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:
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[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
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[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
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[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
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[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
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Edit this bug report at http://bugs.php.net/?id=16949&edit=1
