vSlider - in WordPress I'm using the function supplied in the FAQs:
<?php if(function_exists('vslider')){ vslider('vslider_options'); } ?>
And I'm trying to do this. So its knows to get the post-ID as its name. But its not working.
<?php if(function_exists('vslider')){ vslider('<?php the_ID(); ?>'); } ?>
You cannot nest <?php ?> inside an already open <?php ?>. That is unsupported and syntactically invalid. Just call the function in place.
Apparently, the_ID() is one of those Wordpress functions which prints to the output buffer without returning its value. To get the id returned where it can be useful in a function, use get_the_ID() instead.
<?php if(function_exists('vslider')){ vslider(get_the_ID()); } ?>
The syntax issue becomes more obvious when expressed as properly indented code.
<?php
if (function_exists('vslider')){
vslider(get_the_ID());
}
?>
Yes you can do this by removing the php tags
<?php if(function_exists('vslider')){ vslider('<?php the_ID(); ?>'); } ?>
should be
<?php if(function_exists('vslider')){ vslider(the_ID()); } ?>
and your the_ID() function should return a string at the end
Related
How can I assign the value 1990 to born_year in $u0? I get an error massage saying Invalid text string.
<?php if($u0->i_sex=='98') { ?>
$u0->born_year =1990;
<?php } ?>
All PHP inside the " <?php ?>" tag
<?php
if($u0->i_sex=='98') {
$u0->born_year =1990;
}
?>
You are using PHP outside of the PHP tags.
<?php is used at the start of your PHP code.
?> is used at the end of your PHP code.
Try:
<?php
if($u0->i_sex=='98') {
$u0->born_year =1990;
}
?>
I am making static php and i want somthing like this (i am a php beginner :):
From this:
<?php $titleid="example title"; ?>
To this:
<?php $title="<?php echo $titleid; ?>"; ?>
To get this:
<h1><?php echo $title; ?></h1>
And then, the expected output is:
<h1>example title</h1>
I use this cause the variable $titleid is in other php file.
In this line
<?php $title="<?php echo $titleid; ?>"; ?>
You should not use the echo, since at that time you don't want any output. Also avoid the double <?php ?> brackets.
Just write that line as
<?php $title=$titleid; ?>
First, you open <?php tag only once in file. if you want get example title in your pages, you define this content one variable and then get result with echo :
Ex
<?php
$title = "<h1>example title</h1>";
echo $title;
The php variable is not displaying the result.
<h2 class="bold"><? echo ${"application->client->s_MEMBERCLUB_STATUS_{$i}_NAME"}; ?> Member</h2>
You can use
<?php echo $something; ?>
or
<?= $something; ?>
Which is the shorthand.
If you want to use the original code, although it is discouraged, you can check out how to enable it here:
PHP tags
I collaborate with a web-programmer on a php project based in kirby cms, and he wants to open and close every line as such:
<main>
<?php /*php code here/* ?>
<?php /*more php here*/ ?>
...
Trying to follow this style, I found some errors in my code. The first is that it seems I canNOT do this in the middle of an array as such:
BAD CODE
<?php $oo = array( ?>
<?php 'h' => 100, ?>
<?php 'v' => 100, ?>
<?php ); ?>
but I can do it in the middle of a foreach loop as such:
<?php foreach ($p as $subp): ?>
<div id='<?= $subp->title() ?>'>
<?php endforeach; ?>
Are there any other cases such as array in which I canNOT do this?
/edit
According to the answer, there can only be tag-breaks within 'foreach', 'while', or 'if' blocks.
How about a 'foreach', 'while' or 'if' within a function? is that 'legal'?:
<?php
function myFunction($arg){
if($arg === 'this'): ?>
<?= '<p>yep</p>' ?>
<?php else: ?>
<?= '<p>nop</p>' ?>
<?php endif;
};
?>
And how about nesting if within foreach within function?:
<?php
function myFunction($arr){
foreach($arr as $val): ?>
<p><?= $val ?> <p>
<?php if($val === 'this'): ?>
<?= '<p>yep</p>' ?>
<?php else: ?>
<?= '<p>nop</p>' ?>
<?php endif;
endforeach;
};
?>
edit/
Thank you
you cannot "break out of php" mid statement. wline defining an array for example you cannot close the php tags. The only time when you can "break out" of php is between opening and closing a loop or an if/else statement. This actualy does not break the statement as <?php foreach: ?> is a complete statement whereas <?php foreach{ ?> is not. Here some examples of what you can do:
<?php if($this!=$that): ?>
{something}
<?php endif ?>
<?php foreach($things as $thing): ?>
{something}
<?php endforeach ?>
<?php $while($this): ?>
{something}
<?php endwhile ?>
I think you get the message. you must have complete statements within php tags, without interruptions.
