PHP 4.0 Bug #5614 Updated: Access Violation

ID: 5614 Updated by: fmk Reported By: mbeers@udeco.com Status: Closed Bug Type: MSSQL related Assigned To: Comments: This is fixed in CVS and it will be available from php 4.0.4 - Frank Previous Comments: --------------------------------------------------------------------------- [2000-09-06 18:01:43] sniper@php.net User feedback: ------------ Yes, this still happens in 4.0.2 using the provided win32 binaries. --------------------------------------------------------------------------- [2000-09-04 20:44:22] sniper@php.net Is this still happening when using php4.0.2 ? --Jani --------------------------------------------------------------------------- [2000-07-14 23:10:26] mbeers@udeco.com I am using the Win32 binaries, downloaded on July 14. The error I get is: PHP has encountered an Access Violation at 0150229E. It occurs when I use the mssql_close() function without a link identifier. This is the code which causes the break: <? mssql_connect($SERVER_NAME, $USER_ID, $DB_PASSWORD) or die("unable to connect to server"); mssql_select_db($DB_NAME) or die("unable to connect to db"); $result = mssql_query("select field1, field2 from TestPHP where field1 > 1"); print "num rows=" . mssql_num_rows($result) . "<br>"; while ($data = mssql_fetch_row($result)) { print "field1=$data[0], field2=$data[1]<br>"; } mssql_close(); ?> If I store the link identifier when I call the connect statement ($link = mssql_connect), and pass that link identifier in when I call close (mssql_close($link), all is well. Finally, it happens with any sql statement I execute. Thank you for your hard work. --------------------------------------------------------------------------- Full Bug description available at: http://bugs.php.net/?id=5614