Error code in php [closed] - php

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Closed 8 years ago.
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So on my website im making it where if a member is not sign in they can't see username My account or settings link
this is what i have so far
<?php
if(empty($_SESSION['user'])){
echo "<li>
Register Account
</li>
<li>
Login Account
</li>";
} else {
echo "<li class='dropdown'>";
echo "<a class='dropdown-toggle' data-toggle='dropdown'>
<span class='username'><?php echo $_SESSION['user']['username'];?></span>
<b class='caret'></b>
</a>";
echo "<ul class='dropdown-menu'>
<li><i class='icon-user'></i> My Profile</li>
<li><i class='icon-tasks'></i> My Tasks</li>
<li><i class='icon-calendar'></i> Calendar</li>
<li class='divider'></li>
<li><i class='icon-key'></i> Log Out</li>
</ul>
</li>";
}
?>
This is the error code i receive
Error i get is Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in /var/www/html/index.php on line 92

You have a echo statement in another echo statement!
So change this:
echo "<a class='dropdown-toggle' data-toggle='dropdown'>
<span class='username'><?php echo $_SESSION['user']['username'];?></span>
<b class='caret'></b>
</a>";
To this:
echo "<a class='dropdown-toggle' data-toggle='dropdown'>
<span class='username'>" . $_SESSION['user']['username'] . "</span>
<b class='caret'></b>
</a>";

Related

How to echo a clickable href to another file in php [closed]

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This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 4 years ago.
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Im new to php and i am trying study how to design web page, the code is not mine. My problem is , i have some clickable done using a table for that i have an array
$buttons=array('Home'=>'home.php',
'Contact'=>'contact.php',
'Services'=>'service.php',
'Site map'=> 'map.php');
and there is a function
function DisplayMenu($buttons)
{
echo "<table width='100%' bgcolor
='white' cellpadding='4'
cellspacing='4'\n";
echo "<tr>\n";
$width =100/count($buttons);
while (list($name, $url)=each ($buttons))
{
$this-> DisplayButton($width, $name,
$url);
}
echo "</tr>\n";
echo "</table>\n";
}
and for displaying the button there is also a function
function DisplayButton($width, $name, $url)
{
echo " <td width='$width%'>
<a herf ='".$url."'>
<img src= 'ab.jpg' alt='$name'
border='0'>
</a> <a herf='".$url."'><span
class ='menu'>
".$name."</span></a></td>";
}
My problem is when i click for example the home button its not taking me to the home.php page , can someone help me ?
Just a typo (herf to href)
function DisplayButton($width, $name, $url)
{
echo "
<td width='{$width}%'>
<a href='{$url}'>
<img src='ab.jpg' alt='{$name}' border='0'>
</a>
<a href='{$url}'>
<span class='menu'>{$name}</span>
</a>
</td>
";
}

PHP If Else with Navigation Menu [closed]

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This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 6 years ago.
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I am trying to create an if else statement where if the "if" part is true, a menu icon is displayed that says "Registration" and takes the user to a registration page. If the "else" part is true, the menu will display the user's name that is registered. I have attached basically how I would like it to look, I just do not know how to incorporate my div into a PHP if else statement.
<?PHP
session_start();
if ($_SESSION['email'] == '')
{
"<div id='cssmenu'>
<ul>
<li>
Registration
</li>
</ul>
</div>"
}
else
{
"<div id='cssmenu'>
<ul>
<li>
<a> "Greetings " . $_SESSION['fname'] </a>
</li>
</ul>
</div>"
}
?>
The following should do the trick. As mentioned by #Ghost, you need echo and you needed to change a few quotes to single quotes.
<?PHP
session_start();
if ($_SESSION['email'] == '')
{
echo "<div id='cssmenu'>
<ul>
<li>
<a href='registration.php'> Registration</a>
</li>
</ul>
</div>";
}
else
{
echo "<div id='cssmenu'>
<ul>
<li>
Greetings " . $_SESSION['fname'] . " </a>
</li>
</ul>
</div>";
}
?>
you can use php whitin html
the php code is only considered between php tags so anything else is put directly to browser
as i can see only one part of your html changes so a better design would be
<?php session_start(); ?>
<div id="cssmenu">
<ul>
<li>
<?php if ($_Session['email']=='') {
?>
Registration
<?php }else{ ?>
Greetings -> <?php echo $_SESSION['fname']; ?>
<?php }?>
</li>
</ul>
</div>

Parse error: syntax error, unexpected '<' in C:\xampp\htdocs\eifel\title_bar.php on line 4 [closed]

