Loop in PHP not working

Question

$qPhysician = mysql_query("SELECT * FROM physicians");
$num = mysql_num_rows($qPhysician);
$i=0;
while($i < $num)
{
    "<tr>";
    "<td>" . mysql_result($qPhysician,$i,"lastName") . "</td>";
    "<td>" . mysql_result($qPhysician,$i,"firstName") . "</td>";
    "</tr>";
    $i++;
}

I get blank result.

If I echo $num, I get "19" which is the number of rows in my DB. If I echo $rowPhysician['lastName'] just to test out if I get records, I get at least 1 record of last name. I don't know if there is something wrong with "while". Please help me out.

You really should be using mysql_fetch_array. mysql_num_rows is generally much slower. — cwallenpoole, Jul 17 '11 at 04:37
@cwallenpoole - how would i run that as an alternative? please code :) thanks! — Arnold Porche Villaluz, Jul 17 '11 at 04:49

score 4 · Accepted Answer · answered Jul 17 '11 at 04:20

4

Are you missing the echo to print out the strings?

while($i < $num) {
    echo "tr";
    echo "td" . mysql_result($qPhysician,$i,"lastName") . "/td";
    echo "td" . mysql_result($qPhysician,$i,"firstName") . "/td";
    echo "/tr";
    $i++;
}

answered Jul 17 '11 at 04:20

RHSeeger

16,034
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oh duh! i totally forgot about that echo! thank you so much RHSeeger. Now i'm laughing at myself :) – Arnold Porche Villaluz Jul 17 '11 at 04:27
Didn't you get some errors when checking the page? Looking at the log is my first step to debugging php. – Diff.Thinkr Jul 17 '11 at 04:32

Hnatt · Answer 2 · 2011-07-17T05:03:32.343

0

Anyways echo'ing HTML is bad practice. The more correct output could should look like

...
while($i < $num) :?>
    <tr>
      <td><?php echo mysql_result($qPhysician,$i,"lastName"); ?></td>
      <td><?php echo mysql_result($qPhysician,$i,"firstName") ?></td>
    </tr>
<?php $i++; endwhile;
...

Note that you wrote closing HTML wrong, too.

edited Jul 17 '11 at 05:03

answered Jul 17 '11 at 04:32

Hnatt

5,767
3
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"Bunch of errors" is a bad description. I could help if you tell what error messages you got. – Hnatt Jul 17 '11 at 04:42
Warning: mysql_result() [function.mysql-result]: Unable to jump to row 3 on MySQL result index 6 in C:\wamp\www\iPOC\physicians2.php on line 33 – Arnold Porche Villaluz Jul 17 '11 at 04:48
Have you forgot to place `$qPhysician = mysql_query("SELECT * FROM physicians"); $num = mysql_num_rows($qPhysician); $i=0; ` before loop? – Hnatt Jul 17 '11 at 04:52
those lines are still above the loop – Arnold Porche Villaluz Jul 17 '11 at 04:57
Mah bad, the $i++ part should in the end of the loop. Updated my answer. – Hnatt Jul 17 '11 at 05:04
still error: Parse error: syntax error, unexpected T_INC in line 44 (this is where I have – Arnold Porche Villaluz Jul 17 '11 at 05:09

score 0 · Answer 3 · answered Jul 17 '11 at 05:00

Technically, this isn't an answer, but it is easier to show the code this way:

// first, only take the records you need -- each record adds more time to
// query, so if you only need 2, only select 2.
$query = mysql_query("SELECT lastName, firstName FROM physicians");

// mysql_fetch_assoc returns an associative array of all of the columns
// mysql_fetch_row   returns a numerically indexed array.
// mysql_fetch_array returns an array with both numeric and string indexing.
// they will all return FALSE when there are no more results in the query.
while( $arr = mysql_fetch_assoc( $query ) )
{
    echo "<tr>";
    // Now, use array indexing.
    echo "<td>" . $arr[ "lastName" ] . "</td>";
    echo "<td>" . $arr[ "firstName" ] . "</td>";
    echo "</tr>";
}

It is a very rare circumstance that mysql_result is actually a better option -- usually it is better to just populate a couple of arrays in PHP and be done with the DB.

Loop in PHP not working

3 Answers3

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