PHP while-loop not working with mysql_result - php

Question

I have a Physician Query:
// Primary Physician Query
$qPhysician = mysql_query("SELECT * FROM physicians ORDER BY lastName ASC, firstName ASC");
$rowPhysician = mysql_fetch_array($qPhysician);
// State Query for Physician
$idStatePhysician = $rowPhysician['idstate'];
$qStatePhysician = mysql_query("SELECT * FROM states WHERE idstate=$idStatePhysician");
$rowStatePhysician = mysql_fetch_array($qStatePhysician);
// City Query for Physician
$idCityPhysician = $rowPhysician['idcity'];
$qCityPhysician = mysql_query("SELECT * FROM cities WHERE idcities=$idCityPhysician");
$rowCityPhysician = mysql_fetch_array($qCityPhysician);
I have a while loop to display all physicians row to a table:
$num = mysql_num_rows($qPhysician);
$i=0;
while($i < $num)
{
$idphysicians = $rowPhysician['idphysicians'];
if ($i % 2 == 0){
echo "<tr class='even' onclick=\"DoNav('physicianUpdate.php?idphysicians=$idphysicians');\">";
}
else{
echo "<tr class='odd' onclick=\"DoNav('physicianUpdate.php?idphysicians=$idphysicians');\">";
}
echo "<td>" . mysql_result($qPhysician,$i,"lastName") . "</td>";
echo "<td>" . mysql_result($qPhysician,$i,"firstName") . "</td>";
echo "<td>";
if(isset($rowPhysician['idcity'])){echo mysql_result($qCityPhysician,$i,"name");} else{}
echo "</td>";
$i++;
}
My problem is: I have 3 rows of data from my physicians table. Each has a value for 'idcity' reflecting the idnumber from my City table. However, the 1st row of Data displays the idcity=Name properly, but the 2nd and 3rd row gave an error:
Warning: mysql_result() [function.mysql-result]: Unable to jump to row 1 on MySQL result index 7 in C:\wamp\www\iPOC\physicians.php on line 55
Also, if I have a blank value for idcity on one of the row, it also generates an error.
Please help! Thanks in advance!

Answer 1

The problem is that you're using mysql_result() with a one-way result. The correct fix is to use one of the mysql_fetch_*() functions instead, checking the returned value in your while loop.
while($row = mysql_fetch_array($qPhysician)) {
...
}

Answer 2

Something like this would probably work better:
$qCityPhysician = mysql_query("SELECT * FROM cities WHERE idcities=$idCityPhysician");
$qCityPhysicians = array();
while($row = mysql_fetch_array($qCityPhysician)) {
$qCityPhysicians[$row['idcity']] = $row['name'];
}
$qPhysician = mysql_query("SELECT * FROM physicians ORDER BY lastName ASC, firstName ASC");
$i=0;
while($row = mysql_fetch_array($qPhysician)) {
if ($i % 2 == 0) {
echo "<tr>";
echo "<td>" . $row['lastName'] . "</td>";
echo "<td>" . $row['firstName'] . "</td>";
echo "<td>";
if(isset($row['idcity'])) {
echo $qCityPhysicians[$row['idcity']];
}
echo "</td>";
$i++;
}
}