Mysql/PHP query

Anon

I have two tables that only contain 1 field each that I need. I need to have a form that a number is entered in and when submitted it first verifies that the number exists in the first table, for verification, and then if the same number exists in the second table they are redirected to an access denied page.

Anon

Assuming that the variable that carries a number to be entered in the form is $number:

$query = "select * from table1, table2 where table1.Number = '$number' and table1.Number = table2.Number";
$result = mysql_query($query);
$count = mysql_num_rows($result);
if($count > 0){
You can deny access:
}
else {
what do you want
}

May this help you?

Anon

yes thanks,

however, If the number is in table1 it should check to see if it is also in table2. If it is not in table2 they are authenticated and redirected to a new page, but if it is in table2 than the are declined and redirected to a declined page

Anon

if($count > 0){
You can deny access:
}
else {
what do you want
}

So then change this value that Solomon gave you and change it to.

if($count == 0){
header("location:http://newurl");
}
else {
header("location:http://declinedurl");
}

Anon

I still think I have the query wrong.

there is a member number.
there are two tables member and overdue.
table member is a list of all member numbers
table overdue is a list of all members with an overdue status (thus being declined access)
when they submit number the first check is: is this number a valid member number. if not please reenter number, If so, is this number on the overdue list. If so access denied, if not access granted

Anon

Use a LEFT JOIN.

If the count is 0 the row is not in Table1.

If the count is 1 but the column from Table2 is null the row is in Table1 but not in Table2.

If the column from Table2 is not null, the number is in both tables.

$query = "select table2.mycolumn from table1 left join table2 where table1.Number = '$number' and table1.Number = table2.Number";

Anon

LEFT JOIN should use ON in the join condition instead of WHERE

$query = "select table2.mycolumn from table1 left join table2 ON table1.Number = table2.Number where table1.Number = '$number' ";

Anon

Thanks everyone!