can anyone help me make this work?

[ejwf]

Hi

I am inserting some results from a user input form into mysql database.
The code I have works up to a point but 2 of the values aren't getting put into the database.
By testing the output of certain variables I think I know where the error is, but I cannot see what the problem is - can anyone help......

IF ($_POST[AuthorityLevel] == "Concession User")
{$AuthorityLevel = "2";}
ELSEIF ($_POST[AuthorityLevel] == "Client Administrator")
{$AuthorityLevel = "3";}
ELSE 
{$AuthorityLevel = "4";}

$Client = $_POST[client];
$Section = $_POST[section];
$CName = $_POST[new_name];
$CPhone = $_POST[new_phone];
$CFax = $_POST[new_fax];
$CLogIn = $_POST[new_username];
$CPass = $_POST[new_password1];
$CEmail = $_POST[new_email];

// find ClientId
$query = "SELECT `ClientId` FROM `tblClient` WHERE `CoName` = '$Client'";
$result = mysql_query($query)
	or die(mysql_error());
while($row =  mysql_fetch_array($result))
$ClientId = $row["ClientId"];

// find SectionId
$query = "SELECT SectionId FROM tblSection 
WHERE SectionName = '$Section'";
$result1 = mysql_query($query)
or die(mysql_error());
while($row1 = mysql_fetch_array($result1))
$SectionId = $row1["SectionId"];

// find ConcesRef
$query = "SELECT `ConcesRef` FROM `tblUser` 
WHERE `CLogIn` = '$username'"; 
$result2 = mysql_query($query) 
   or die(mysql_error());
while($row2 = mysql_fetch_array($result2))
$ConcesRef = $row2["ConcesRef"]; 

//test these variables
echo $ClientId;
echo $SectionId;
echo $ConcesRef;

IF ($_POST[new_password1] != $_POST[new_password2])
{ 
echo ' The passwords you gave were not the same, please click the back arrow on your browser and try again.';
}
else
{

$sql = "INSERT INTO tblUser (ConcesRef, SectionRef, ClientRef, CName, CPhone, CFax, CLogIn, CPass, CEmail, AuthorityLevel)
VALUES ('$ConcesRef', '$SectionId', '$ClientId', '$CName', '$CPhone', '$CFax', '$CLogIn', '$CPass', '$CEmail', '$AuthorityLevel')";

$result = mysql_query($sql)
or die(mysql_error());

echo '<div align="center">
  <p>&nbsp;</p>
  <p><font face="Arial, Helvetica, sans-serif"><b><font size="3" color="#FF0000">
  The record has been added.
    </font></b></font></p>
</div>';
}

Out of the variables I am INSERTING, only $SectionId and $ClientId don't work.
I know that $Client and $Section work OK because when I echo them out they come up with the expected results.
So I am guessing that any error lies in the $query sections of the code.
From the echo test of the variables $ClientId and $SectionId don't output anything.
Interestingly the $ConcesRef value goes in OK based on the same query structure........
Can anyone see where I have gone wrong - thanks

[devinemke]

Originally posted by ejwf
...I know that $Client and $Section work OK because when I echo them out they come up with the expected results....

...From the echo test of the variables $ClientId and $SectionId don't output anything....

huh??? don't these two statements contradict each other? are those variable set or not? what error (if any) is mysql spitting back? you may want to try this INSERT syntax:

$sql = "INSERT INTO tblUser SET 
ConcesRef = '$ConcesRef',
SectionRef = '$SectionId',
ClientRef = '$ClientId',
CName = '$CName',
CPhone = '$CPhone',
CFax = '$CFax',
CLogIn = '$CLogIn',
CPass = '$CPass',
CEmail = '$CEmail',
AuthorityLevel =  '$AuthorityLevel'";

[ejwf]

thanks for having a look at this for me........
Firstly to explain my code
$Client is a company name like acme trading inc.
$ClientId is the id number this company is given within the database and that is used to cross reference information between tables.
Same with $Section and $SectionId.
It is the Id variables that do not output anything
I don't get any mysql error message either, anywhere in my code.

I have tried your INSERT code, but still not inserting the ClientId and SectionId information

any ideas?