Getting no value inputs in database

[prcollin]

Ok same code different problem.  When i submit the form that corresponds to this code the auto number goes up and all zeros show up in the respective interger fields but no information is entered into the database other than the autonumber. 

 

<?php

include "clientconnect.php";
include "newclient.html";

mysql_select_db('greencut_customercenter', $con);

$client_id = mysql_real_escape_string($_POST['client_id']);
$client_fname = mysql_real_escape_string($_POST['client_fname']);
$client_lname = mysql_real_escape_string($_POST['client_lname']);
$client_address = mysql_real_escape_string($_POST['client_address']);
$client_city = mysql_real_escape_string($_POST['client_city']);
$client_state = mysql_real_escape_string($_POST['client_state']);
$client_zipcode = mysql_real_escape_string($_POST['client_zipcode']);
$client_phone = mysql_real_escape_string($_POST['client_phone']);
$client_cphone = mysql_real_escape_string($_POST['client_cphone']);
$client_email = mysql_real_escape_string($_POST['client_email']);
$client_website = mysql_real_escape_string($_POST['client_website']);
$client_notes = mysql_real_escape_string($_POST['client_notes']);

$sql = "insert into clients values ('$client_id', '$client_fname', '$client_lname', '$client_address', '$client_city', '$client_state', '$client_zipcode', '$client_phone', '$client_cphone', '$client_email', '$client_website', '$client_notes')";


if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";
mysql_close($con);
?>

[blueman378]

have you checked if your post values are what you expect?

[$username]

Might want to try something like this

 

$sql = "insert into clients (clientid) values ('$client_id'')";

make sure the field is listed than the value

 

Brett

 

[prcollin]

Might want to try something like this

 

$sql = "insert into clients (clientid) values ('$client_id'')";

make sure the field is listed than the value

 

Brett

 

 

 

getting an error that says "duplicate entry for key 2" no clue what that means

[JonnoTheDev]

Remove the single quotes around your php variables!

 

values ($client_id, $client_fname

 

or

 

values ('".$client_id."', '".$client_fname."'

[JonnoTheDev]

also are you supplyinmg the correct number of fields for values. You said you had an auto incremental key so there should be:

values ('', '".$field1."'

 

It is much better writing:

 

SET name='".$name."', address='".$address."' etc

[MFHJoe]

Remove the single quotes around your php variables!

Usually that'd be a good idea but not here. $sql is actually in double quotes which allow variables to be used inside them. Without the single quotes, $sql would look like:

$sql = "insert into clients values (1, Firstname, Lastname, etc....

Which would result in a mySQL error as the values need to be denoted by single quotes.

 

[JonnoTheDev]

Just a style preference I have

values ('".$client_id."', '".$client_fname."'

 

I hate this sort of query using the "VALUE" syntax, especially if you modify the structure of your table and forget about queries written like this. SET is much more efficient as the order of fields makes no difference.

[prcollin]

get error check sal syntax near ') at line one

values ($client_id, $client_fname

 

 

get duplicate entry still and no values input.  

values ('".$client_id."', '".$client_fname."'

 

number of fields and name of fields are correct

[prcollin]

now i get this error

 

Error: Duplicate entry '' for key 2

[JonnoTheDev]

My code sinppets are not complete and wont work. They are just examples to get you on the right track.

 

If you have a duplicate key then get rid of the bad value and make sure that you definately have an auto incremental field for your primary key an no other uniques indexes on fields taht may product the error.

 

If there are values in your variables. I assume that you have tested this then this code will work (I have assumed the names of your database fields so make sure that you change them in the following SQL):

 

$sql = "INSERT INTO clients SET client_id='".$client_id."', client_fname='".$client_fname."', client_lname='".$client_lname."', client_address='".$client_address."', client_city='".$client_city."', client_state='".$client_state."', client_zipcode='".$client_zipcode."', client_phone='".$client_phone."', client_cphone='".$client_cphone."', client_email='".$client_email."', client_website='".$client_website."', client_notes='".$client_notes."')";

[prcollin]

My code sinppets are not complete and wont work. They are just examples to get you on the right track.

 

If you have a duplicate key then get rid of the bad value and make sure that you definately have an auto incremental field for your primary key an no other uniques indexes on fields taht may product the error.

 

If there are values in your variables. I assume that you have tested this then this code will work (I have assumed the names of your database fields so make sure that you change them in the following SQL):

 

$sql = "INSERT INTO clients SET client_id='".$client_id."', client_fname='".$client_fname."', client_lname='".$client_lname."', client_address='".$client_address."', client_city='".$client_city."', client_state='".$client_state."', client_zipcode='".$client_zipcode."', client_phone='".$client_phone."', client_cphone='".$client_cphone."', client_email='".$client_email."', client_website='".$client_website."', client_notes='".$client_notes."')";

 

no matter how i change the format around, mess with the db it says duplicate entry for key 2

any ideas still?

[blueman378]

do a database structure export and post it here (in code tags) and we may be able bo help