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math equation in php

Posted by phillipmulvaney (phillipmulvaney), 29 April 2004
I am working on a php page which extracts "quantity" from database "a".
I then want to subtract 1 from quantity and then send the new quantity back to the database.


$db_name = "harro";
$table_name = "a";
$connection = @mysql_connect("localhost", "harro", "eMHiGiTU") or
die(mysql_error());
$db = @mysql_select_db($db_name, $connection) or die(mysql_error());

// select quantity from database a
$sql = "SELECT quantity FROM $table_name WHERE barcode = '$_POST[barcode]'";

$result = @mysql_query($sql,$connection) or die(mysql_error());

$num = mysql_num_rows($result);
$sum = 0;

// want to subtract 1 from quantity and post back to database

quantiy --;
     "update a set quantity ='$sum' where   barcode = '$_POST[barcode]'";
     

}



Posted by admin (Graham Ellis), 29 April 2004
Perhaps I'm missing something here, but I think your query might be as simple as:

// mysql_connect as in your example
// mysql_select_db as in your example
mysql_query("update a set quanitiy = quantity - 1 where barcode = '$_POST[barcode]'");

Suggestions:

a) Try out the SQL command to see if it does what you want using the mysql client before you wrap it up in PHP.  That way, you're developing the code in two easy steps

b) Do NOT put the @ in front of the MySQL calls until your code is tested and working - as it stands, you're supressing error and warning messages which will be very useful in helping you find out where problems lie

Posted by phillipmulvaney (phillipmulvaney), 29 April 2004
cheers that was it.
think im just tired have been working on a website for days and had tried everything for the last 2 hours and nothing would work.
thanks again,
phillip



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