MySqli Select is not working - php

Question

I couldn't figure out why it has some problem like this. The error displayed mysqli_fetch_assoc() expects parameter 1 to be mysqli_result.
My code is like this. I don't know which one is wrong.
Have a look at my con.php.
<?php
$conn = mysqli_connect('localhost', 'username', 'password');
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
?>
and this is my page.php
<?php
include 'con.php';
$link = mysqli_connect('localhost', 'username', 'password', 'username') or die("Error " . mysqli_error($link));
$query = " SELECT thread_id, thread_name, thread_date
FROM forum_thread
ORDER BY thread_date";
$result = mysqli_query($link,$query);
while ($row = mysqli_fetch_assoc($result)) {
$thread_id = $row ['thread_id'];
$thread_name = $row['thread_name'];
$thread_date = $row['thread_date'];
echo "$thread_id, $thread_name, $thread_date";
Any ideas? Appreciate any answers from you. Cheers!

Answer 1

I think I figured it out:
Change $result = mysqli_query($link,$query);
to $result = mysqli_query($conn,$query); since $conn is your DB connection, not $link
as per what you posted above
<?php
$conn = mysqli_connect('localhost', 'username', 'password', 'database_name');
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
?>
and remove:
$link = mysqli_connect('localhost', 'username', 'password', 'username') or die("Error " . mysqli_error($link));
since you're already loading your DB con with include 'con.php';
Or this way:
<?php
$conn = mysqli_connect('localhost', 'username', 'password', 'database_name') or die("Error " . mysqli_error($conn));
$query = " SELECT thread_id, thread_name, thread_date
FROM forum_thread
ORDER BY thread_date";
$result = mysqli_query($conn,$query);
while ($row = mysqli_fetch_assoc($result)) {
$thread_id = $row ['thread_id'];
$thread_name = $row['thread_name'];
$thread_date = $row['thread_date'];
echo "$thread_id, $thread_name, $thread_date";

Answer 2

The error message indicates the query did fail for some reason. Instead of just using $result test to see if it's valid and if not output the error to the screen or a log file:
if ($result) {
// continue processing $result
}
else {
echo mysqli_error($link);
}
Edit:
The 4th parameter in the mysqli_connect should be the name of the database, "abc" or whatever.

Answer 3

Try to use following code
$mysqli = new mysqli("localhost", "user", "password", "database");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli- >connect_error;
}
Here when you try to get connection object itself you can identify the problem if something goes wrong.