examples that I've searched for all have this function listed, but it never gets executed. Is this a problem with the web hosting service? Can someone test this code on their server just to see if the geocoding array output actually gets printed out as a string? Of course, I am trying to assign the output to a variable, but there is no output here in this test file....
<html>
<head>
<title>Test File</title>
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false">
</script>
</head>
<body>
<?
$adr = 'Sydney+NSW';
echo $adr;
$url = "http://maps.googleapis.com/maps/api/geocode/json?address=$adr&sensor=false";
echo '<p>'.$url.'</p>';
echo file_get_contents($url);
print '<p>'.file_get_contents($url).'</p>';
$jsonData = file_get_contents($url);
echo $jsonData;
?>
</body>
</html>
Check file_get_contents PHP Manual return value. If the value is FALSE then it could not read the file. If the value is NULL then the function itself is disabled.
To learn more what might gone wrong with the file_get_contents operation you must enable error reporting and the display of errors to actually read them.
# Enable Error Reporting and Display:
error_reporting(~0);
ini_set('display_errors', 1);
You can get more details about the why the call is failing by checking the INI values on your server. One value the directly effects the file_get_contents function is allow_url_fopen. You can do this by running the following code. You should note, that if it reports that fopen is not allowed, then you'll have to ask your provider to change this setting on your server in order for any code that require this function to work with URLs.
<html>
<head>
<title>Test File</title>
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false">
</script>
</head>
<body>
<?php
# Enable Error Reporting and Display:
error_reporting(~0);
ini_set('display_errors', 1);
$adr = 'Sydney+NSW';
echo $adr;
$url = "http://maps.googleapis.com/maps/api/geocode/json?address=$adr&sensor=false";
echo '<p>', $url, '</p>';
$jsonData = file_get_contents($url);
echo '<pre>', htmlspecialchars(substr($jsonData, 128)), sprintf(' ... (%d)', strlen((string)$jsonData)), '</pre>';
# Output information about allow_url_fopen:
if (ini_get('allow_url_fopen') == 1) {
echo '<p style="color: #0A0;">fopen is allowed on this host.</p>';
} else {
echo '<p style="color: #A00;">fopen is not allowed on this host.</p>';
}
# Decide what to do based on return value:
if ($jsonData === FALSE) {
echo "Failed to open the URL ", htmlspecialchars($url);
} elseif ($jsonData === NULL) {
echo "Function is disabled.";
} else {
echo '<pre>', htmlspecialchars($jsonData), '</pre>';
}
?>
</body>
</html>
If all of this fails, it might be due to the use of short open tags, <?. The example code in this answer has been therefore changed to make use of <?php to work correctly as this is guaranteed to work on in all version of PHP, no matter what configuration options are set. To do so for your own script, just replace <? or <?php.
If PHP's allow_url_fopen ini directive is set to true, and if curl doesn't work either (see this answer for an example of how to use it instead of file_get_contents), then the problem could be that your server has a firewall preventing scripts from getting the contents of arbitrary urls (which could potentially allow malicious code to fetch things).
I had this problem, and found that the solution for me was to edit the firewall settings to explicitly allow requests to the domain (or IP address) in question.
If it is a local file, you have to wrap it in htmlspecialchars like so:
$myfile = htmlspecialchars(file_get_contents($file_name));
Then it works
Wrap your $adr in urlencode().
I was having this problem and this solved it for me.
//JUST ADD urlencode();
$url = urlencode("http://maps.googleapis.com/maps/api/geocode/json?address=$adr&sensor=false");
<html>
<head>
<title>Test File</title>
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false">
</script>
</head>
<body>
<?php
$adr = 'Sydney+NSW';
echo $adr;
$url = "http://maps.googleapis.com/maps/api/geocode/json?address=$adr&sensor=false";
echo '<p>'.$url.'</p>';
echo file_get_contents($url);
print '<p>'.file_get_contents($url).'</p>';
$jsonData = file_get_contents($url);
echo $jsonData;
?>
</body>
</html>
The error may be that you need to change the permission of folder and file which you are going to access. If like GoDaddy service you can access the file and change the permission or by ssh use the command like:
sudo chmod 775 file.jpeg
and then you can access if the above mentioned problems are not your case.