I have my php code that will hold and insert date to mysql:
from an input textbox
$id2 = $_POST['id']; Random Number Generator How can I generate it always on 5 digit?
$random = rand(); Insert code:
$sql3="Insert into warranty(serial_no,a_id) values ('$random','$id2')"; $result3=mysql_query($sql3); How can I deny insert when $id2 doesn't have a value?
Just use empty() function to determine, if given variable has value or not (manual page). Like this:
if (!empty($id2)) { $sql3="Insert into warranty(serial_no,a_id) values ('$random','$id2')"; $result3=mysql_query($sql3); } But passing $_POST value directly to SQL query is huge* security issue (SQL Injection to be clear), os use at least mysql_real_escape_string():
$sql3="Insert into warranty(serial_no,a_id) values ('$random','" . mysql_real_escape_string($id2) . "')"; For more info about this, read How can I prevent SQL injection in PHP?.
EDIT: At first, I miss your second question, so moved from comments - to generate 5 digits random number, use simple:
$fiveDigitsRand = rand(10000, 99999); 
rand(10000, 99999);put below code and remove existing code.You don't have need of $id2 variable if you are getting value using $_POST method. You can do it directly by following code.
$random = rand(); if(isset($_POST['id'])) { $sql3="Insert into warranty(serial_no,a_id) values ('$random','".$_POST['id']."')"; $result3=mysql_query($sql3); } else { echo "No ID Found"; //you can do any action here. } $random = rand(10000,99999);
if(!empty(trim($id2)))
{ $sql3="Insert into warranty(serial_no,a_id) values ('$random','$id2')"; $result3=mysql_query($sql3); } 
if(isset($id2)) { $sql3="Insert into warranty(serial_no,a_id) values ('$random','$id2')"; $result3=mysql_query($sql3); } 