（更新）php在注册的同时创建两个东西？

Question

I am going to create a register page to do the homework,and i can't found that simple answer of it.Here is the problem. i want to insert the new loginid and password in the a table call customerlist and create the new table named of the new loginid.

Here is the code:

$loginid = $_POST['loginid'];
$pw = $_POST['pw'];
$pw2 = $_POST['pw2'];

if($loginid != null && $pw != null && $pw2 != null && $pw == $pw2)
{ $sql = "insert into customerlist(LoginID,Password) values ('$loginid','$pw')";
  $sql="create table {$loginid} 
                (
  TradeID(50) VARCHAR(20),
  Trademode VARCHAR(10),
  Tradedate data,
  MTRdate   VARCHAR(50) ,
  Postaddress VARCHAR(255),
  Tel VARCHAR(15),
  Receivables int,
  Status VARCHAR(10)
                 );";



        if(mysql_query($sql))
        {
                echo 'success!';
                echo '<meta http-equiv=REFRESH CONTENT=2;url=index.php>';
        }
        else
        {
                echo 'fail!';
                echo '<meta http-equiv=REFRESH CONTENT=2;url=index.php>';
        }
}
else
{
        echo 'get out!';
        echo '<meta http-equiv=REFRESH CONTENT=2;url=index.php>';
}
?>

update

<?php

$con = mysql_connect("localhost","root","password");
mysql_select_db("babytradeapps");

$loginid = $_POST['loginid'];
$pw = $_POST['pw'];
$pw2 = $_POST['pw2'];
$name = $_POST['name'];
$telephone = $_POST['telephone'];
$idcard = $_POST['idcard'];
$email = $_POST['email'];
$bankac = $_POST['bankac'];
$plan = $_POST['plan'];
$date = $_POST['date'];

if($loginid != null && $pw != null && $pw2 != null && $pw == $pw2)
{ $sql = "insert into customerlist(LoginID,Password,Name,Phone,IDCARDNO,Email,BankACNO,Plan,Date) values ('$loginid','$pw','$name','$telephone','$idcard','$email','bankac','plan','date')";
  $sql2 = "insert into overall(LoginID,TradeID,Trademode,Tradedate,MTRdate,Postaddress,Tel,Receivables,Status) values ('$loginid','example','example','example','example','example','example','example','example')";
        if(mysql_query($sql)&& mysql_query($sql2))
        {
                echo 'success!';
                echo '<meta http-equiv=REFRESH CONTENT=2;url=index.php>';
        }
        else
        {
                echo 'fail!';
                echo '<meta http-equiv=REFRESH CONTENT=2;url=index.php>';
        }
}
else
{
        echo 'get out!';
        echo '<meta http-equiv=REFRESH CONTENT=2;url=index.php>';
}
?>

i have changed, but why the result is fail??? whats wrong of it?

Answer 1

You are overwriting your $sql variable with the insert statement right in the next line with the create table statement, so only the second command will be executed. Use a different variable for this and make two queries.

 $sql = "insert into customerlist(LoginID,Password) values ('$loginid','$pw')";
 $sql2="create table {$loginid} 
                 (
   TradeID(50) VARCHAR(20),
   Trademode VARCHAR(10),
   Tradedate data,
   MTRdate   VARCHAR(50) ,
   Postaddress VARCHAR(255),
   Tel VARCHAR(15),
   Receivables int,
   Status VARCHAR(10)
                  );";
 if(mysql_query($sql) && mysql_query($sql2))

Also heed the warning of zerkms, creating so many tables is not a good database layout.

Another note: You are using mysql_* functions that are becoming deprecated and will be removed from future versions of PHP. use mysqli_* or PDO instead.