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This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 8 years ago.
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I have a development server for PHP that runs XAMPP. The problem that I have is a parse error. The exact parse error that XAMPP is giving is:
"Parse error: syntax error, unexpected '<' in C:\xampp\htdocs\eifel\title_bar.php on line 4"
I don't see any missing curly braces, nor semi-colons; but there is something that Notepad++ is giving me. There are wavy red lines under the file extension '.php'. The code for the class is below:
The code for the 'title_bar.php' class:
<div>
<?php
if(loggedin()){
<a href='index.php'>Home</a>
<a href='messages.php'>Messages</a>
<a href='logout.php'>Log Out</a>
}else{
echo "Not Logged In";
}
}
<a href='index.php'>Home</a>
<a href='login.php'>Login</a>
<a href='register.php'>Register</a>
</div>
Well that's not PHP. You have PHP and HTML and only PHP code should be between <?php and ?> tags:
<?php
if(loggedin()){
?>
<a href='index.php'>Home</a>
<a href='messages.php'>Messages</a>
<a href='logout.php'>Log Out</a>
<?php
}else{
echo "Not Logged In";
}
?>
<a href='index.php'>Home</a>
<a href='login.php'>Login</a>
<a href='register.php'>Register</a>
</div>
Or similar to what has already been said, you could use:
<div>
<?php
if(loggedin()){
echo "<a href='index.php'>Home</a>
<a href='messages.php'>Messages</a>
<a href='logout.php'>Log Out</a>";
}else{
echo "Not Logged In";
echo "<a href='index.php'>Home</a>
<a href='login.php'>Login</a>
<a href='register.php'>Register</a>";
}
?>
</div>
Make sure you echo or print any HTML elements in PHP otherwise you will get errors as HTML is not PHP.

Code not working in PHP if else statement [closed]

Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 8 years ago.
Improve this question
I am trying to create a nav bar where if you are a guest it displays a lock icon and 'Log in' and when you are logged in it displays a lock icon and 'Logout'.
Here is the code for my header:
<section class="top-bar-section">
<ul class="left">
<li><img src="images/icons/home_icon.png"> Home</li>
<li><img src="images/icons/about_icon.png"> About</li>
<li><img src="images/icons/pages_icon.png"> Showcase</li>
<li><img src="images/icons/videos_icon.png"> Videos</li>
</ul>
<ul class="right">
<?php
if (!empty($UserName))
{
echo <li><img src="../images/icons/login_icon.png"> Log in</li>;
}
else
{
echo <li><img src="../images/icons/login_icon.png"> Logout</li>;
}
?>
</ul>
</section>
the section I am working on is under <ul class="right">
and the error I am getting when testing the webpage is
`Parse error: syntax error, unexpected '<' in
C:\wamp\www...\header.php on line 16
PS: Line 16 is where the first 'echo' is
HELP! Please :) Thanks
You need to put the string in the echo statement in quotes (I'd suggest single-quotes, since you've got doubles in the text).
try like this you are missing quote on echo
if (!empty($UserName)) {
echo '<li><img src="../images/icons/login_icon.png"> Log in</li>';
}
else {
echo '<li><img src="../images/icons/login_icon.png"> Logout</li>';
}
The echo requires string like this
echo "Hello World";
So your echo statements must be like this:
echo '<li><img src="../images/icons/login_icon.png"> Log in</li>';
echo '<li><img src="../images/icons/login_icon.png"> Logout</li>';
If you want to use double quotes you should escape quotes that are in the string
echo "\""; /* would echo " */

Unexpected T_ECHO, expecting ',' or ';' [closed]

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Questions asking for code must demonstrate a minimal understanding of the problem being solved. Include attempted solutions, why they didn't work, and the expected results. See also: Stack Overflow question checklist
Closed 9 years ago.
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So, I've got this bit of code, for which I'm getting the following error.
Parse error, unexpected T_STRING, expecting ',' or ';'
I'm sure this is a very, very simple fix, but I'm still a bit new to the world of PHP. Any thoughts?
echo '
<li>
<a href="'.get_permalink().'">
<img src="'echo get_post_meta(get_the_ID(), 'video_tour_url', true);'">
<div class="galDiv">
<div class="boatTitle">'.get_the_title().'</div>
<div class="boatPrice">'.currency ().$price.'</div>
<div class="boatPower"> '.get_post_meta(get_the_ID(), '_map_ar_address', true).'</div>
</div>
</a>
</li>';
Your problem is with the following code:
<img src="'echo get_post_meta(get_the_ID(), 'video_tour_url', true);'">
Like the rest of your code it should use the concantination operator . and doesn't need an echo statement.
<img src="'.get_post_meta(get_the_ID(), 'video_tour_url', true).'">
You need a period in this code on the third line. and remove the echo:
<img src="'.get_post_meta(get_the_ID()
^
Don't echo HTML. Instead use inline PHP if possible.
<li>
<a href="<?php print get_permalink(); ?>">
<img src="<?php print get_post_meta(get_the_ID(), 'video_tour_url', true); ?>">
<div class="galDiv">
<div class="boatTitle"><?php print get_the_title(); ?></div>
<div class="boatPrice"><?php print currency() . $price; ?></div>
<div class="boatPower"><?php print get_post_meta(get_the_ID(), '_map_ar_address', true); ?></div>
</div>
</a>
</li>
looping? sure.
<?php foreach($widgets as $widget): ?>
<?php endforeach; ?>

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