prcollin Posted May 27, 2008 Share Posted May 27, 2008 Can someone please help me with this. I thought I was doing well then I got a couple errors thrown at me Parse error: syntax error, unexpected T_STRING in /home/f8888r/public_html/p8888t/login.php on line 17 <?php $host="localhost"; $user="fin888r_pr8888"; $passwd="ni888888"; $database="fin888r_U88rs"; $con = mysql_connect("$host", "$user", "$passwd", "$database"); if (!$con) ( die('Could not connect: ' . mysql_error()) ) mysql_select_db("findzer_Users", $con); $sql=\ "insert into newuser ('user_name', 'user_pass', 'full_name', 'user_email', 'confirm_email', 'alt_email', 'user_city') Values ('$_post[user_name]','$_post[user_pass]','$_post[full_name]','$_post[user_email]','$_post[co nfir_memail]','$_post[alt_email]','$_post[user_city]"; if (!mysql_query($sql,$con)) ( die('Error: ' . mysql_error()); ) echo "1 record added"; mysql_close($con) ?> is there something I am typing wrong? I also tried it without the ' ' in the insert into part Quote Link to comment Share on other sites More sharing options...
DarkWater Posted May 27, 2008 Share Posted May 27, 2008 Whenever using arrays in double quotes, they must be enclosed in { }. {$_POST['key']}. Quote Link to comment Share on other sites More sharing options...
jonsjava Posted May 27, 2008 Share Posted May 27, 2008 <?php $host="localhost"; $user="fin888r_pr8888"; $passwd="ni888888"; $database="fin888r_U88rs"; $con = mysql_connect("$host", "$user", "$passwd", "$database"); if (!$con) { die("Could not connect:" . mysql_error()); } mysql_select_db("findzer_Users", $con); $sql= "insert into newuser ('user_name', 'user_pass', 'full_name', 'user_email', 'confirm_email', 'alt_email', 'user_city') Values ('{$_post['user_name']}','{$_post['user_pass']}','{$_post['full_name']}','{$_post['user_email']}','{$_post['co nfir_memail']}','{$_post['alt_email']}','{$_post['user_city']}"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ?> Quote Link to comment Share on other sites More sharing options...
prcollin Posted May 27, 2008 Author Share Posted May 27, 2008 <?php $host="localhost"; $user="fin888r_pr8888"; $passwd="ni888888"; $database="fin888r_U88rs"; $con = mysql_connect("$host", "$user", "$passwd", "$database"); if (!$con) { die("Could not connect:" . mysql_error()); } mysql_select_db("findzer_Users", $con); $sql= "insert into newuser ('user_name', 'user_pass', 'full_name', 'user_email', 'confirm_email', 'alt_email', 'user_city') Values ('{$_post['user_name']}','{$_post['user_pass']}','{$_post['full_name']}','{$_post['user_email']}','{$_post['co nfir_memail']}','{$_post['alt_email']}','{$_post['user_city']}"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ?> when I use this code it give me Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''user_name', 'user_pass', 'full_name', 'user_email', 'confirm_email', 'alt_e' at line 1 Quote Link to comment Share on other sites More sharing options...
jonsjava Posted May 27, 2008 Share Posted May 27, 2008 change the query to: insert into newuser ('user_name', 'user_pass', 'full_name', 'user_email', 'confirm_email', 'alt_email', 'user_city') Values('{$_post['user_name']}','{$_post['user_pass']}','{$_post['full_name']}','{$_post['user_email']}','{$_post['confir_memail']}','{$_post['alt_email']}','{$_post['user_city']}" Quote Link to comment Share on other sites More sharing options...
prcollin Posted May 27, 2008 Author Share Posted May 27, 2008 now i get Parse error: syntax error, unexpected T_STRING in /home/findzer/public_html/philsshit/login.php on line 21 sorry guys i dont have phpedit at work so i cant check all this on my own that is why I am asking for your help so much. THANKS ALOT AHEAD OF TIME!!! Quote Link to comment Share on other sites More sharing options...
jonsjava Posted May 27, 2008 Share Posted May 27, 2008 I'm putting it through Zend Studion 5.5.1, and getting a clean bill of health. I'm guessing you aren't sending the full code. Quote Link to comment Share on other sites More sharing options...
