You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 7 |
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 7 |
tyler.watkins |
Apr 10 2012, 03:24 PM
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#1
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Novice Group: Members Posts: 27 Joined: 6-March 12 Member No.: 16,654 |
<?php
$mysql_host = "********"; $mysql_database = "********"; $mysql_user = "*************"; $mysql_password = "*********"; $fname=$_POST['fname']; $lname=$_POST['lname']; $user=$_POST['username']; $password=$_POST['password']; $verify=$_POST['verify']; $email=$_POST['email']; $country=$_POST['country']; $month=$_POST['month']; $day=$_POST['day']; $year=$_POST['year']; $con = mysql_connect( $mysql_host, $mysql_user, $mysql_password ); if ( !$con ) { die( 'Could not connect: ' . mysql_error() ); } mysql_select_db( "$mysql_database",$con ); $table = "CREATE TABLE `$user` ( `fname` varchar(30), `lname` varchar(30), `username` varchar(30), `password` varchar(30), `email' varchar(30), 'country' varchar(30), 'month' varchar(30), 'day' int(2), 'year' int(4), PRIMARY KEY '$user' )" ; echo "Table Created!"; mysql_query($table,$con) or die (mysql_error()); mysql_close($con); ?> Above is my php code. I am trying to make a login system. I keep getting this error message "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 7" I have no idea what this means because i have no code on line 7. Any help is greatly appreciated. Also if you have any suggestions, please pm me. The form for this php code is on the page "http://blocks.netne.net/register.html " Please visit and give me any suggestions as to what to do! Thanks! |
tyler.watkins |
Apr 12 2012, 05:38 AM
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#2
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Novice Group: Members Posts: 27 Joined: 6-March 12 Member No.: 16,654 |
Thanks!
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Ephraim F. Moya |
Apr 12 2012, 10:51 AM
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#3
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Advanced Member Group: Members Posts: 167 Joined: 2-September 07 From: New Mexico Member No.: 3,702 |
Thanks! I made a mistake in the query string for reading a row. FOR should be WHERE. $query = "SELECT * FROM `table2` WHERE `id` = '7'"; and then I noticed that you're using mysqli This fits in the $resultArray = mysqli_query( $con, "{$query}", MYSQLI_STORE_RESULT ); instruction like this. Of course there are supporting instructions all around these. |
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