mysql fetch array is not showing result - php

Question

So I have been working on this code for a while. I believe I am really close. My if statement that is inside my while loop isn't showing any data in the area it's suppose to show. I know mysql is old and deprecated. I am going to change it once I figure this out.
$result = mysql_query("SELECT * FROM inventoryTable",$db);
$result2 = mysql_query("SELECT * FROM users WHERE username='$username' and sub = 'yes'",$db);
echo "<TABLE style=\"background-color: #FFFFFF; border: 10px solid A4A4A4;\">";
echo"<TR><TD>"."<B>Title</B>"."</td>";
echo"<TD>"."<B>Authors First Name</B>"."</td>";
echo"<TD>"."<B>Authors Last Name</B>"."</td>";
echo"<TD>"."<B>ISBN</B>"."</td>";
echo"<TD>"."<B>Publisher</B>"."</td>";
echo"<TD>"."<B>Course Number</B>"."</td>";
echo"<TD>"."<B>Source</B>"."</td></TR>";
while ($myrow = mysql_fetch_array($result))
{
echo "<TR><TD>".$myrow["title"]."</td>";
echo "<TD>".$myrow["authorsFirst"]."</td>";
echo"<TD>".$myrow["authorsLast"]."</td>";
echo "<TD>".$myrow["ISBN"]."</td>";
echo "<TD>".$myrow["publisher"]."</td>";
echo "<TD>".$myrow["courseNum"]."</td>";
while ($subResults = mysql_fetch_row($result2))
{
If($subResults == 'yes' )
{
echo "<td>".$myrow["source"]."</td>";
} else {
echo "<TD>"."Please subscribe to View"."</td>";
}
echo "</TABLE>";
}
}
?>
This is the part of my code that isn't showing any results.
while
while ($myrow = mysql_fetch_array($result))
{
echo "<TR><TD>".$myrow["title"]."</td>";
echo "<TD>".$myrow["authorsFirst"]."</td>";
echo"<TD>".$myrow["authorsLast"]."</td>";
echo "<TD>".$myrow["ISBN"]."</td>";
echo "<TD>".$myrow["publisher"]."</td>";
echo "<TD>".$myrow["courseNum"]."</td>";
while ($subResults = mysql_fetch_row($result2))
{
If($subResults == 'yes' )
{
echo "<td>".$myrow["source"]."</td>";
} else {
echo "<TD>"."Please subscribe to View"."</td>";
}
echo "</TABLE>";
}
}
I want my session user to be able to see the source from my inventory table if they have a yes in the sub field. If they do not have a yes in the sub field, they will see please subscribe to view. Am i doing the mysql_fetch incorrectly or is there a problem because I have 2 while loops going on at once.

Answer 1

you need to have "==" to compare two values, otherwise you assign the second value to the first variable:
...If($username == $subResults)...
or to use a strict comparison of type and content, use "==="
If($username === $subResults)
also I am thinking the code should be
...If($subResults ==="yes"){echo"....///desired content";}else{echo"...//alternate content";}...
and you are missing the echo statement and closing </td> in the code
"<td>".$myrow["source"];
should be
echo"<td>".$myrow["source"]."</td>";
in fact - aren't you missing the closing td's in all of the cells?