Hi Leelakh,
Here is your answer.
For no 1
<?php
$connection = @mysql_connect("localhost" , "user_id" , "password") or die
("Unable To Connect To The Database");
$db_name = "database name";
$db_table = "table name";
$db = @mysql_select_db($db_name , $connection) or die ("Unable To Select
Database");
$chk_query = mysql_query("SELECT * FROM $db_table WHERE tname = '$_POST[name]'
AND taddress = '$_POST[address]' AND tcity = '$_POST[city]'", $connection ) or
die (mysql_error());
if (mysql_affected_rows() > '0')
{
echo "Sorry Already Have Infomation";
}
else
{
//here type your database insert query
}
?>
For no 2
//pls insert this script on your form page
<?php
session_start();
$_SESSION['security_no'] = rand();
echo $_SESSION['security_no'];
?>
after copying this make new fileld with name what ever you want.
//pls insert this script on the page where your databases insert query is
working or form submition page.
<?php
session_start();
if ($_POST['insert new created fileld name here'] == $_SESSION['security_no'])
{
enter here the database insert query
}
else
{
echo "pls re-enter the code";
}
?>
Cheers
Leelakh Ran <[EMAIL PROTECTED]> wrote: Dear
members,
I have one form to accept name, address and city of the user. Sometimes some
users sending same data several times. So I want to solve this peoblem by
following two ways.
1) comparing input data with database table. It is the fields tname,
taddress and tcity from table1 on db1 mysql database. If the user inputs
name+address+city = tname+taddress+tcity, it should not accept.
How I can write code for this comparison.
2) I want to check inputs by a security code. If each time form loads the
sucurity code should be different and if the user not feeding the code
correctly, form feedbacks should not accept.
How it can do?
If you can give answers for any of these query, it will be highly
appreceable.
With thanks and regards,
Leelakh Ran.
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