PHP/SQL Simple click counter button

Question

Tried to make a simple click counter, doesnt work. Whats wrong with my trash code?

addLike.php

 <?php

$requestLikes = "SELECT Likes FROM test1";
$Likes = mysql_query($requestLikes);

$insertToLikes = "INSERT INTO test1 (Likes) VALUES (" . $Likes + 1 . ")";
mysql_query($insertToLikes);
$Likes = mysql_query($requestLikes);

?>

HTML

    <button action="/DEMO/PHP/addLike.php" method="post">Yeet
    </button>0

    <?php 
    echo '<h1>' . $Likes . '</h1>'; 
    ?>

Result

www.mySite/DEMO/PHP/addLike.php

Fatal error: Uncaught Error: Call to undefined function mysql_query() in /storage/ssd4/786/4680786/public_html/DEMO/PHP/addLike.php:4 Stack trace: #0 {main} thrown in /storage/ssd4/786/4680786/public_html/DEMO/PHP/addLike.php on line 4

and yes i have established a connection to the database EDIT: Dont mind the random "0" after the button tag

$Likes is assigned to the result of mysql_query, which returns an object. Then you do $Likes + 1, but $Likes is not a number so you can't add 1 to it. — Al.G.
– Al.G., Commented Feb 24, 2018 at 16:42

Richard · Accepted Answer · 2018-02-24 16:45:40Z

3

 <?php
$conn = mysqli_connect("127.0.0.1", "my_user", "my_password", "my_db");

$requestLikes = "SELECT Likes FROM test1";
$LikesQ = mysqli_query($conn, $requestLikes);
$Likes = mysqli_fetch_array($LikesQ);
$x = $Likes[0] + 1;
$insertToLikes = "INSERT INTO test1 (Likes) VALUES ('$x')";
mysqli_query($conn, $insertToLikes);
$y = mysqli_query($conn, $requestLikes);
$output = mysqli_fetch_array($y);
echo "<h1>$output[0]</h1>";

?>

Use MYSQLI please

$insertToLikes = "INSERT INTO test1 (Likes) VALUES (" . $Likes + 1 . ")";

You are vulnerable to MySQL injection, always use prepared statements or mysqli_real_escape_string

edited Feb 24, 2018 at 16:45

answered Feb 24, 2018 at 16:42

Richard

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6 Comments

Phil Over a year ago

This. mysql_ was deprecated due to it being insecure, mysqli_ is a speedier and more secure version that functions in a very similar way (to explain why Use MYSQLI please)

M. Eriksson Over a year ago

You should stick to Prepared Statements. Even mysqli_real_escape_string() has some security issues in certain situations. stackoverflow.com/questions/2353666/…

Jaypaque Over a year ago

Now allow me a one more question. is mysqli_fetch_array somehow related to php arrays?

Richard Over a year ago

@Jaypaque yes it brings the array, for example $test = mysqli_query($conn, "SELECT a,b,c") and when you use $array = mysqli_fetch_array($test) you get the array. this can be displayed like this echo $array[1] or echo $array['b'] both gives you b

Jaypaque Over a year ago

Getting "mysqli_fetch_array() expects parameter 1 to be mysqli_result" error and have no idea what it wants of me

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