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bigdoggy59

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  1. Until this situation occurred, I'd have said the same thing. What's odd is that if you try to go to the page directly and don't log in, you get a 403-Access Denied. When the page has been sent as e-mail, it still asks for the password, but then shows the page whether a login is entered or not.
  2. Environment: Web site hosted on a Linux box. Using CPanel's File Manager to password-protect a folder for our Sales group. Problem: One of our sales guys logged into the Sales area, opened a report, then clicked "send as e-mail" to a customer. The customer receives the e-mail, gets a popup asking for username and password (as it should be). She clicks "Cancel" then gets to see the report anyway (not good). She tries again and intentionally fails the log in - and gets to see the report anyway (even more not good). Then I get notified. Question: Is there any way in PHP to keep a password-protected page from being accessed from an e-mail? Do I need to ignore the File Manager and build (or use) a PHP authentication class? Honestly, I didn't realize this was a security risk until now and would like to plug it before someone decides to send something more dangerous than a price sheet.
  3. This is the strangest thing I've ever seen... Building an HTML form from a query I have: echo $nsn; echo "<input type='hidden' name='NSN[$idx]' value='$nsn'/>"; echo "</td>\n"; Looking at the source in the browser I see: <input type='hidden' name='NSN[0]' value='8465014652096'/> <input type='hidden' name='NSN[1]' value='GSA-700'/> On the next page, a print_r($NSN) returns: Array ( [0] => GSA-700 [1] => 8465014652096 ) 1 Note that the index numbers=>values have been re-ordered. Why is this happening and how do I stop it? None of the other arrays in the app are doing this, only this one. Thanks
  4. I took this right out of the MySQL Manual. Hope it helps. If there is no matching record for the right table in the ON or USING part in a LEFT JOIN, a row with all columns set to NULL is used for the right table. You can use this fact to find records in a table that have no counterpart in another table: mysql> select table1.* from table1 LEFT JOIN table2 ON table1.id=table2.id where table2.id is NULL; This example finds all rows in table1 with an id value that is not present in table2 (that is, all rows in table1 with no corresponding row in table2). This assumes that table2.id is declared NOT NULL, of course. See section 12.5.5 How MySQL Optimizes LEFT JOIN and RIGHT JOIN.
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