Re: INSERT Question

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... PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/

-----Original Message----- From: Charles Kline [mailto:ckline@rightcode.net] Sent: Saturday, March 29, 2003 9:11 PM To: Peter Lovatt Cc: php-db@lists.php.net 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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php

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