Php code prints nothing while using with a HTML table

Question

What is the problem with my code?

$sql = "select username,num from login order by num desc";
$result = $conn->query($sql);
$rnk=1;
if ($result->num_rows > 0) {
    while($row = mysqli_fetch_array($result)){
        echo "<tr><td>";
        echo $row["username"];
        echo "</td><td>";
        echo $row["num"];
        echo "</td><td>";
        echo $rnk;
        echo "</td></tr>"
    }
} else {
    echo "0 results found";
}
$conn->close();

I had placed the above code after providing necessary connection codes between table element in my html page with .php extension. When i call the page it shows only a blank page. Its not even showing the table headers.

even echo "<tr><td>" . $row["username"]. "</td><td>" . $row["num"]. " " . $rnk. "</td></tr>"; doesnt works — Gautam Krishna R
– Gautam Krishna R, Commented Jun 22, 2015 at 5:11
@Hanky웃Panky is trying to say that you forgot ; on that line. — Logan Wayne
– Logan Wayne, Commented Jun 22, 2015 at 5:13

Gautam Krishna R · Accepted Answer · 2015-06-22 05:30:42Z

1

Try this one

$sql = "select username,num from login order by num desc";
$result = $conn->query($sql);
$rnk=1;
if ($result->num_rows > 0) {
 echo "<table>";
 while($row = mysqli_fetch_array($result)){
 echo "<tr><td>".($row['username'])."</td>";
 echo "<td>".($row['num'])."</td>";
 echo "<td>".($rnk)."</td></tr>";
 }

    echo "</table>";
} else {
    echo "0 results found";
}
$conn->close();

edited Jun 22, 2015 at 5:30

Gautam Krishna R

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answered Jun 22, 2015 at 5:13

Nipun Tyagi

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Comments

Cody Caughlan · Accepted Answer · 2015-06-22 05:18:33Z

0

Crank up PHP error reporting so you can see what the real problem is.

http://php.net/manual/en/errorfunc.configuration.php#ini.error-reporting

Also look in the Apache error.log to see what, if any, errors are being logged. Assuming you are using Apache

answered Jun 22, 2015 at 5:18

Cody Caughlan

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