Bug #16949 Updated: mysql_query SELECT returns Resource ID
ID: 16949 Updated by: [EMAIL PROTECTED] Reported By: [EMAIL PROTECTED] Status: Bogus Bug Type: *General Issues Operating System: ? PHP Version: 4.1.2 New Comment: maybe because it's spelled mysql_QUERY and not mysql_EURY. I asked you kindly no to use the bug reporting form for support questions and instead move to the mailing lists. Here's the link again: http://www.php.net/support.php -daniel Previous Comments: [2002-05-01 18:48:26] [EMAIL PROTECTED] I didn't post here intending to ask for help, I was posting a bug because $sql = mysql_query(SELECT * FROM users); works just fine but $sql = mysql_eury(SELECT * FROM users WHERE name = '$user'); doesn't work I've tried this on completely diff servers and got the same result [2002-05-01 16:37:57] [EMAIL PROTECTED] I give some help anyway. Vipervirus, mysql_query returns the connection id. You must use mysql_fetch_array() to get the values. Please have a look at the example provided in the manual: http://www.php.net/manual/en/ref.mysql.php or http://www.php.net/manual/en/function.mysql-fetch-array.php which has some shorter examples. good luck. Next time please consult the PHP General mailinglist: http://www.php.net/support.php [2002-05-01 16:23:26] [EMAIL PROTECTED] The bug system is not the appropriate forum for asking support questions. For a list of a range of more appropriate places to ask for help using PHP, please visit http://www.php.net/support.php [2002-05-01 16:23:24] [EMAIL PROTECTED] I've been trying for a week now to fix this, finally gave up on it: $sql = mysql_query(SELECT pass FROM users WHERE name = '$user'); echo $sql . br; when ran, $sql = Resource ID #2 instead of the actual value, I couldn't get any help from #php on irc.gamesnet.net [2002-05-01 16:22:30] [EMAIL PROTECTED] I've been trying for a week now to fix this, finally gave up on it: $sql = mysql_query(SELECT pass FROM users WHERE name = '$user'); echo $sql[pass] . br; when ran, $sql = Resource ID #2 instead of the actual value, I couldn't get any help from #php on irc.gamesnet.net -- Edit this bug report at http://bugs.php.net/?id=16949edit=1
Bug #16949 Updated: mysql_query SELECT returns Resource ID
ID: 16949 Updated by: [EMAIL PROTECTED] Reported By: [EMAIL PROTECTED] -Status: Open +Status: Bogus Bug Type: *General Issues Operating System: ? PHP Version: 4.1.2 New Comment: The bug system is not the appropriate forum for asking support questions. For a list of a range of more appropriate places to ask for help using PHP, please visit http://www.php.net/support.php Previous Comments: [2002-05-01 16:23:24] [EMAIL PROTECTED] I've been trying for a week now to fix this, finally gave up on it: $sql = mysql_query(SELECT pass FROM users WHERE name = '$user'); echo $sql . br; when ran, $sql = Resource ID #2 instead of the actual value, I couldn't get any help from #php on irc.gamesnet.net [2002-05-01 16:22:30] [EMAIL PROTECTED] I've been trying for a week now to fix this, finally gave up on it: $sql = mysql_query(SELECT pass FROM users WHERE name = '$user'); echo $sql[pass] . br; when ran, $sql = Resource ID #2 instead of the actual value, I couldn't get any help from #php on irc.gamesnet.net -- Edit this bug report at http://bugs.php.net/?id=16949edit=1
Bug #16949 Updated: mysql_query SELECT returns Resource ID
ID: 16949 Updated by: [EMAIL PROTECTED] Reported By: [EMAIL PROTECTED] Status: Bogus Bug Type: *General Issues Operating System: ? PHP Version: 4.1.2 New Comment: I give some help anyway. Vipervirus, mysql_query returns the connection id. You must use mysql_fetch_array() to get the values. Please have a look at the example provided in the manual: http://www.php.net/manual/en/ref.mysql.php or http://www.php.net/manual/en/function.mysql-fetch-array.php which has some shorter examples. good luck. Next time please consult the PHP General mailinglist: http://www.php.net/support.php Previous Comments: [2002-05-01 16:23:26] [EMAIL PROTECTED] The bug system is not the appropriate forum for asking support questions. For a list of a range of more appropriate places to ask for help using PHP, please visit http://www.php.net/support.php [2002-05-01 16:23:24] [EMAIL PROTECTED] I've been trying for a week now to fix this, finally gave up on it: $sql = mysql_query(SELECT pass FROM users WHERE name = '$user'); echo $sql . br; when ran, $sql = Resource ID #2 instead of the actual value, I couldn't get any help from #php on irc.gamesnet.net [2002-05-01 16:22:30] [EMAIL PROTECTED] I've been trying for a week now to fix this, finally gave up on it: $sql = mysql_query(SELECT pass FROM users WHERE name = '$user'); echo $sql[pass] . br; when ran, $sql = Resource ID #2 instead of the actual value, I couldn't get any help from #php on irc.gamesnet.net -- Edit this bug report at http://bugs.php.net/?id=16949edit=1
Bug #16949 Updated: mysql_query SELECT returns Resource ID
ID: 16949 Updated by: [EMAIL PROTECTED] Reported By: [EMAIL PROTECTED] Status: Bogus Bug Type: *General Issues Operating System: ? PHP Version: 4.1.2 New Comment: I didn't post here intending to ask for help, I was posting a bug because $sql = mysql_query(SELECT * FROM users); works just fine but $sql = mysql_eury(SELECT * FROM users WHERE name = '$user'); doesn't work I've tried this on completely diff servers and got the same result Previous Comments: [2002-05-01 16:37:57] [EMAIL PROTECTED] I give some help anyway. Vipervirus, mysql_query returns the connection id. You must use mysql_fetch_array() to get the values. Please have a look at the example provided in the manual: http://www.php.net/manual/en/ref.mysql.php or http://www.php.net/manual/en/function.mysql-fetch-array.php which has some shorter examples. good luck. Next time please consult the PHP General mailinglist: http://www.php.net/support.php [2002-05-01 16:23:26] [EMAIL PROTECTED] The bug system is not the appropriate forum for asking support questions. For a list of a range of more appropriate places to ask for help using PHP, please visit http://www.php.net/support.php [2002-05-01 16:23:24] [EMAIL PROTECTED] I've been trying for a week now to fix this, finally gave up on it: $sql = mysql_query(SELECT pass FROM users WHERE name = '$user'); echo $sql . br; when ran, $sql = Resource ID #2 instead of the actual value, I couldn't get any help from #php on irc.gamesnet.net [2002-05-01 16:22:30] [EMAIL PROTECTED] I've been trying for a week now to fix this, finally gave up on it: $sql = mysql_query(SELECT pass FROM users WHERE name = '$user'); echo $sql[pass] . br; when ran, $sql = Resource ID #2 instead of the actual value, I couldn't get any help from #php on irc.gamesnet.net -- Edit this bug report at http://bugs.php.net/?id=16949edit=1
