Problem... Please help.

I keep getting the Error at line 25 with this code.

<?
$connection = mysql_connect("localhost","root","") or die("Couldn't connect to DB");
mysql_select_db("abraham",$connection) or die("Couldn't select DB");
$select = "SELECT username, password FROM users WHERE username='$username' AND password='$password'";
$getnum = mysql_query($select,$connection);
$number = mysql_num_rows($getnum);
if($number == 1)
{
$decide = "SELECT * FROM info WHERE id='$id'";
$information = mysql_query($decide);
$infoarray = mysql_fetch_array($information);
$ssn1 = $infoarray["ssn"];
$LastName1 = $infoarray["LastName"];
$FirstName1 = $infoarray["FirstName"];
$Address2 = $infoarray["address1"];
$City1 = $infoarray["city"];
$State1 = $infoarray["state"];
$zip1 = $infoarray["zip"];
$hphone1 = $infoarray["hphone"];
$eligible1 = $infoarray["eligible"];
$felony1 = $infoarray["felony"];
if((!$ssn1) && (!$LastName1) && (!$FirstName1) && (!$Address2) && (!$City1) && (!$State1) && (!$zip1) && (!$hphone1) && (!$eligible1) && (!$felony1))
{
$infoquery = "INSERT INTO info (ssn, LastName, FirstName, initial, othername, email, address1, city, address2, state, zip, hphone, aphone1, aphone2, eligible, emergencynum, felony, felexplain, referred, refname, refcomment) VALUES ('$ssn', '$lastname', '$firstname', '$initial', '$othername', '$email', '$address1', '$city', '$address2', '$state', '$zip', '$hphone', '$aphone1', '$aphone2', '$eligible', '$emergencynum', '$felony', '$felexplain', '$referred', '$refname', '$refcomment') WHERE id='$id'";
mysql_query($infoquery,$connection) or die("Couldn't Insert At Line 25!");
echo $infoquery;
}
else
{
$infoquery="UPDATE info SET ssn='$ssn', LastName='$lastname', FirstName='$firstname', initial='$initial', othername='$othername', email='$email', address1='$address1', city='$city', address2='$address2', state='$state', zip='$zip', hphone='$hphone', aphone1='$aphone1', aphone2='$aphone', elgible='$eligible', emergencynum='$emergencynum', felony='$felony', felexplain='$felonyexp' WHERE username='$username'";
mysql_query($infoquery) or die("Couldn't update at line 30");
}

}
else
{
echo "You are not an authorized user. Please <a href="main.html">Click Here</A> to login.";
}
?>

Thanks,


------------------
Hartmann

OK. Is that a question, or are you just telling us so we know?

use this in your die statement to find out what is going wrong

or die("Error: " . mysql_error());

That'll give you a more detailed error message so you can figure out what's going on. I'm guessing it's something concerning quotes in the variables you're trying to insert. if so, look into addslashes().

Also, you might want to look into the REPLACE command for MySQL. It'll make new row or replace the data in the row if it already exists. That can simplify your page a lot.

Also, look into the list() function
<BLOCKQUOTE><font size="1" face="Verdana,Arial,Helvetica">code:</font><HR><pre>
$query = "SELECT id,ssn,name FROM your_table";
$result = mysql_query($query);
if(list($id,$ssn,$name) = mysql_fetch_row($result))
{
echo "Id = $id<BR>n";
echo "SSN = $ssn<BR>n";
echo "Name = $name<BR>n";
}
else
{
echo "Nothing found<BR>n";
}
[/code]
Use something like that instead of your long list of $var = $info_array["whatever"]; You should be able to get rid of all of that by using the REPLACE command, though.

Have fun.

Oh...one more thing. If you post this in the PHP forum you'll get a faster answers, probably....

---John Holmes...

------------------
*************************************************************
* The manual can probably answer 90% of your questions...
*
* PHP Manual. www.php.net/manual
* MySQL Manual: www.mysql.com/documentation/mysql/bychapter
*************************************************************

Ok I got the problem fixed... But how would I use list to make $variable=$array["info"]; throughout that many variables?
Thanks,

------------------
Hartmann

[This message has been edited by Hartmann (edited November 25, 2000).]