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I have this code :
<div class="gigel"><?php
// In case there is opening and closing shortcode.
echo do_shortcode('[woocs show_flags=1 width='300px' flag_position='right']');
?></div>
This is shortcode:
[woocs show_flags=1 width='300px' flag_position='right']
This is the error:
Parse error: syntax error, unexpected '300' (T_LNUMBER) in /home/dacproie/public_html/test2/wp/wp-content/themes/mix/header.php on line 189
How can I solve this problem?
Thanks in advance!
just change below code with
<div class="gigel"><?php
// In case there is opening and closing shortcode.
echo do_shortcode("[woocs show_flags=1 width='300px' flag_position='right']");
?></div>
i hope this is working for you.
please, try to clean your code
<?php
$forHtml = do_shortcode("[woocs show_flags=1 width='300px' flag_position='right']");
?>
<!-- separate your layers -->
<div class="gigel"><?php
<?php print ("%s", $forHtml); ?>
?></div>
Always try to use double quotes when you have to use do_shortcode() function.
So use of double quotes instead of single quotes fix your problem and you need to modify it as like :
<div class="gigel">
<?php
echo do_shortcode("[woocs show_flags=1 width='300px' flag_position='right']");
?>
</div>
Related
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Closed 6 years ago.
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I am trying to add an image file to my PHP so when I echo, the picture will appear alongside the message. However I am getting a syntax error on line 3. Any help would be appreciated.
<?php
echo "President has been killed";
<IMG SRC = "D:/User Data\Documents/Sheridan/Summer Year 3/Enterprise Java Development/Projects/PhpAssignment/skull.png;"/>
?>
<?php
echo 'President has been killed
<img src="D:/User Data\Documents/Sheridan/Summer Year 3/Enterprise Java Development/Projects/PhpAssignment/skull.png;" /> ';
?>
Note the change in quotes -- single and double -- and placement of the semi-colon.
or
<?php
echo "President has been killed";
?>
<img src="D:/User Data\Documents/Sheridan/Summer Year 3/Enterprise Java Development/Projects/PhpAssignment/skull.png;" />
Move the image outside the php. Using this depends on overall markup though.
The semi-colon ends, or stops the echo statement.
Of course there's an syntax error. You are trying to output HTML inside PHP block:
Change your code to:
<?php
echo "President has been killed";
?>
<IMG SRC = "D:/User Data\Documents/Sheridan/Summer Year 3/Enterprise Java Development/Projects/PhpAssignment/skull.png;"/>
<?php
echo "President has been killed";
<IMG SRC = "D:/User Data\Documents/Sheridan/Summer Year 3/Enterprise Java Development/Projects/PhpAssignment/skull.png;"/>
?>
firstly you don't need the entire location, place the image in the folder in your root called images then place the image in there.
<?php
echo '<p>President has been killed<p><img src = "D:/User Data\Documents/Sheridan/Summer Year 3/Enterprise Java Development/Projects/PhpAssignment/skull.png"/>';
?>
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parse error unexpected endif at line i don't know why
<form name="contactform" method="post" action="check.php" class="form-horizontal" role="form">
<?php if(array_key_exists('errors', $_SESSION)); ?>
<div class="alert alert-danger">
<?php implode('<br>', $_SESSION['errors']); ?>
</div>
<?php unset($_SESSION['errors']); ?>
<?php endif;?>
Change
<?php if(array_key_exists('errors', $_SESSION)); ?>
to:
<?php if(array_key_exists('errors', $_SESSION)): ?>
Reference
Explanation:
In your code, you are putting semi-colon ; after if statement.
It means you are closing if control structure, hence, endif below has no meaning.
If you put :, the code between if and endif will be considered as body of control structure and hence will not produce error.
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Closed 8 years ago.
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I'm having an issue and am out of ideas.
I'm trying to get a parameter from the URL, but PHP insists in saying that the variable is not defined.
The URL is
http://localhost/trendwatcher/index.php?date=2014-10-18
And the script is something like
<?php include "header.php"; ?>
<section id="stats">
Showing trends for <?php echo $GET_["date"]; ?>
</section>
And finally, the error:
Showing trends for
Notice: Undefined variable: GET_ in C:\xampp\htdocs\trendwatcher\index.php on line 4
Does anyone have any ideas?
Thanks!
Fix your code from:
<section id="stats">
Showing trends for <?php echo $GET_["date"]; ?>
</section>
to
<section id="stats">
Showing trends for <?php echo $_GET["date"]; ?>
</section>
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Closed 8 years ago.
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I have a problem with this code:
<?php echo 'Tweet
<script>!function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0],p=/^http:/.test(d.location)?'http':'https';if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src=p+'://platform.twitter.com/widgets.js';fjs.parentNode.insertBefore(js,fjs);}}(document, 'script', 'twitter-wjs');</script>'; ?>
I get the error:
Parse error: parse error, expecting ','' or';''
You either need to escape all of the single quotes in that string or don't use PHP to output that code (recommended):
<?php
//some php code
?>
Tweet
<script>!function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0],p=/^http:/.test(d.location)?''http':'https';if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src=p+'://platform.twitter.com/widgets.js';fjs.parentNode.insertBefore(js,fjs);}}(document, 'script', 'twitter-wjs');</script>
<?php
// more PHP code
?>
Escaped quotes:
<?php echo 'Tweet
<script>!function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0],p=/^http:/.test(d.location)?\'http\':\'https\';if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src=p+\'://platform.twitter.com/widgets.js\';fjs.parentNode.insertBefore(js,fjs);}}(document, \'script\', \'twitter-wjs\');</script>'; ?>
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i have this simple typo issue here i guess.
How can i resolve it?
<?php the_content("<br /> <span class='custom-more'>Read More: " . get_the_title('', '', false) "</span>"); ?>
I need to make a read more button in WordPress and i need to wrap the read more inside a span class.
I get this error message:
Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in
How am i supposed to resolve this?
Thanks
You are missing the concatenation operator . between get_the_title function and </span>
<?
php the_content("<br /> <span class='custom-more'>Read More: ".get_the_title('', '', false)."</span>");
?>
You are missing the concatenate (.)
Correct code will be
php the_content("<br /> <span class='custom-more'>Read More: ".get_the_title('', '', false)."</span>");