Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 8860721
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 14, 20262026-06-14T15:19:17+00:00 2026-06-14T15:19:17+00:00

I’d like to navigate images using both keyboard and mouse (clicking left and right

  • 0

I’d like to navigate images using both keyboard and mouse (clicking left and right arrow images).

I’m using jQuery to do this, but the shared imgIndex seems to be off from the .keydown function and the .click function. Whenever .keydown function -- or ++ the imgIndex, isn’t that changed index also used in the click function? So shouldn’t they always be on the same index?

keydown function:

<script type="text/javascript">
var imgArray = [<?php echo implode(',', getImages($site)) ?>];

$(document).ready(function() {      

    var img = document.getElementById("showimg");
    img.src = imgArray[<?php echo $imgid ?>];
    var imgIndex = <?php echo $imgid ?>;
    alert(imgIndex);

    $(document).keydown(function (e) {
        var key = e.which;
        var rightarrow = 39;
        var leftarrow = 37;
        var random = 82;
    
        if (key == rightarrow) 
        {
            imgIndex++;
            if (imgIndex > imgArray.length-1) 
            {
                imgIndex = 0;

            }
            img.src = imgArray[imgIndex];
        }
        if (key == leftarrow) 
        {
            if (imgIndex == 0) 
            {
                imgIndex = imgArray.length;
            }
            
            img.src = imgArray[--imgIndex];
        }   
    });
    

click function: Connected to left and right clickable images

enter image description here

    $("#next").click(function() {
        imgIndex++;
            if (imgIndex > imgArray.length-1) 
            {
                imgIndex = 0;
            }
            img.src = imgArray[imgIndex];
    });
    $("#prev").click(function() {
        if (imgIndex == 0) 
            {
                imgIndex = imgArray.length;
            }           
            img.src = imgArray[--imgIndex];
    });
});

</script>

Just so you have some visibility into the getImages PHP function:

<?php
function  getImages($siteParam) {
include 'dbconnect.php';
if ($siteParam == 'artwork') { 
    $table = "artwork"; 
}       
else { 
    $table = "comics"; 
}   
    
$catResult = $mysqli->query("SELECT id, title, path, thumb, views, catidFK FROM $table");   
$img = array();
while($row = $catResult->fetch_assoc()) 
{
    $img[] = "'" . $row['path'] . "'";
}
return $img;
}
?>

Much appreciated!

Snapshot of where the script is on "view image.php"

enter image description here

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-06-14T15:19:19+00:00Added an answer on June 14, 2026 at 3:19 pm

    I think the problem is you are defining the imgIndex as a local variable inside the ‘ready’ handler, and try to use as a global. In this case both prev/next handlers will get their own copies of imgIndex.

    Check this fiddle: http://jsfiddle.net/BuddhiP/f2WzJ/

    var imgArray = ['img1', 'img2', 'img3', 'img4', 'img5'];
    var imgIndex = 3;
    
    $(document).ready(function() {
        var $img = $("#imgIndex");
        $img.text(imgIndex);
        //alert(imgIndex);
    
        $(document).keydown(function(e) {
            var key = e.which;
            var rightarrow = 39;
            var leftarrow = 37;
            var random = 82;
    
            if (key == rightarrow) {
                imgIndex++;
                if (imgIndex > imgArray.length - 1) {
                    imgIndex = 0;
    
                }
                //img.src = imgArray[imgIndex];
                 $img.text(imgIndex);
            }
            if (key == leftarrow) {
                if (imgIndex == 0) {
                    imgIndex = imgArray.length;
                }
    
                //img.src = imgArray[--imgIndex];
                $img.text(--imgIndex);
            }
        });
    
        $("#next").click(function() {
            imgIndex++;
            if (imgIndex > imgArray.length - 1) {
                imgIndex = 0;
            }
            $img.text(imgIndex);
            //img.src = imgArray[imgIndex];
        });
        $("#prev").click(function() {
            if (imgIndex == 0) {
                imgIndex = imgArray.length;
            }
            $img.text(--imgIndex);
            //img.src = imgArray[--imgIndex];
        });
    });​
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I have a string like this: La Torre Eiffel paragonata all&#8217;Everest What PHP function
For some reason, after submitting a string like this Jack’s Spindle from a text
this is what i have right now Drawing an RSS feed into the php,
I am reading a book about Javascript and jQuery and using one of the
This could be a duplicate question, but I have no idea what search terms
I'm parsing an RSS feed that has an &#8217; in it. SimpleXML turns this
I want to count how many characters a certain string has in PHP, but
link Im having trouble converting the html entites into html characters, (&# 8217;) i
That's pretty much it. I'm using Nokogiri to scrape a web page what has
I have a jquery bug and I've been looking for hours now, I can't

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.