Posted by finbogey jones on 12/06/06 18:08

Jerry Stuckle wrote:
> finbogey jones wrote:
>> OK, total noob here trying to teach himself PHP/mySQL by reading books
>> and playing with code from the web and from trying to write increasingly
>> difficult (for me) code. Here's what I'm trying to do:
>>
>> I have a db with a few tables. I have managed to write a simple query to
>> one of these tables that then displays all the fields of each record
>> that matches the query. Now I want to take it one step further, I want
>> to create a drop down selection box where the user can select any of
>> these records. I managed to create the box, but can only display one
>> field of the records (so if the fields were firstname and lastname, I
>> have only figured out how to show one of the two in the dropdown).
>>
>> Any hints would be appreciated,or pointers to good web resources for
>> examples.
>>
>> Here's the query code. I have deleted the form part as it is crap and
>> I'd like to see how it is properly done.
>> -----------------------------
>> $query = "SELECT ID, first, last FROM master WHERE fhl is null";
>> $result = mysql_query($query) or die('Error, query failed');
>>
>> $tsv = array();
>> $html = array();
>>
>> while($row = mysql_fetch_array($result, MYSQL_NUM))
>> {
>> $tsv[] = implode("\t", $row);
>> $html[] = "<tr><td>" .implode("</td><td>", $row) .
>> "</td></tr>";
>> }
>>
>> $tsv = implode("\r\n", $tsv);
>> $html = "<table>" . implode("\r\n", $html) . "</table>";
>
> Well, first of all I don't just implode the data; I prefer to access the
> individual columns myself.
>
> But you can do something like:
>
> <select name="myselect">
> <?php
> while ($row = mysql_fetch_array($result)) {
> echo "<option value=$row['id']>" . $row[firstname] . " "
> .$row[lastname] . "</option>\n";
> ?>
> </select>
> Of course you'll need to adjust based on your particular column names in
> your database.
>
Thanks for the info, I'll try that out tonight.

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