P.S. Also avoid using the shorthand <? instead of <?php at all cost, moving your project to a different hosting or an upgrade of your hosting might break your project as per default short tags are not activated. <?= ?> shorthand is safe as this is unaffected by the setting for newer php versions.
P.P.S Do not listen to the guy who wants php in one line, this will make your code hard to read and maintain. Stand strong and write beautiful code :)
UPDATE: (after the update on the question from #Jaume Mal)
I did not mean the examples in my answer as exclusive but as examples of statements that are complete vs statements that are incomplete. (I also forgot to mention closing php tags mid fuction, wich also work but I despise and woudl strongly advise against.) So for example <?php function foo(){ is a complete statement of starting a function but (as the other cases with loops etc..) it needs a closing statement, in this case }. This is true for if / else or foreach and so on:
<?php if($this){ ?>
some code
<?php } ?>
is a valid code, as the code pieces within the php tags are complete statements.
My WordPress options panel has a section where the user can paste their logo's URL to show up in the header. If the input is blank, I want the Blog's title to show up instead on my header. The ID of the input is "nl_logo", so I added an if statement in my header.
<?php if ("nl_logo") { ?>
<img src="<?php echo get_option('nl_logo'); ?>">
<?php } else { ?>
<h1><?php bloginfo('name'); ?></h1>
<?php } ?>
The first part of the if statement works. However, anything below else doesn't work when I have no URL saved in my input. So, if the input is empty, how do I display something else with PHP? Or is there a different and better way to do this? For example, creating a function and calling the results to display with a simple line of PHP?
Try this... also try to understand it.
Keeping with the established coding style:
<?php $nlLogo = get_option('nl_logo'); ?>
<?php if (empty($nlLogo)) { ?>
<h1><?php echo(bloginfo('name')); ?></h1>
<?php } else { ?>
<img src="<?php echo($nlLogo); ?>">
<?php } ?>
That should atleast be valid PHP now. I don't know if the functions you are using are correct, but if they are this should work.
Here is a cleaner way to do it...
<?php
$nlLogo = get_option('nl_logo');
if (empty($nlLogo)) {
echo('<h1>'.bloginfo('name').'</h1>');
} else {
echo('<img src="'.$nlLogo.'">');
}
?>
Option three because I'm feeling "teachy" using a ternary. Probably a little long for this to be the best choice, but it is another option.
<?php
$nlLogo = get_option('nl_logo');
echo(empty($nlLogo) ? '<h1>'.bloginfo('name').'</h1>' : '<img src="'.$nlLogo.'">');
?>
Note I switched the if / else because I'm using empty and it just feels cleaner to do it this way instead of using !empty()
<h1><?php bloginfo('name'); ?></h1>
Should be
<h1><?php echo bloginfo('name'); ?></h1>
You could try the empty method to test if a string is empty. In context:
if(!empty($nl_logo)) {
// stuff
} else {
// other stuff
}
This does not make sense:
if ("nl_logo")
You're basically saying if the string exists then proceed, which it does, of course.
It makes more sense to take the input string:
$input = $_POST["postedInput"];
As an example, doesn't really matter as long as you know how you got the value into $input.
Now, you can use the ternary operator to determine whether you want to use the user's input or if you want to use the default title:
$img = $input == "" ? bloginfo($input) : get_option($input);
Depending on what your functions do.. if get_options returns a string then this will work.
Anyway, don't mis PHP and HTML, it will make things complicated and you'll be locked into bad design.
Note:
Make sure to check if the input is actually received, using isset.
You are testing for string "nl_logo" to be true. That returns always true.
try changing your code like this:
<?php
$nl_logo = get_option('nl_logo');
if (!empty($nl_logo)):
?>
<img src="<?php echo $nl_logo; ?>">
<?php else: ?>
<h1><?php bloginfo('name'); ?></h1>
<?php endif; ?>
Don't close the php tags.
Whatever html you need to write, just write it as echo statements within one big php block.
This should fix your problem