Mysql: Warning: mysql_fetch_array() expects parameter 1 to be resource - php

Question

I earlier made a database in mysql and now i am trying to list all the values from it in a table, but I get the following error Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean on line: while ($row=mysql_fetch_array($result))
Here is my code:
$con=mysql_connect("localhost","root","");
if (!$con) {
die("Error: " . mysql_error);
}
mysql_select_db("my_db",$con);
$result = mysql_query("SELECT * FROM Users");
echo "<table border='1'>
<tr>
<th>Username</th>
<th>First Name</th>
<th>Last Name</th>
<th>Age</th>
<th>Email adress</th>
</tr>";
while($row=mysql_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['username'] . "</td>";
echo "<td>" . $row['firstname'] . "</td>";
echo "<td>" . $row['lastname'] . "</td>";
echo "<td>" . $row['age'] . "</td>";
echo "<td>" . $row['emailadress'] . "</td>";
echo"</tr>";
}
echo "</table>";
mysql_close($con);
I read other similar question but diden't get an answer.

Answer 1

Your code doesn't evaluate the response from MySQL about selecting the database or running your query. The error indicates that your query didn't succeed (hence mysql_query returns FALSE) - which means one or both of the above didn't work.
Test for errors when you select the database and use die(mysql_error()); to see why these calls are failing.

Answer 2

Make sure this line is actually getting the rows from the table:
$result = mysql_query("SELECT * FROM Users");
Try doing an error check in that query:
$result = mysql_query("SELECT * FROM Users") or die(mysql_error());
Possibly your table is called "users" and not "Users" (note the caps), so change accordingly.
Regards,
Richi