mysql_error()
returns a string depending on what is duplicate
error number is 1062
is there a way how I can reference this error number?
something like if mysql_error()=='1062'
or am I thinkinking in the wrong direction

    You would use [man]mysql_errno/man. That said, have you considered using the MySQLi extension or the PDO extension instead?

       
if(!$sql) 
        { 
        echo "Error-No. " . mysql_errno()." - " .mysql_error();
        }

      for some reason I dont get the number
      I get the error description ok

      All I want to avoid is posting data twice.
      If the user is using back and forward buttons on the browser

        Unfortunately, if you are not getting the error number with mysql_errno() when you should, then it implies that you have discovered a bug. What version of PHP and MySQL are you using, and on which operating system?

          Apache 2.2.14
          PHP 5.3.1
          MySql mysql Ver 14.14 Distrib 5.1.42, for Win32 (ia32)
          Win XP Pro SP3

            Do some more tests
            with
            echo mysql_errno();

            I find it not very probable there is such a bug in any of the common mysql functions.
            Because they are used by million and million users and tester in the world.
            And even long before PHP 5.3.1 was released, you can count many many used
            the pre-release from SVN, development version.

              ok I did a little more testing
              error number is displayed once (only the first time) I can do my error handling based on that.
              however if I dont do the proper error handling and just disregard everything and move with the browsers back and forward buttons the error number disappears and only the error warning text is displayed.

                Note that this function only returns the error code from the most recently executed MySQL function (not including mysql_error() and mysql_errno()),
                so if you want to use it,
                make sure you check the value before calling another MySQL function.

                Return Values

                Returns the error number from the last MySQL function,
                or 0 (zero) if no error occurred.

                So, this function [man]mysql_errno[/man] should be put right after
                the mysql function you want to catch an error from.
                And I would use it preferably with an exit(),
                to stop further execution and messing up my database

                For example:

                <?php
// more here

$result = mysql_query("SELECT * FROM nonexistenttable");
      // test eventual error in this query
if( mysql_errno() ) exit( mysql_errno() .' :: '. mysql_error() );

// more here
?>

                  I only want to catch the 1062 duplicate entry one
                  then send the user back to the page he came from
                  there another query picks up the next number automatically
                  and the form can be resubmitted

                  $sql=mysql_query( "INSERT INTO......
 if (mysql_errno($sql)==1062)
        {
        echo "Somebody was faster."  . header("Refresh: 3; URL=pageyoucamefrom.php");//    
                  
        }
        elseif(!$sql)       
                  
        { 
        echo "Error-No. " . mysql_errno()." - " .mysql_error();

                  but there is something wrong in the statement only the elseif is executed

                    You should not put the $result ($sql) in mysql_errno/mysql_error

                    mysql_errno ([ resource $link_identifier ] )

                    Just put nothing.
                    There can be cases when you need to put
                    mysql_errno($link);
                    where $link is your mysql connection.

                    I would do something like this, for your idea:

                    $sql=mysql_query( "INSERT INTO......");

$eno = mysql_errno();
if($eno==1062)
{
    echo "Somebody was faster."  . header("Refresh: 3; URL=pageyoucamefrom.php");
	exit();   
                    
}
elseif($eno!=0)       
                    
{
    echo "Error-No. " . $eno . " - " .mysql_error(); 
}

// else do nothing, errno=0
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