Trying to check for duplicates before entering data.

Keaner

Ok, I'm not sure if any of this is right but for some reason it keeps saying:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/html/enter.php on line 32

All I'm trying to do is check for a duplicate serial number. Can anybody tell me if I'm going about it the wrong way or if its my poor syntax?

<?
include("header.inc");
include("menu.inc");
include ("details.inc");

//Post all values into variables from the enterForm.php form
$serial = $_POST['serial'];
$asset = $_POST['asset'];
$name = $_POST['name'];
$location = $_POST['location'];
$description = $_POST['description'];
$doe = date("F j, Y");
$warranty = $_POST['warranty'];

//Check to see if the user left the item name blank
if($name!="") {

//If fields are left blank, fill them with hyphons
   if($serial=="") {
      $serial="-";
   }
   if($asset=="") {
      $asset="-";
   }
   if($description=="") {
      $description="-";
   }
   if($warranty=="") {
      $warranty="-";
   }

   While ($rs = mysql_fetch_array($result)) {
      if ($rs['serialNum']==$serial) {
         printf("The serial number entered already exists");
      }
   }

   if ($rs['serialNum']!=$serial) {
    //Insert the data into the database and display to the user that this has been done
   $query = "INSERT INTO assets SET serialNum = '$serial',assetNum = '$asset',itemName = '$name',location = '$location',description = '$description',entryDate = '$doe',warranty = '$warranty'";
   mysql_query($query);
   }
   ?>

   <TR ALIGN="CENTER">
      <TD>The data has been entered successfully</TD>
   </TR>
<?
}

//If the user left the item name blank, infor the user it is required to complete this field
Elseif($name=="") {
   ?>

   <TR ALIGN="CENTER">
      <TD>You must enter a name and location</TD>
   </TR>

<?
}

include("footer.inc");
?>

YappyDog

Where is your query to check for duplicates?

Where are you putting anything into $result?

Keaner

Here is how I'm trying to find the duplicate:

While ($rs = mysql_fetch_array($result)) { 
      if ($rs['serialNum']==$serial) { 
         printf("The serial number entered already exists"); 
      } 
   } 

   if ($rs['serialNum']!=$serial) { 
    //Insert the data into the database and display to the user that this has been done 
   $query = "INSERT INTO assets SET serialNum = '$serial',assetNum = '$asset',itemName = '$name',location = '$location',description = '$description',entryDate = '$doe',warranty = '$warranty'"; 
   mysql_query($query); 
   } 
   ?>

Now I don't know for sure but in throey shouldn't this loop through the database until it finds the duplicate then go out of the loop and bypass the INSERT statement. If it doesn't find a duplicate it will process the INSERT statement. But for some reason I keep getting a fetch array error.

YappyDog

It will only loop if there is something in $result, which would normally be put there by some sort of a SELECT statement that looks for the duplicates.

Keaner

Ok, so how do i go about makeing a select statement that checks for duplicants?

YappyDog

Look for the stuff you are trying to insert. If you find a match, you have a duplicate.

SELECT serialNum FROM assets
WHERE serialNum = '$serial'
AND assetNum = '$asset'
AND itemName = '$name'
AND location = '$location'
AND description = '$description'
AND entryDate = '$doe'
AND warranty = '$warranty'

Keaner

Blah, you have me lost with this. Right now my variables are posted as the variables you listed. So will what you typed automatically check the posted vaiables with the database records for me?

Then can I just go ahead with those same variables and insert them?

YappyDog

You need to look at the manuals in the section about SELECTing information from a database.

http://us2.php.net/manual/en/ref.mysql.php

<?php
/ Connecting, selecting database /
$link = mysql_connect("mysql_host", "mysql_user", "mysql_password")
or die("Could not connect : " . mysql_error());
print "Connected successfully";
mysql_select_db("my_database") or die("Could not select database");

/* Performing SQL query */
$query = "SELECT * FROM my_table";
$result = mysql_query($query) or die("Query failed : " . mysql_error());

/* Printing results in HTML */
print "<table>\n";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
    print "\t<tr>\n";
    foreach ($line as $col_value) {
        print "\t\t<td>$col_value</td>\n";
    }
    print "\t</tr>\n";
}
print "</table>\n";

/* Free resultset */
mysql_free_result($result);

/* Closing connection */
mysql_close($link);