Question

I try next script:

// Insert data into mysql 

$qry="INSERT INTO $tbl_name1 (ID, REFERENCE, CODE, NAME) VALUES (UUID(), '$REFERENCE', '$CODE', '$NAME')";
$result=mysql_query($qry);

$qry2="INSERT INTO $tbl_name2 (PRODUCT) VALUES ('$ID')"; <--- Here is a problem
$result=mysql_query($qry2)

I do not know how two insert the same UUID in two tables simultanoiusly. Please help me!
I will appreciate much your support!

DONE!!!
THE WORKING SCRIPT:

$q = "SELECT UUID() AS uid";
$res = mysql_query($q) or die('q error: '.mysql_error());
$row = mysql_fetch_assoc($res);

// Insert data into mysql 
$qry="INSERT INTO $tbl_name1 (ID, REFERENCE, CODE, NAME) VALUES ('".$row['uid']."', '$REFERENCE', '$CODE', '$NAME')";
$result=mysql_query($qry) or die('err 034r '.mysql_error());

$qry2="INSERT INTO $tbl_name2 (PRODUCT) VALUES ('".$row['uid']."')";
$result=mysql_query($qry2) or die('gg2345  '.mysql_error());

Answer 1

Just do SELECT UUID() before you send the INSERTs and put the values into the statements in PHP. Something like this (untested):

$result = mysql_query("SELECT UUID() AS UUID") or die('SQL error: ' . mysql_error());
$row = mysql_fetch_assoc($result);
$UUID = $row["UUID"];

$qry="INSERT INTO $tbl_name1 (ID, REFERENCE, CODE, NAME) VALUES ('$UUID', '$REFERENCE',   '$CODE', '$NAME')";
$result=mysql_query($qry);

$qry2="INSERT INTO $tbl_name2 (PRODUCT) VALUES ('$UUID ')"; <--- Here is a problem
$result=mysql_query($qry2)

Another way would be the use of a user-defined variable (see SQL Fiddle):

SET @UUID = (SELECT UUID() AS UUID);
INSERT INTO test1 VALUES(@UUID, "foo");
INSERT INTO test1 VALUES(@UUID, "bar");