Simple PHP Question List Box

aidanmcg33

Hi i have a form with a drop down menu on it where the user can select a team from a list. how do i get this list to be automatically on page load from a mysql table called teams on the page load? I have tried several things before but i always seem to get the php code just appearing on the page rather than going off and processing the code. and i have used the <?php ?> tags thanks

aidanmcg33

I have actually solved the problem of the code appearing on the page and solved a few syntax errors but it still isnt functioning as planned.

<?php
/ sql statement to get details from table /
$query="SELECT Team FROM Teams ORDER BY Team";

					$result = mysql_query($query);
					echo '<select name="TeamAffil" value=">Select Team</option>';

					while($nt=mysql_fetch_array($result)){

					echo "<option value=$nt[id]>$nt[Team]</option>";

					}

					echo "</select>";


				?>

bradgrafelman

Couple of things:

Array indeces in your case should be strings. $row[name] does not have a string as an index, but a constant. PHP will detect this (assuming there actually isn't a constant named 'name') and throw an E_WARNING and then assume it is a string. So, use quotes: $row['name'].
Arrays, more specifically items in arrays referenced by indeces, should not simply appear inside a double-quoted string. If they do, they should be wrapped in braces ( {} ). Instead, I usually escape out of the string and concatenate the array in the middle, like so:
```
echo 'Your name is: ' . $row['name'] . '! Hello there.';
```
Why is there text inside of the SELECT tags? Did you mean for "Select Team" to be the first option? If so, you forgot to add an OPTION tag before that text..

aidanmcg33

I just seen that code on the internet in a guide to drop down boxes and tried to use it for my case. All i want it to do is connect to the database and populate the list box with each name in the table.

ahundiak

You need to query the id as well as the team.
$query="SELECT id,Team FROM Teams ORDER BY Team";

And make sure there is no case sensitive issues here. Dump out $nt just to make sure the indexes are being named as expected.

aidanmcg33

I am not that familiar with php yet and I know that in VB, i need to have a connection string to connect to the database is there any thing like this required in php?

aidanmcg33

I have been tryin to cut out the code and just get it to put out a list on the page without building up the list box to see if its bringin in the data ok and im getting an error on this line:

while($nt=mysql_fetch_array($result)){

Mysql_fetch_array(): supplied argument is not a valid MySQL result resource in register.php on line 80

ahundiak

The "not a valid MySQL" message indicates that the query was either was invalid or that a connection was not established. Probably the latter in your case.

Take the time to go through a simple php-sql tutorial. There are dozens of them available. Or else hire someone for the task.

aidanmcg33

I have been able to get all the values appearing on my form, but i am having problems with the line : echo "<option value=$nt[id]>$nt['County']</option>";

Parse error: parse error, expecting T_STRING' orT_VARIABLE' or `T_NUM_STRING'

Ive been trying to create this line with the joining you mentioned before with the my name is . $ name etc but not having any luck

ahundiak

Using quotes around array index names $nt['County'] is what is causing the problem. PHP is a bit funny that way. Normally you do want to use quotes except when you inside of a "string". As previously pointed out, you kight try something like:

echo "<option value={$nt['id']}>{$nt['County']}</option>";

aidanmcg33

thank you. I actually managed to get it going just in a round about kinda way. I want this list box to trigger the contents of a second list box. I reckon the best way to do this would be to have a javascript function that when called, runs some php similar to what i used above. I know that i just add in onChange attribute to the line where i build up the list box. Would you know what i would need to do in the javascript function that would run the php? thanks

ahundiak

That's bit outside of my area. Understand that javascript runs on the client while php runs on the server. So you cannot directly run php from inside of javascript. There are plenty of tutorials that talk about this.