[SOLVED] Trouble Calling From Database

[refiking]

Here is the other insert code:

 

$store = "INSERT INTO Records(analyst_id , record_id , b_first_name , b_last_name , c_first_name ,

c_last_name , address, city , state , zip , loan_amount , rate , cap , mo_pay , new_rate , new_mopay ,

pay_change , change_date , ltv , lender , telephone , ipaddy , timestamp)

VALUES('$analyst_id','$record_id','$bfn','$name','$cfn','$cln','$add','$city','$state','$zip','$la','$iir',

'$adjc','$mpay','$nr','$nmpay','$change','$cpdate','$ltv','$lender','$tel','$ipaddy','$dun')";

 

[Yesideez]

You've got a lot of fields there!

 

Can you check in the `Records` table and check if the records ma123, ma124, and ma125 aren't already there? That could be one reason why the INSERT query is failing. I've had problems with inserting before where I've not been specifying enough fields but I think it depends on how the table is set up but I really am not sure.

[refiking]

I only have 2 set up so far. ma1 and ma2.

[Yesideez]

I'm thinking you may have to add more data into that record and not just the analyst_id. From what I can see `analyst_id` is the main index and you're basically adding another record with no data!

[refiking]

There are more fields than analyst_id.

 

I think there are 5 in all.

[refiking]

bump

[refiking]

Got it working now!  Here's what I did:

 

$que = mysql_query("SELECT `analyst_id` FROM `Analysts` WHERE username = '$username'");

while($row = mysql_fetch_assoc($que)){

$analyst_id = $row['analyst_id'];

echo $analyst_id;

}

 

Then, I did this:

 

$store = "INSERT INTO Records(analyst_id , record_id , b_first_name , b_last_name , c_first_name ,

c_last_name , address, city , state , zip , loan_amount , rate , cap , mo_pay , new_rate , new_mopay ,

pay_change , change_date , ltv , lender , telephone , ipaddy , timestamp)

VALUES('$analyst_id','$record_id','$bfn','$name','$cfn','$cln','$add','$city','$state','$zip','$la','$iir',

'$adjc','$mpay','$nr','$nmpay','$change','$cpdate','$ltv','$lender','$tel','$ipaddy','$dun')";