php mysqli not working

Question

I am making a chat application and this is the part that checks for new additions.

<?php
$servername = "*";
$username = "*";
$password = "****";
$dbname = "*"; 

$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (mysqli_connect_error($conn)) {
    die("Connection failed: " . mysqli_connect_error($conn));
} 
$id = $_GET['id'];
$sql = "SELECT position, user, comment,time FROM chat WHERE position > $id";
$result = mysqli_query($conn,$sql);
if (mysqli_num_rows() > 0) {
    // output data of each row
    while($row = mysqli_fetch_assoc($result)) {
        $row2 = mysqli_fetch_assoc(mysqli_query($conn,"SELECT * FROM login WHERE username=".$row['user']));
        $userImage = $row2["avatar"];
        echo "<div class='container-fluid well'><p class='left'>"."<img class='img-circle zoom' src='/profile_images/".
        $userImage
        ."' style='width:32px;height:32px;'>".$row["user"]. ": " . $row["comment"]. "</p><h4><small class='right'> at ".$row['time']."</small></h4></div>";
    }
}
mysqli_close($conn);
?>

it was working until I changed the line

$row2 = mysqli_fetch_assoc(mysqli_query($conn,"SELECT * FROM login WHERE username=".$row['user']));

Help would be appreciated.

Update:
this is my html: There is more. but this is the most important

<div id='newchat'></div>
<script>
$(document).ready(function(){
    getChat();
    var id = <?php echo $id ?>;
    function getChat(){
        setTimeout(getChat,1000);
        $.get("getChat.php?id="+id,function( text ) {
            if (text != ""){
                id ++;
                $( "#newchat" ).prepend( text );
            }
        });
    }
});
</script>

I know the error but i won't tell you. Why don't you check with `mysqli_error()`? — Hanky Panky, Feb 16 '16 at 08:02
You should escape your username... au this code just burns my eyes... not only is it vunerable for sql injection but also to xss attacks. Please read `mysqli_real_escape_string` and `htlmspecialchars()` functions. — Niki van Stein, Feb 16 '16 at 08:04
Not related, but you could get everything in just 1 query using a `JOIN` instead of making database calls in a loop. — jeroen, Feb 16 '16 at 08:10

score 2 · Answer 1 · answered Feb 16 '16 at 08:03

2

you forget 2 '

$row2 = mysqli_fetch_assoc(mysqli_query($conn,"SELECT * FROM login WHERE
username='".$row['user']."'"));

answered Feb 16 '16 at 08:03

Gouda Elalfy

6,888
1
26
38

score 2 · Accepted Answer · edited May 23 '17 at 12:15

2

Try this query :

$row2 = mysqli_fetch_assoc(mysqli_query($conn,"SELECT * FROM login WHERE username='{$row['user']}'"));

Side note : Your query is unsafe. Read this How can I prevent SQL injection in PHP?.

edited May 23 '17 at 12:15

Community

1
1

answered Feb 16 '16 at 08:08

Mr. Engineer

3,522
4
17
34

I am using it for testing I will change it later. – DrevanTonder Feb 16 '16 at 08:11
I saw your HTML. Change your function like this : `var id = ;getChat(id);` – Mr. Engineer Feb 16 '16 at 08:23
Dont use `echo` inside while loop. Concate a string and `echo` it out side the loop. And put `exit` after `mysqli_close($conn);`. – Mr. Engineer Feb 16 '16 at 09:03
it gets stuck at `if (mysqli_num_rows() > 0) `. – DrevanTonder Feb 16 '16 at 09:19

score 0 · Answer 3 · answered Feb 16 '16 at 08:15

Simply use this query:

"SELECT * FROM login WHERE username='".$row['user']."'"

instead of

"SELECT * FROM login WHERE username=".$row['user']

and if you want a simple query then use:

$usr = $row['user'];
$row2 = mysqli_fetch_assoc(mysqli_query($conn,"SELECT * FROM 
login WHERE username=$usr"));

It'll definitely work.

score 0 · Answer 4 · answered Feb 16 '16 at 08:24

0

Write your query as below:-

$sql = "SELECT * FROM login WHERE username='{$row['user']}'";

mysqli_fetch_assoc(mysqli_query($conn,$sql));

Hope it will help you :)

answered Feb 16 '16 at 08:24

Ravi Hirani

6,511
1
27
42

php mysqli not working

4 Answers4