insert problem during a for loop

Turkey

OK, here's my problem - the values in this for loop are positively there in each loop (had echo on to test) - but PHP (or mysql) refuses to insert anything after the first loop.

$pkgresult = mysql_query("select from packages order by pkg_name");
if (!$result)
echo "<br>could not fetch the packages and their prices.";
else
{
for ($count = 0; $count < 5; $count++)
{
$row = mysql_fetch_row($pkgresult);
$item = $row[0];
$pkgvar = $row[5];
if ($pkgvar == "fxtra")
$price = $row[1] $howmany;
else
$price = $row[1];

       if ($item = substr($$pkgvar,0))
         {
           $result =  mysql_query("insert into contract_items  (contr_num, merch_id, contr_yr, contr_mo, contr_day, contr_item, item_price, rep_id) values ('$contrnum',  '$merchid', '$year', '$month', '$day', '$item', '$price', '$repid') ");

if ($inserter)
echo "<br>updated ok.";
else
echo "<br>uhoh! - did not update! substr($$pkgvar,0)";
}
}
}

So, can anyone out there tell me what is wrong here? Thanks.

Turkey

tomhath

Did you mean to make an assignment here? Or did you mean == ?

Where does $inserter get set?

Why not check mysql_error to see what want wrong?

...
if ($item = substr($$pkgvar,0)) 
{ 
$result = mysql_query("insert into contract_items (contr_num, merch_id, contr_yr, contr_mo, contr_day, contr_item, item_price, rep_id) values ('$contrnum', '$merchid', '$year', '$month', '$day', '$item', '$price', '$repid') "); 
if ($inserter) 
...

Turkey

tomhath,

thanks for your reply --

I have tried coding that utilizes assignment or otherwise. I have tried skinning this cat from several directions. The variable values are echoed back to me when the if condition is true -- and the values properly do not echo when the if condition is not met.

I have even tested using stated values instead of variables within repititious code blocks. And it is only with the first "if" condition being true that the insert will work on me.

The crux of the problem lies in the failure to insert any records beyond the first insert. This really has me baffled.

ahundiak

Read tomhath's reply most closely.
What does mysql_error() say?

Put this code at the top of your page:

function db_query($sql)
{
  $result = mysql_query($sql);
  if (mysql_errno())
  {
    die(mysql_error() . '<br />' . $sql);
  }
  return $result;
}

Then replace all mysql_query calls with db_query. It will save endless hours of query debugging.

Turkey

ahundiak (and tomhath),

I inserted the following code:

echo mysql_errno();
echo "<br>";
echo mysql_error();

On each if condition that was true, the following was printed.
1062
Duplicate entry '85' for key 1 -- I had not realized that I had assigned a key field to this table.

regarding "Then replace all mysql_query calls with db_query. It will save endless hours of query debugging." -- I already have 2-300 files using mysql_query..

Thanks tons though, - problem solved.

Turkey
BTW - saw not that we could mark a thread as completed - but can' find where I do that.