Okay. Will do. Thanks both of you for your time and help. It is appreciated!

- Charles

On Saturday, March 29, 2003, at 09:21 PM, John W. Holmes wrote:

The only addition I have is to adapt the code to create a single INSERT
query in the format I gave earlier. It will run quicker overall.

---John W. Holmes...

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-----Original Message-----
From: Charles Kline [mailto:[EMAIL PROTECTED]
Sent: Saturday, March 29, 2003 9:11 PM
To: Peter Lovatt
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] INSERT Question

Big help! Thanks much.

Now... to figure out how to create that kind of form in HTML_QuickForm
(hehehe). If I can't then I just make it by hand, but now I know the
method! Phew... always more to learn.

- Charles


On Saturday, March 29, 2003, at 08:51 PM, Peter Lovatt wrote:


Hi

you need to create 5 selects, all named investigator[] with index
1-5
(or
0-4), with the option set to empty for no selection.

this will return an array
investigator[1] => '1064618047'
investigator_yesno[1] => 'Y'

investigator[2] => '1649815377'
investigator_yesno[2] => 'N'

for example


<form> <table> <tr> <td> <select name="investigator[1]"> <option value="">Choose person</option> <option value="1064618047">Paul A</option> <option value="1655387822">Katrina A</option> <option value="1649815377">David A</option> </select> is a primary investigator? <input name="investigator_yesno[1]" value="Y" type="radio" checked="checked" />Yes <input name="investigator_yesno[1]" value="N" type="radio" />No </td> </tr> <tr> <td> <select name="investigator[2]"> <option value="">Choose person</option> <option value="1064618047">Paul A</option> <option value="1655387822">Katrina A</option> <option value="1649815377">David A</option> </select> is a primary investigator? <input name="investigator_yesno[2]" value="Y" type="radio" checked="checked" />Yes <input name="investigator_yesno[2]" value="N" type="radio" />No </td> </tr> ...more here .... </table> </form>



on the script receiving the data you  then need a for loop to insert
the
records, one for each person


for ($i = 1; $i <= 4; $i++) { //only do it if there is a value for $investigator[$i], ie there is
a
selection
if($investigator[$i]) {
    $query = 'INSERT INTO tbl_report_people
                (record_id
                , person_id
                , investigator_yesno
                )
                VALUES
                ("'.$record_id[$i].'"
                , "'.$investigator[$i].'"
                , "'.$investigator_yesno[$i].'"
                )
                                ';
    $mysql_result = mysql_query($query, $mysql_link);

}//end if

}//end for



hope this helps

Peter







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