[RESOLVED] Populating Drop-Down From A Query

scriderpsd

Hi All, I'm trying to populate a drop-down box from the results of a query. I end up with the first option coming through and then blanks in the rest. Something like this...

<option value=Acevedo>Acevedo</option>
<option value=></option>
<option value=></option>
<option value=></option>
<option value=></option>

Here's my code.

	
		$sql="SELECT last_name FROM active ORDER BY last_name";
		$conn = db_trans_connect();
		$result = mysql_query($sql,$conn);




echo '<TD>&nbsp;</TD></TR>';
echo '<tr><td></td><td align=left><span>';
echo '&nbsp;<label for="lname">LAST NAME</label>&nbsp;';
echo '<select name="lname_select" id="lname_select">';
		for ($i = 0; $i < mysql_num_rows($result); $i++) {
			$row_array=mysql_fetch_array($result);
			echo("\n<option value=" . $row_array[$i] . ">" . $row_array[$i]);		
			echo("</option>");
			}

echo '</select>';
echo '</span></td></tr>';

Anyone able to see where my error is at???

Thanks,

Scott

bpat1434

This is very inefficient and can better be made as a while loop:

 /* Innefficent code....
for ($i = 0; $i < mysql_num_rows($result); $i++) {
            $row_array=mysql_fetch_array($result);
            echo("\n<option value=" . $row_array[$i] . ">" . $row_array[$i]);        

            echo("</option>");
            } */

/* REPLACE WITH THE FOLLOWING */
while($row = mysql_fetch_array($result))
{
    echo "\n<option value='". $row['last_name']."'>".$row['last_name']."</option>";
}

scriderpsd

Thanks so much bpat143. That fixed it up and it's working as expected. Now, I have a couple more questions just for knowledge's sake.

Why is the while loop more effecient than the for loop in this instance? I was under the impression that the for loop was the effecient way to traverse an array. Obviously I was wrong. Can you or anyone else give a little explanation.

Second, even though it is inefficient, I cannot see why my for loop was not working. Anyone?

Thanks again,

Scott

bpat1434

Ah junk.... can't delete the post....

bpat1434

Typically, you want a while() loop when itterating through arrays. They are just faster than for() loops, and less confusing most of the time. It just works a lot better as a while() than a for() since you have less room for error most of the time. That's not to say it's the only way to go, but typically it's better.

I would guess your for() loop wasn't working because if you look at your array, all the keys will be "0". You incremented your key instead of just leaving it set at "0". It's an honest mistake that everyone makes. One way you could have solved this on your own is to print out the array structure so you can see how it looks. Then think about your code and what it's doing. If you had seen the array, you'd see that it's only one record long, with the key of "0". While you incremented $i properly to account for another row, you also lost the proper key. So when $i was 1, there was no array key 1, so you get a blank option.

Hope that makes sense.

pashedu

I have a problem related to the one above:
My code looks like this:

function getReferers(){
echo '<select name="refby" style="font: 11px tahoma black;">';
$qry = "select * from customer";
$rslt = mysql_query($qry) or die(mysql_error());
$row = mysql_fetch_array($rslt);
while($row = mysql_fetch_array($rslt)){
echo "<option value=".$row['cusid']." selected>".$row['lname'].", ".$row['fname']."</option>";
}
echo '</select>';

I have like 200 records, how can I select in the drop-down-box record 100 when the page is loaded???
I know that the <select> has the option "selected" but it does not work!
Thanks

Piranha

bpat, how effective is foreach compared to a while-loop?

bpat1434

I have a problem related to the one above:
My code looks like this:

function getReferers(){
echo '<select name="refby" style="font: 11px tahoma black;">';
$qry = "select * from customer";
$rslt = mysql_query($qry) or die(mysql_error());
$row = mysql_fetch_array($rslt);
while($row = mysql_fetch_array($rslt)){
echo "<option value=".$row['cusid']." selected>".$row['lname'].", ".$row['fname']."</option>";
}
echo '</select>';

I have like 200 records, how can I select in the drop-down-box record 100 when the page is loaded???
I know that the <select> has the option "selected" but it does not work!
Thanks

Well, if you're grabbing everything back from the row you could use the row ID (if you have one) to specify which is the 100th. THen based upon that put the "selected" text inside the <option> tag.

while($row = mysql_fetch_array($result))
  echo '<option value="'.$row['cusid'].'"'.($row['id']==100 ? ' selected':'').'>'.$row['lname'].', '.$row['fname'].'</option>';

As it stands now, you're "selecting" each and every record.

bpat, how effective is foreach compared to a while-loop?

Supposedly foreach is faster when dealing with arrays. But it all depends on what you want to do, and what information about the array you need. You could always benchmark your own tests to see which types of loop do better with which types of arrays 😉