ADD multi fields from drop down list of table

Barb54

Hi,
I have a form for adding records to a table. It connects first to a category table to determine what to display for the next list. This next list is yet another table so now only items in the selected catergory display. This all works fine.

This 2nd table holds information in 4 fields that is specific to the row the user will select from this list. I don't want to "display" those fields but I DO need to ADD those fields to the records table when the form is processed. So I am thinking they should be stashed in a variable somehow, somewhere once the selection has been made, then get added as "hidden" values? These fields are crucial as they are used to produce reports from the records table.

I have searched here to try to find an answer on how to accomplish this. Nothing I have seen so far does this quite this way. Can this be done? I see some using "." when looking at other tables but don't really understand why, how, when on this.

If you like I can post example code, but it's basically the standard "connect to a table as a drop down" code at this point.

once catCode is selected all types in that code will fill Table 2 dropdown
Table 2 - type table - (display only):typeName
(Store for adding): typeCode, amt_type, amt, amt_count

Syntax seems to be the biggest issue here for me. Any help would be greatly appreciated.

Thanks,
Barb

barand

The usual way would be to store the code as the option value and display the name. When processing the form, get the other details with a query using the code value.

If you do want all 4, make the option value something like

echo"<option value='$typeCode|$amt_type|$amt|$amt_count'>" $typeName </option>";

and, on processing, separate out with explode()

hth

Barb54

Thanks Baarand,

I will give that a try. I already have a $fieldlist for exploding the vars so I'm hoping just adding the others to the list in the selection will take care of it? If it doesn't work, I will come back for more help. If it does, I'll mark this resolved.

Thanks again,
Barb

Barb54

Here is some original code, vars are a bit different but much the same idea:

$sql="SELECT * FROM pointsTable 
WHERE "`catCode` = 'A'"
ORDER BY point_id ASC";  

$result=mysql_query($sql,$dblink);  

echo "<ol>";  

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

    echo '<option value='.$row['point_type' , 'pointamt' , 'amt_type' , 'pointCount' ].'>' . $row['pointName'] .  

'</option>';  

}  // end while($row = mysql_fetch_array($result)) 
echo "</ol>"; 

?>  

</SELECT></P>

I don't know if this is correct or not. I decided to just SET the catCode = "A" instead of using a dropdown for this. It seemed easier for this form. So now based on the catCode, only Points with that code should show up in the dropdown list.

I have other fields in the form that are set at the beginning of the script such as recID, date, nick(based on sessionID, etc) all of which go in as "hidden on the form.

I'm confused on how to set the fields for $results(or is it $row) picked up from the drop down table. Each field has a similar yet just a tad different name in the records table than what they have in the Points table.

I don't want these tables linked for viewing data. I know someone will probably suggest that but I have various reasons for this choice. They are only used to help get the correct fields into the records table with as little decision making by the user as possible.

Thanks again in advance,
Barb

leatherback

Hi Barb,

You say that the data extracted from the second table, depend on the choice of the user. Couldn't you just select those 4 fields from the table once the user has made his/her selection, and has submitted the form.?

Then after extracting that data from the table, you insert everything into the final table?

Second point: If those 4 fields are the only fields in the second table, have you considered just using an ID colum to link to that table, instead of copying all the information when some data is inserted by a user?

Barb54

Couldn't you just select those 4 fields from the table once the user has made his/her selection, and has submitted the form.?

Once the user has made their selection, I want those extra fields to be posted when the form is submitted. I just get lost when I get to this point and don't know how to do that. I know, I know, it's probably staring me in the face and I just can't see it, lol.

If those 4 fields are the only fields in the second table

No, there are other fields in that table that will not be added to the records table. I don't want my data to be "linked" back to this drop down table. The Points Table and Cat tables are only used for the insertion of data to make it easier for the user, nothing more.

HELP? lol Thanks
Barb

Barb54

Hi,
Day #3 and still no where, lol. I can use the 2nd table to reference itself to determine one of the next fields that needs to go in but I don't think this is good. I just add another drop down list for the same table2 but reference a different field filter. I think this is sloppy and user still has to select the 1 item that will show up in that "so called" list, lol. And I still have yet another field that needs to go in.

Can someone explain the use of an array to catch these fields and then how I would "explode" them so they can then each be individually inserted to the correct fields in table3 during @?

I understand that I can ref all fields in the table in the SELECT statement. I also 'think' I understand how to pick all 4 up in the option select value = statment. I'm starting to get frustrated now :-( I learn these things as I knock up against them, lol.

SPLAT............................ lol

I'll give my famous "Southern Fried Chicken" recipe to the first one that can help, lol

barand

$row['point_type' , 'pointamt' , 'amt_type' , 'pointCount' ].

You can't do that. Specify each field separately. ie

$row['point_type'] , $row['pointamt'] , $row['amt_type'] , $row['pointCount' ]