PHP - onClick not working and calling "echo" Video and Image SRC using PHP tag

Question

I have a problem because I cannot call my data that I want to retrieve on a video src and image src. here is my code. image and video src are not showing. can someone give me the correct syntax, please? thanks

few questions, do I need to concatenate something? do my rows[''] are concatenate correctly?

but my main problem here is I cannot click or change any videos that are retrieved or stored. how can I fix the onClick? thanks

    <?php 


echo '<li>
<a href="javascript:void();" onClick="document.getElementById("vid_frame").src="images/promvid/pal/<?php row['videos']  ?>">

<span class="vid-thumb">
<img width=72 src="images/promvid/philippines.jpg"/<?php row['image']  ?>
</span>

<div class="desc">Philippines<?php  row['title']  ?>
</div></a></li>';

     ?>

MY ERROR

Parse error: syntax error, unexpected 'videos' (T_STRING), expecting ',' or ';

@digijay can u show me the proper code please i think there's a lot more problem on that — , Jan 03 '19 at 06:40
@AkhilAravind can you please correct my cod sir thanks and ill try it — , Jan 03 '19 at 06:40
Do not use PHP tags within PHP tags. You can fix it yourself. Further more variables need to be echoed or they won't be output. — Pinke Helga, Jan 03 '19 at 06:44
While there still are parse errors, of course PHP is not executed at all. — Pinke Helga, Jan 03 '19 at 06:47

score 1 · Accepted Answer · answered Jan 03 '19 at 06:47

You cannot use PHP tags inside PHP tags you need to concatenate the string part with variable using echo.

 <?php 


    echo '<li>
    <a href="javascript:void();" onClick="document.getElementById("vid_frame").src="images/promvid/pal/'.$row['videos'].'">

    <span class="vid-thumb">
    <img width=72 src="images/promvid/philippines.jpg"/'.$row['image'].' </span>

    <div class="desc">Philippines'.$row['title'].'    </div></a></li>';

         ?>

Ramya · Answer 2 · 2019-01-03T06:56:54.603

1

This below code may help you.

In php we need use the variable by using $

echo '<li>
        <a href="javascript:void();" 
            onClick="document.getElementById("vid_frame").src="images/promvid/pal/"'.$row["videos"].'>

            <span class="vid-thumb">
                <img width=72 src="images/promvid/philippines.jpg/"'.$row['image'].'>
            </span>

            <div class="desc">Philippines'.$row['title'].'</div>
        </a>
    </li>';

edited Jan 03 '19 at 06:56

answered Jan 03 '19 at 06:48

Ramya

199
10

still there is error, please check answer before posting – Akhil Aravind Jan 03 '19 at 06:52

Pinke Helga · Answer 3 · 2019-01-03T09:30:40.170

You are using bad practice and run into errors. You cannot have <?php ?> tags within <?php ?> tags.

Write clean code and make good practice to a habit.

PHP is an embedded language. Do not try to generate all HTML by PHP echo. Embed peaces of PHP code into a surrounding HTML template. You can even close PHP tags within a block of loop constructs.
Variables standing alone in a PHP tag are not output, except you are using shot-open tags <?=$variable?>. Short-open tags should not be used since most server configurations do not enable them.
Inline JavaScript is old school and considered to be bad practice. First of all inline on-handlers might be removed from the standards some day. Use event listeners instead.
You have block elements div inside the inline element a. This is valid only in HTML5. Ensure that your document type explicitly is HTML5 by prepending <!DOCTYPE html> as the very first line in the document.
Links doing nothing has been widely discussed if the href attribute should be # or javascript:void(0);. I tend to the latter. void is a keyword, however, the function style is fine. Regardless to the code style there has to be an argument.
- even better: Do not use non-linking links at all. Use links with a valid fallback URL instead. In the event handler you can call the preventDefault() method wich will prevent the href action, i.e. location change.

PHP site:

<head>
  <title>Non-Empty Title</title>

<body>

  <ul>

  <?php
    // fake query result
    $rows =
    [
      [
        'videos'              => 'phillippines.mpeg4',
        'image'               => 'philippines.jpg' ,
        'title'               => 'Philippines'     ,
        'noscript-frame-page' => 'philippines.html',
      ],
      [
        'videos'              => 'usa.mpeg4',
        'image'               => 'usa.jpg' ,
        'title'               => 'USA'     ,
        'noscript-frame-page' => 'usa.html',
      ]
    ];

    // fake fetch row
    foreach ($rows as $row)
    {
  ?>

    <li>
      <a class="video-ref"
         href="<?php echo $row['noscript-frame-page'];?>"
         data-video-src="images/promvid/pal/<?php echo $row['videos'];?>"
         target="video-frame"
      >
        <span class="vid-thumb">
          <img width=72 src="images/promvid/"<?php echo $row['image'];?>
        </span>

        <div class="desc"><?php echo $row['title'];?>
        </div>
      </a>
    </li>

  <?php
    }
  ?>

  </ul>

  <noscript>
    <iframe id="video-frame" name="video-frame" src="start-video-page.html"></iframe>
  </noscript>

  <div id="debug">DEBUG OUTPUT</div>

  <script src="my-script.js"></script>

my-script.js

document.addEventListener('DOMContentLoaded', evt =>
{
  "use strict";
  const VideoLinkListener = evt =>
  {
    const videoSrc = evt.currentTarget.getAttribute('data-video-src');
    if(videoSrc)
    {
      document.getElementById('debug').innerHTML = videoSrc;  // document.getElementById('vid_frame').src=videoSrc;

      // do not change location to link's href
      evt.preventDefault();
    }
  };

  document.querySelectorAll('a.video-ref').forEach( link => link.addEventListener('click', VideoLinkListener) );
});

PHP - onClick not working and calling "echo" Video and Image SRC using PHP tag

3 Answers3