prcollin Posted May 27, 2008 Author Share Posted May 27, 2008 I'm putting it through Zend Studion 5.5.1, and getting a clean bill of health. I'm guessing you aren't sending the full code. this is my exact code <?php $host="lxxxxt"; $user="fixxxx_pxxxxxn"; $passwd="nxxxxx7"; $database="fixxxxr_Uxxxs"; $con = mysql_connect("$host", "$user", "$passwd", "$database"); if (!$con) { die("Could not connect:" . mysql_error()); } mysql_select_db("findzer_Users", $con); $sql= "insert into newuser ('user_name', 'user_pass', 'full_name', 'user_email', 'confirm_email', 'alt_email', 'user_city') Values('{$_post['user_name']}','{$_post['user_pass']}','{$_post['full_name']}','{$_post['user_email']}','{$_post['confir_memail']}','{$_post['alt_email']}','{$_post['user_city']}" Values ('{$_post['user_name']}','{$_post['user_pass']}','{$_post['full_name']}','{$_post['user_email']}','{$_post['co nfir_memail']}','{$_post['alt_email']}','{$_post['user_city']}"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ?> and the form that feeds this script the information is <html> <body> <form action="login.php" method="post"> Username: <input type="text" name="user_name"> Password: <input type="password" name="user_pass"> Full Name: <input type="text" name="full_name"> E-Mail : <input type="text" name="user_email"> Confirm E-mail: <input type="text" name="confirm_email"> Alt Email: <input type="text" name="alt_email"> City: <input type="text" name="user_city"> <input type="submit" /> Thats all of it Quote Link to comment Share on other sites More sharing options...
jonsjava Posted May 27, 2008 Share Posted May 27, 2008 cleaned up your code: <?php $host="lxxxxt"; $user="fixxxx_pxxxxxn"; $passwd="nxxxxx7"; $database="fixxxxr_Uxxxs"; $con = mysql_connect("$host", "$user", "$passwd", "$database"); if (!$con) { die("Could not connect:" . mysql_error()); } mysql_select_db("findzer_Users", $con); $sql= "insert into newuser ('user_name', 'user_pass', 'full_name', 'user_email', 'confirm_email', 'alt_email', 'user_city') Values('{$_post['user_name']}','{$_post['user_pass']}','{$_post['full_name']}','{$_post['user_email']}','{$_post['confir_memail']}','{$_post['alt_email']}','{$_post['user_city']} Values('{$_post['user_name']}','{$_post['user_pass']}','{$_post['full_name']}','{$_post['user_email']}','{$_post['confir_memail']}','{$_post['alt_email']}','{$_post['user_city']}"; if (!mysql_query($sql,$con)) { die("Error: ". mysql_error()); } echo "1 record added"; mysql_close($con) ?> Quote Link to comment Share on other sites More sharing options...
prcollin Posted May 27, 2008 Author Share Posted May 27, 2008 now getting Error: 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 ''user_name', 'user_pass', 'full_name', 'user_email', 'confirm_email', 'alt_email' at line 1 i am using mysql version 4.1.22-standard Quote Link to comment Share on other sites More sharing options...
jonsjava Posted May 27, 2008 Share Posted May 27, 2008 oops, my fault. should have error checked the sql too: <?php $host="lxxxxt"; $user="fixxxx_pxxxxxn"; $passwd="nxxxxx7"; $database="fixxxxr_Uxxxs"; $con = mysql_connect("$host", "$user", "$passwd", "$database"); if (!$con) { die("Could not connect:" . mysql_error()); } mysql_select_db("findzer_Users", $con); $sql= "insert into newuser (`user_name`, `user_pass`, `full_name`, `user_email`, `confirm_email`, `alt_email`, `user_city`) Values('{$_post['user_name']}','{$_post['user_pass']}','{$_post['full_name']}','{$_post['user_email']}','{$_post['confir_memail']}','{$_post['alt_email']}','{$_post['user_city']} Values('{$_post['user_name']}','{$_post['user_pass']}','{$_post['full_name']}','{$_post['user_email']}','{$_post['confir_memail']}','{$_post['alt_email']}','{$_post['user_city']}"; if (!mysql_query($sql,$con)) { die("Error: ". mysql_error()); } echo "1 record added"; mysql_close($con) ?> Quote Link to comment Share on other sites More sharing options...
prcollin Posted May 28, 2008 Author Share Posted May 28, 2008 oops, my fault. should have error checked the sql too: <?php $host="lxxxxt"; $user="fixxxx_pxxxxxn"; $passwd="nxxxxx7"; $database="fixxxxr_Uxxxs"; $con = mysql_connect("$host", "$user", "$passwd", "$database"); if (!$con) { die("Could not connect:" . mysql_error()); } mysql_select_db("findzer_Users", $con); $sql= "insert into newuser (`user_name`, `user_pass`, `full_name`, `user_email`, `confirm_email`, `alt_email`, `user_city`) Values('{$_post['user_name']}','{$_post['user_pass']}','{$_post['full_name']}','{$_post['user_email']}','{$_post['confir_memail']}','{$_post['alt_email']}','{$_post['user_city']} Values('{$_post['user_name']}','{$_post['user_pass']}','{$_post['full_name']}','{$_post['user_email']}','{$_post['confir_memail']}','{$_post['alt_email']}','{$_post['user_city']}"; if (!mysql_query($sql,$con)) { die("Error: ". mysql_error()); } echo "1 record added"; mysql_close($c ?> now i get Error: 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 1 Quote Link to comment Share on other sites More sharing options...
prcollin Posted May 28, 2008 Author Share Posted May 28, 2008 bump Quote Link to comment Share on other sites More sharing options...
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