mysql fetch array is not showing result

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.

Apart from the =/== issue – $username likely contains a string value, yes? But what you are comparing it to is an array. — C3roe
– C3roe, Commented Apr 9, 2016 at 3:52
And of course your inner while loops only once, on the first iteration of the outer while loop. After that, mysql_fetch_array($result2) will continue to return false, so the loop body isn’t executed any more. You’ll need to use mysql_data_seek to reset the record pointer for that result set, if you want to loop over it again. — C3roe
– C3roe, Commented Apr 9, 2016 at 3:55
And the MySQL extension has been deprecated for long time now already, and has been completely removed in PHP 7. You should be using MySQLi or PDO instead. — C3roe
– C3roe, Commented Apr 9, 2016 at 3:57
@CBroe - the OP knows that and posted this"... I know mysql is old and deprecated. I am going to change it once I figure this out. ..." indicating that will be fixed in due course — gavgrif
– gavgrif, Commented Apr 9, 2016 at 3:58
$username is from the session. mysql_data_seek requires 2 parameters. What should I use as an offset if I am searching the table? — Cory
– Cory, Commented Apr 9, 2016 at 4:58

gavgrif · Accepted Answer · 2016-04-09 03:59:17Z

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?

edited Apr 9, 2016 at 3:59

answered Apr 9, 2016 at 3:41

gavgrif

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8 Comments

Cory Over a year ago

Nothing shows under code"<td>".$myrow["source"];code still. The row is just blank but every other row loads data from my inventory table correctly.

gavgrif Over a year ago

answer amended - look at the last half :)

Cory Over a year ago

I am getting an error now. The error says syntax error, unexpected '"<TD>"' (T_CONSTANT_ENCAPSED_STRING), expecting ',' or ';' on line 42. Line 42 is code<TD>".$myrow["authorsFirst"]."</td>"code

gavgrif Over a year ago

that would proabably be becuase the preceding TD isn't being closed - you need to go through and close all TD's that have been opened as well as the TR. also - i just noticed that your border styling is wrong too - you need a "#" before the border color declaration ... "#A4A4A4"...

Cory Over a year ago

I am no longer getting an error. "Please subscribe to View" is showing up 7 times though. I thought it was showing up 7 times because there may have been 7 users in my users table. There's 10 so it can't be that.

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