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



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