Is it correct to use php inside jquery?

I want to change jquery command through PHP.

So for example:

$envelope<?php echo'1'?>.toggleClass( 'card-out' );

Is it really for use? I tried this, and I think it did not work, but I do not know if there was an error in this line or in the rest of the code.

this is the main part of my code:

<div class="library-card envelope<?php echo'2'?>">
    <div class="front">
        <p class="stamp">Jacob Haase<br> 18.05.2000</p>
    </div>
    <div class="card">sdsdsd</div>
    <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
    <script type="text/javascript">
        var $envelope = $( '.library-card.envelope' ),
            $card = $envelope.find( '.card' ),
            $front = $envelope.find( '.front' );

        $front.on( 'click', function( event ) {
            $envelope<?php echo'2'?>.toggleClass( 'card-out' );
        });

        $card.on( 'click', function( event ) {
            $envelope<?php echo'2'?>.toggleClass( 'card-out' );
        });
    </script>

</div>

Thank you!

+4
source share
2 answers

This is great if you use jQuery in a .php file.

In this line:

$envelope<?php echo'1'?>.toggleClass( 'card-out' );

It will $envelope1.toggleClass('card-out');, but I'm not sure if you have the jQuery $ envelope1 variable. But maybe you are trying to do like this:

$('.envelope1').toggleClass('card-out');

If so, use:

$(".envelope"+<?php echo'1';?>).toggleClass( 'card-out' );
+3
source

PHP- click, , , PHP- , .

javascript jquery, PHP, .

PHP jQuery script?

$(".envelope"+<?php echo'1';?>).toggleClass( 'card-out' );

http://wptricks.co.uk/include-php-variables-into-jquery/

+1

Source: https://habr.com/ru/post/1599860/


All Articles