As many people have told me and I have just now got it through my thick head try either echoing the query or use
$result = mysql_query($sql) or die(mysql_error()); You will find out what is being passed (or not passed) through your variables or find out what error you are getting. HTH Jennifer "Neil" <[EMAIL PROTECTED]> wrote in message 002501c1ecc2$8b1e16a0$1bc5fea9@neilmar">news:002501c1ecc2$8b1e16a0$1bc5fea9@neilmar... Help. Help Please, Please tell me what I am doing wrong here...I have a form to input data but it is not posting it....here is the code. It seems to work no errors but a query show no new additions to the table. $sql = "INSERT INTO employees (first,last,address,date_of_birth,sex,age,telephone, parent_guardian_name,parent_guardian_workplace,work_telephone,emergency_info rmation_contact, list_physical_conditions_if_any,all_schools_previously_attended,examinations _passed,older_brothers,younger_brothers,older_sisters,younger_sisters, names_sisters_or_brothers_attending_school) VALUES ('$first','$last','$address','$date_of_birth',$sex','$age','$telephone','$pa rent_guardian_name','$parent_guardian_workplace', '$work_telephone','$emergency_information_contact','$list_physical_condition s_if_any','$all_schools_previously_attended','$examinations_passed','$older_ brothers','$younger_brothers','$older_sisters','$younger_sisters','$names_si sters_or_brothers_attending_school')"; $result = mysql_query($sql); echo "Thank you! Information entered.\n"; } else{ // display form ?> Neil --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.351 / Virus Database: 197 - Release Date: 4/19/2002 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
