problem using anchor button in server(cent os) [closed] - php

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In my project i am using anchor button based on if else condition. In windows it's working perfectly.
HTML
<?php
$actionbutton = $row['status'] == 'success' ?
anchor('Users/getpdf/'.$row['appno'],'Generate Pdf') :
anchor('Users/validate/'.($row['appno']),'Pending') ;
?>
<td align="center"><?php $actionbutton;?></td>
after uploading in server (cents os) the button not displayed. how to solve this error?

On your first command, it assigns the value to your $actionbutton variable.
Your 2nd command, I think you want to echo the $actionbutton.
Maybe you want this <?=$variable?>.
So, in your case <?=$actionbutton?>
Codeigniter Documentation 👈 about alternative syntax.

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Why I can't run a plugin anycomment on wordpress [closed]

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Closed 1 year ago.
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Now I try to install a plugin "anycomment" on my site. In single.php I inserted this:
<?php do_shortcode('[anycomment]') ?>
But it's not working. Why?
I attached screens with plugin:
First
Second screen
Can you help me with it? Thanks advance
Try
<?php echo do_shortcode('[anycomment]'); ?>

Image Source to Include a PHP Input [closed]

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Closed 6 years ago.
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I am trying to create an image that changes dependent on the genre grabbed from an icecast server, I am pretty sure I have the base code correct I think I've just incorrectly inputted the PHP variable.
<?php
$stats = $core->radioInfo( "http://http://sc.onlyhabbo.net:8124/status-json.xsl" );
?>
<img src=http://www.habbo.com/habbo-imaging/avatarimage?user=<?php
echo $stats['genre'];
?>&action=std&direction=2&head_direction=2&gesture=sml&size=m&img_format=gif/>
is the full code. Have I inputted the PHP variable incorrectly
Where are the quotes in your Html?
<img src="http://www.habbo.com/habbo-imaging/avatarimage?user=<?php
echo $stats['genre'];
?>&action=std&direction=2&head_direction=2&gesture=sml&size=m&img_format=gif"/>
UPDATE EVERYBODY
This is now resolved, I decided to go down the CURL route for this, and at first it didn't work until my host raised our CloudLinux Process Limit. I am unsure what the actual issue with this code was, but the CURL route works fine. Thank you for any answers

What is the "returnto" in this piece of code [closed]

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Closed 7 years ago.
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header("Location: {$TBDEV['baseurl']}/login.php?returnto=" . urlencode($_SERVER["REQUEST_URI"]));
Is it a variable? PHP reserved word? something to do with HTML?
It's a $_GET parameter. When you submit the code, the page receiving it will be able to use $_GET['returnto'] to return you to the page you're currently on.
Take some time to learn about $_GET

Php help find where is syntax error [closed]

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Closed 8 years ago.
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foreach($db->fetch_array("SELECT id_categories FROM csn_categories_join_kartes where id_kartes===".$card['id']."") as $kat){
echo (kat['id_categories']);
}
table cols and values are all matched, something is wrong in this part of code
I tried adding $ before kat and using only one "=", sill doesnt work
NEW LINK
http://pastebin.com/RPK7vEaJ
this
where id_kartes===".$card['id']."
would be
where id_kartes=".$card['id']."
and missing $
echo $kat['id_categories'];
so full code :-
foreach($db->fetch_array("SELECT id_categories FROM csn_categories_join_kartes where id_kartes='".$card['id']."'") as $kat){
echo $kat['id_categories'];
}
best practice if you store your query result in a variable and loop over this variable.
foreach($db->fetch_array("SELECT id_categories FROM csn_categories_join_kartes where id_kartes=".$card['id']."") as $kat)

PHP - Using $_POST after a link [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 form with a textbox with name="name".
In my code, I use a direct image hosted on another website in the format:
$grav_url = "http://yourwebsitehere.com/avatarimage.php?username=YourUsername";
I want the YourUsername part to be replaced with the input of the textbox.
For this to work, i'm trying the following code:
$grav_url = "http://yourwebsitehere.com/avatarimage.php?username="+ $_POST['name'];
What am I doing wrong? PHP noob here
in php . is used for concatenate string not +
here try this
$grav_url = "http://yourwebsitehere.com/avatarimage.php?username=".$_POST['name'];

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