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PHP parse/syntax errors; and how to solve them
(20 answers)
Reference - What does this error mean in PHP?
(38 answers)
Closed 4 years ago.
I have been attempting to create a user login script in PHP, however I am met with an error telling that $_SESSION["username"] is undefined. I have session_start() inside of $_SERVER["DOCUMENT_ROOT"] . "/assets/php/main.header.php", however the variable still shows as undefined.
<?php
if (!isset($_SESSION["username"])) {
header("Location: /en/");
}
?>
<!DOCTYPE html>
<html>
<head>
<?php
include($_SERVER["DOCUMENT_ROOT"] . "/assets/php/main.header.php");
?>
<title>Profile | Project Club</title>
</head>
<body>
<h1>Profile</h1>
<p><?php echo $_SESSION["username"]?></p>
<?php
include($_SERVER["DOCUMENT_ROOT"] . "/assets/php/main.footer.php");
?>
</body>
</html>
Use session_start() at the beginning of the code.
Related
This question already has answers here:
PHP parse/syntax errors; and how to solve them
(20 answers)
Closed 4 years ago.
When I run this code, it always shows:
Parse error: syntax error, unexpected end of file in /opt/lampp/htdocs/japan_top10_09112018/index.php on line 30
<?php
echo "<title>Top 10 Things to Do in Japan</title>";
require_once('templates/header.php');
?>
<?php
// Ask teacher: does this code is right to show errors or necessary?
ini_set('display_errors', 1);
error_reporting(E_ALL | E_STRICT);
?>
<?php
require_once('database/data.php');
?>
<div class = "container">
<h1>Top 10 things to do in Japan</h1>
<?php
foreach($data as $info) {
print_r($info["Picture"]);
echo "<hr>";
// echo '<img src="$info["Picture"]" class="img-thumbnail">';
?>
<?php
require_once('templates/footer.php');
?>
The line 30 is the last line.
You are forgetting to add a } to close your foreach
This question already has answers here:
PHP parse/syntax errors; and how to solve them
(20 answers)
Closed 4 years ago.
Trying to set up templates on a site I am working on, and I want to change the linked stylesheet document instead of using PHP within the CSS. Would this be possible? Thanks.
<?php
require 'includes/templates.php';
if($_COOKIE['template'] == $blocky) {
echo "<link rel="stylesheet" href="/testpages/css/blocky.css">";
}
else {
echo "<link rel="stylesheet" href="/testpages/css/default.css">";
}
?>
Yes that would possible, but make sure to escape your quotes
<?php
require 'includes/templates.php';
if($_COOKIE['template'] == $blocky) {
echo "<link rel=\"stylesheet\" href=\"/testpages/css/blocky.css\">";
}
else {
echo "<link rel=\"stylesheet\" href=\"/testpages/css/default.css\">";
}
?>
This question already has answers here:
Reference - What does this error mean in PHP?
(38 answers)
How do I get PHP errors to display?
(27 answers)
How can I get useful error messages in PHP?
(41 answers)
Closed 5 years ago.
Learning PHP and MySQL. Made a short script that retrieves contents from the database, and populates it in a tag in my html header.
index.php:
<?php
$link = mysqli_connect('localhost', 'root', 'abdullah');
if (!$link) {
echo "Could not connect";
exit();
}
if (!mysqli_select_db($link, 'chitra')){
echo "Could not find database";
exit();
}
$query = 'SELECT * FROM photos;';
$result = mysqli_query($link, $query);
while ($row = mysqli_fetch_array($result)){
$categories[] = $row['category'];
}
include 'main.html';
?>
main.html:
<!DOCTYPE html>
<html>
<head>
<title>Chitra</title>
<link rel="stylesheet" type="text/css" href="index.css">
</head>
<body>
<header>
<div id="header_username">username</div>
<div id="header_cat">
<select>
<?php foreach ($categories as $category): ?>
<option><?php echo htmlspecialchars($category, ENT_QUOTES, 'UTF-8'); ?></option>
<?php endforeach; ?>
</select>
</div>
</header>
<main>
TODO
</main>
</body>
</html>
This code was initially working. I started working on the body for a while, until suddenly, it started returning me a blank white page. I removed my main content to start over from scratch, but it still give me the same result.
I put echo statemnts throughout the PHP file, everything is executing fine until right before the include statement. I checked my Apache, MySQL, PHP installations, checked that everything is up and running. Everything is. I can't figure out the issue.
You can not use raw PHP functions etc in HTML files,
try to change the extension to .php maybe this helps you.
This question already has answers here:
PHP parse/syntax errors; and how to solve them
(20 answers)
Closed 5 years ago.
I am trying to echo multiple images but i get the error:
"PHP Parse error: syntax error, unexpected '?>' in
/public_html/View/Pages/Home.php on line 17".
Can you give me any guidance on how to fix it? code below
<?php
for($i=0;$i<$length;$i++) {
echo ?><img class="meme-image" src="<?php $meme["$imd_id" == "$i"]->$path?>"><?php
}
?>
Try By this.
<?php
for($i=0;$i<$length;$i++) {
echo '<img class="meme-image" src='.$meme[$imd_id == $i]->$path.' ">';
}
?>
Note : You do not require to close php tag. instead of that insert this in php echo.and also you must read this For best practice.
This question already has answers here:
PHP parse/syntax errors; and how to solve them
(20 answers)
Closed 6 years ago.
I have 35.php that includes common menu for my website. 36.php page should display the menu by using include()
Why doesn't this work?
My error:
35.php is shown below;
<?php
echo "Home
PHP
HTML
CSS <br />";
?>
36.php is shown below;
enter code here
<html>
<body>
<?php include("35.php"); ?>
<p>About Menu is imported from 35.php file</p>
</body>
</html>
You missused the ". They include the string but you also used them for the href. But that closes the string
echo "<a href="
This is where the String ends for PHP and it wants either a ; fo finish the statement or a . to connect it to another string.
So this is what you could to to solve it:
echo 'Home
PHP
HTML
CSS <br />';
With switching the "to a ' The string symbol is now the ' and therefor the " can be used int he string normaly.
Otherwise you could also escape the " like this \" inside the string:
echo "Home ...