Form population

[Seapep]

Hi Folks,

 

I am learning php and mysql so please bear with me and again if this is not in the right place please let me know.

 

Anyways, I have a form setup that will be used to add events for a local sports team. I have 3 tables.

 

Table1(Level) is working fine and populating the drop down form as I need it to.

 

Table 2 is the Events. This is used to track all the event names that are used through the organization.

 

Table 3 is the Schedule. This show all the events that are scheduled.

 

What I need the form to do is allow the client to choose the event type and the level type for the registration process.

 

Can I do this with mutliple dropdown lists with mysql/php? Here is what I have that is working.

 

mysql_select_db($db_database)
or die("Unable to select database: " . mysql_error());
$sql = "SELECT\n"
    . "levels.LevelID,\n"
    . "levels.LevelName\n"
    . "FROM\n"
    . "levels\n";

$result = mysql_query($sql) or die(mysql_error());

$options="";

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

$LevelID=$row["LevelID"];
$LevelName=$row["LevelName"];
$options.="<OPTION VALUE=\"$LevelID\">".$LevelName.'</option>';


}

?>



<style type="text/css">
<!--
body p {
color: #F00;
}
-->
</style>



<table width="500" border="1" align="center" cellpadding="0" cellspacing="1" >
<tr>
<td>
<form name="form1" method="post" action="scheduleinsert_ac.php">
<table width="100%" border="1" cellspacing="1" cellpadding="3">
<tr>
<td colspan="3"><div align="center">
  <p><strong>Myers Schedule Form</strong></p>
</div></td>
</tr>
<tr>
<td>Event  Name</td>
<td><div align="center">:</div></td>
<td></td>
</tr>

<tr>
  <td>Level</td>
  <td><div align="center">:</div></td>
  <td><select name="LevelID">
    <OPTION VALUE="0">Choose
      <?php echo $options;?>
      </select></td>
</tr>

[BLaZuRE]

Yes, why wouldn't you?  You can put multiple drop-downs in HTML, so what's the problem?  Maybe I'm not understanding what you're trying to ask.

[Seapep]

I guess the issue is where would I put the 2nd query which would populate the 2nd drop down list? I have seen a lot of examples but they all are linking back to each other via a relationship.

 

Whereas what I have is 2 drop down lists querying 2 different tables. I tried to copy what I have with the corresponding fields and all I get is a blank field.

[fortnox007]

just apart frm that i would add another table named category (like weekly monthly children elder etc) that way in the end you can out put everything in nice categories. back to your thing.

 

I think this is what you are looking for:

ie.

SELECT table1.column1, table2.column2 FROM table1, table2 WHERE table1.column1 = table2.column1;

source: http://articles.techrepublic.com.com/5100-10878_11-1050307.html

 

-also nice for people to make a category first.  so you first show them the existing categories, and give thme the opportunity to add one which is ofcourse pretty simple (ID, category_name, category_description)

[BLaZuRE]

You can also do something like:

$query1 = 'SELECT field1, field2 FROM table1';

while($row=mysql_fetch_array($query1)
{
     //Code to echo form options
}

$query2 = 'SELECT field1, field2 FROM table2';

while($row=mysql_fetch_array($query2)
{
     //Code to echo form options for events
}

 

You don't need to prefix fieldX with tableX (tableX.fieldX) if you're only using one table in the query.

[Seapep]

So something lilke this

 

<?php

require_once 'dblogin.php';

$db_server = mysql_connect($db_hostname, $db_username, $db_password);

if (!$db_server) die("Unable to connect to MySQL: " . mysql_error());

 

mysql_select_db($db_database)

or die("Unable to select database: " . mysql_error());

$sql = "SELECT\n"

    . "levels.LevelID,\n"

    . "levels.LevelName\n"

    . "FROM\n"

    . "levels\n";

 

$result = mysql_query($sql) or die(mysql_error());

 

$options="";

 

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

 

$LevelID=$row["LevelID"];

$LevelName=$row["LevelName"];

$options.="<OPTION VALUE=\"$LevelID\">".$LevelName.'</option>';

 

 

$sql2 = "SELECT\n"

    . "events.EventID,\n"

    . "events.EventName\n"

    . "FROM\n"

    . "events\n";

 

$result2 = mysql_query($sql2) or die(mysql_error());

 

$options2="";

 

while($row2 = mysql_fetch_array($result2))

 

$EventID=$row2["EventID"];

$EventName=$row2["EventName"];

$options2.="<OPTION VALUE=\"$EventID\">".$EventName.'</option>';

}

 

?>

 

 

<table width="500" border="1" align="center" cellpadding="0" cellspacing="1" >

<tr>

<td>

<form name="form1" method="post" action="scheduleinsert_ac.php">

<table width="100%" border="1" cellspacing="1" cellpadding="3">

<tr>

<td colspan="3"><div align="center">

  <p><strong>Schedule Form</strong></p>

</div></td>

</tr>

<tr>

<td>Event  Name</td>

<td><div align="center">:</div></td>

<td><select name="EventID">

    <OPTION VALUE="0">Choose

      <?php echo $options2;?>

      </select></td>

</tr>

 

<tr>

  <td>Level</td>

  <td><div align="center">:</div></td>

  <td><select name="LevelID">

    <OPTION VALUE="0">Choose

      <?php echo $options;?>

      </select></td>

</tr>

[Seapep]

Thanks for the help guys. I got it to work, in the above code I was missing a set of { } brackets.

 

Cheers,