Using PHP variable within ACF repeater field - php

When using ACF (Advanced Custom Fields) repeater fields, the markup should look like this to initiate the loop.
<? if( have_rows('my-repeating-field) ): ?>
<? endif ?>
However, I would like to make this dynamic, meaning I use another PHP variable as part of the if() code, something like as follows:
<? $variable = get_field('my-variable-field'); ?>
<? if( have_rows("'" . $variable . "-repeating-field" . "'") ): ?>
<? endif ?>
However, this is not working. Is there a way I can do this with ACF/PHP? Seems pretty simple to me?

The problem is that you're concatenating the string in a way which would cause the end value being passed to have_rows to have quotes inside of it. You only need to concatenate the value and the end of the string, and that should be enough.
have_rows($variable . '-repeating-field')

Related

Using Meta Box Plugin to assign URL to a variable

I'm a little lost here, hoping that someone can help. I'm using the Meta Box plugin for WordPress, and I'm trying to create a process for the user to select an option from a predefined list, and then assign a URL to that option as a link. Im trying to define the URL in a variable, and then call it in a function, but I'm still a little green on PHP syntax. this is my code now:
<?php
$article_url= rwmb_meta('orion_2016_article_url', 'type=URL');
if (rwmb_meta('orion_2016_article_source') != '') {
echo '<a href= ("$article_url") target=blank>';
echo rwmb_meta('orion_2016_article_source');
echo '</a>';} ?> on <?php the_date(); ?>
Since the options are already predefined, it seems like assigning a random URL to one of the options should be pretty simple. Hopefully this makes sense!
You need to to place variables you wish to echo inside double quotes or simply concatenate strings using . as in my example. Note that I didn't check the plugin's specific syntax, only general PHP syntax.
<?php
$article_url= rwmb_meta( 'orion_2016_article_url', 'type=URL' );
if (rwmb_meta('orion_2016_article_source') != '') {
echo '' . rwmb_meta( 'orion_2016_article_source' ); . '';
} ?> on <?php the_date(); ?>

How to use php variable inside another one

I'm trying to find the category ID of a post and then use it in the short code below. To find the category id I'm using <?php the_category_ID(); ?>, but i'm not sure how to use the output from category id and replace it in include_categories=. i.e. i want to replace number 4 with the current category ID i got from <?php the_category_ID(); ?>.
<?php echo do_shortcode('[include_categories="4"]'); ?>
i tired doing the following, but it didn't work. Any ideas will be appreciated.
<?php echo do_shortcode('[include_categories="<?php the_category_ID(); ?>"]'); ?>
I know i can't use php inside another php code, but i'm not sure how to place the category id between the quotation.
Thanks.
A couple of things: (1) you can't have nested <?php ?> tags, (2) the_category_ID() has been deprecated since WordPress 0.71. You'll want to use get_the_category() instead.
The correct solution would be:
<?php echo do_shortcode('[include_categories="' . get_the_category() . '"]'); ?>
The dots . "concatenate" the string with the returned values of get_the_category(). You can read more about string operators in the PHP docs.
Try with -
<?php echo do_shortcode('[include_categories="'.the_category_ID().'"]'); ?>
Use this to add php verial in short code:
<?php $category=the_category_ID();
echo do_shortcode("[include_categories=$category]"); ?>

Hide HTML with PHP conditional - Not working

I am very new to PHP and I am trying to get a piece of PHP code to run inside of my HTML file.
I have a drop down menu. If users select one item, it should display additional fields. So, I want them to only display if they select that item from the drop down menu. I am trying to select it based on the value for that drop down item. I have not declared any PHP values in a PHP script. This is all in HTML.
I know that with jquery you have to pull in the jquery library before running the script. Do I need to do this with PHP also?
Here is the code that I am trying to run:
Thank you in advance!
<?php
if ($dropmenuValue == "specificParameter") {
?>
<div>
-conditional content-
</div>
<?php
}
?>
<?php
if ($value ) {
?>
<div>
-conditional content-
</div>
<?php
}
?>
If you are setting the $values variable like this
$value = true
Then you can do
<? if($value){ ?>
<div>Content Here</div>
<? } ?>
Is the page a .php page?
A multi conditional statement is done in a different manner. The manner being:
<?php if($val): ?>
<div>
-conditional content-
</div>
<?php endif; ?>
the problem is that your comparing a boolean to a String.That's why your code won't work I would suggest something like If($value){ or if($value==true){.
you want to do if($value == true). The quotes you have around true does not make it a boolean variable.
Everything else in your code looks find.
tldr: correct: if($value == true) ----------
incorrect: if($value == "true")
<?php if ($value ):?>
<div>
-conditional content-
</div>
<?php endif;?>
Here are a few pointers.
$value is a boolean value. It is best to evaluate using if($value)
The PHP if syntax used above is a much cleaner way.
You will be able to see where your if statement begins and ends.
Enjoy!

Simple php syntax error, needs fixing

I have a code that can get me the category_id of an item. This is the code:
<?php echo lavada_category_id() ; ?>
I want to know how I can add this code. Inside this, I want to replace the number 2 in here;
<?php lavada_query_item("category=2");?>
with:
<?php echo lavada_category_id() ; ?>
I know you cannot do like this
<?php lavada_query_item("category=<?php echo lavada_category_id() ; ?>");?>
But how can I do it?
Why not store it into a variable and then use that variable?
<?php
$catID = lavada_category_id();
lavada_query_item("category={$catID}");
?>
OR if you just want category ID to be passed into lavada_query_item do this:
lavada_query_item($catID);
The syntax error that you have is that you can not use <?php within <?php
You just need to concatenate the string like this:
<?php lavada_query_item("category=". lavada_category_id() );?>
I think this is what you are looking for:
<?php lavada_query_item(lavada_category_id());?>
The value returned from thelavada_category_id() function will be passed into the lavada_query_item() function.

Is this possible to use a variable for an url in php?

I'm calling a function like this:
<?php print get_thumbnail('http://url.com/?skin=rss'); ?>
Being a php newbie, I'm wondering if there is a way to change the http://url.com part based on a custom metadata I have set up in Wordpress. So I guess it would look something like this:
<?php print get_thumbnail('<?=$video_src?>/?skin=rss'); ?>
Is something like this possible?
Yes, you have the right idea, you just don't need to re-open PHP tags since you're already inside some. You can use . to concatenate (join together) the value of $video_src and "?skin=rss".
<?php print get_thumbnail($video_src . "?skin=rss"); ?>
Try this:
<?php print get_thumbnail($video_src . '/?skin=rss'); ?>
Keep in mind that <?= $foo ?> is shorthand for <?php echo $foo; ?>. <?= ?> won't be expanded in strings, but you can achieve something similar using double quoted strings:
<?php print get_thumbnail("$video_src/?skin=rss"); ?>
Yes, except within PHP, you don't need to enter the PHP tags again.
<?php print get_thumbnail($video_src . '/?skin=rss'); ?>

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