By Sayyed Rizwan

2019-11-08 14:15:19 8 Comments

I can only like the first post not second or third or other...

The image is only changing on the first post.. and other they are not working...

        $sqkl = "SELECT id FROM favourite WHERE userid='$mainuserid' AND photoid='$photoid' ";
        $rersult = $conn->query($sqkl);

        if ($rersult->num_rows > 0) {
            // output data of each row
            while($rkow = $rersult->fetch_assoc()) {

           <img id="fav" style="float:right;" id="fav" onclick="myFav()" src="images/fav.png" height="22" width="22"> </div>

        } else {
           ?>  <img id="fav" style="float:right;" onclick="myFav()" src="images/unfav.png" height="22" width="22"> </div> <?php

  var image =  document.getElementById("fav");

    function myFav() {
          if (image.getAttribute('src') == "images/fav.png")


             }else if (image.getAttribute('src') == "images/unfav.png"){


@Qirel 2019-11-09 11:16:38

You should instead of using IDs, use onclick="myFav(this)", thereby passing that exact element into the function. You can use that argument directly instead.

Your query is also vulnerable to SQL injection, and you should use a prepared statement instead.

It would also appear to me as though you're running this query inside another query-loop - if that's the case, you should instead get this query integrated into the outer query, as a query within a loop is bad for performance.

$stmt = $conn->prepare("SELECT id FROM favourite WHERE userid=? AND photoid=?");
$stmt->bind_param("ii", $mainuserid, $photoid);
if ($stmt->fetch()) {
    <img style="float:right;" onclick="toggleFavorite(this)" src="images/fav.png" height="22" width="22"> </div>
} else {
    <img style="float:right;" onclick="toggleFavorite(this)" src="images/unfav.png" height="22" width="22"> </div>
    function toggleFavorite(element) {
        if (element.getAttribute("src") === "images/fav.png") {
            element.src = "images/unfav.png";
        } else {
            element.src = "images/fav.png";

@Sayyed Rizwan 2019-11-09 12:19:06

lovely thank you..... it worked... you used element to define in JS thankss...

@Anis R. 2019-11-08 14:21:41

In your PHP's while loop, you are adding several images with the same ID of "fav". However, as IDs must be unique, your Javascript code:

var image =  document.getElementById("fav");

Will always return the first image, because it assumes there is only one document with this ID.

What you can do instead is to use a class name of "fav" (rather than ID), and then use the below code to get a list of all images:

var images =  document.getElementsByClassName("fav");

@Anis R. 2019-11-08 14:33:06

Why the downvote?

@Qirel 2019-11-08 14:58:10

I haven't voted up or down, but using a class and then using document.getElementsByClassName("fav") would probably break the logic - you would need to use this element. And since there can be more than 1 row here, we're probably looking at a flawed PHP logic too - either its a query within a loop, or its X rows for favorites, and 1 row if no favorite exist.

@Sayyed Rizwan 2019-11-09 07:28:59

not its not working ...... any other idea

@Christoffer 2019-11-08 14:20:01

The reason your code is not working, is because all images have the same ID. ID's are unique identifiers, and document.getElementById("fav"); will only select the first element with that ID.
You can use the this keyword in the myFav function to reference the clicked image

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