How can i run a mysql query on button click

Question

I'm new to PHP so this might be something obvious I missed. Im trying to make a button which increments a value in my database:

<?php
        $alist = mysqli_query($conn, "SELECT * FROM `posts` ORDER BY `posts`.`id` DESC");
        $results = mysqli_num_rows($alist);

        if ($results > 0){
            while($row = mysqli_fetch_array($alist)) {
                echo $row['uid']. " says: ".$row['postText']."  <button onclick=".mysqli_query($conn, "UPDATE `posts` SET `postLikes` = postLikes+1 WHERE uid = ".$row['uid'])." name='likebtn'>👍</button>".$row['postLikes']."<br>";
            }
        }
        ?>

The part that makes the button is on line 6 I just want to find how I can use the mysqli_query on button click by the way, I already tried this: "https://stackoverflow.com/questions/3862462/php-mysql-run-query-on-button-press-click" but with no result

Thanks in advance

You can call AJAX Request on Button click. (onClick) function not working directly in server script, So you need to write ajax function and call PHP Query. — Chandrakant Devani
– Chandrakant Devani, Commented Jul 7, 2020 at 20:23

Marko Kurtović · Accepted Answer · 2020-07-07 20:29:37Z

1

if your current page is called first.php, put inside button

<button><a href="first.php?p=like"></a></button>

<?php
if ( isset($_GET['p']) && $_GET['p']=="like") {

do your query

}

?>

if you dont want to reload then you need ajax,

hope this was helpful. :)

edited Jul 7, 2020 at 20:29

answered Jul 7, 2020 at 20:24

Marko Kurtović

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Rupesh Chaudhari · Accepted Answer · 2020-07-07 20:21:59Z

1

You have a syntax error here echo $row['uid']. " says: ".$row['postText']." <button onclick=".mysqli_query($conn, "UPDATE posts SET postLikes = postLikes+1 WHERE uid = ".$row['uid'])." name='likebtn'>👍</button>".$row['postLikes']."<br>";

you can either do this

$q = mysqli_query($conn, "UPDATE posts SET postLikes = postLikes+1 WHERE uid = ".$row['uid']);

echo $row['uid']. " says: ".$row['postText']." <button onclick=".$q." name='likebtn'>👍</button>".$row['postLikes']."<br>";

answered Jul 7, 2020 at 20:21

Rupesh Chaudhari

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