parse error, sql Database and php

Question

i am getting this error after uploading my page to server :

Parse error: syntax error, unexpected T_LOGICAL_OR in /data/19/1/86/59/1901711/user/xxxx/htdocs/Maxxxxxxt/Mxxxxxxe/Pxx/txxx/index.php on line 11

on line i have this database statement , which is working fine when i am using it at localhost.

$objConnect = mysql_connect('localhost','root',''); or  die(mysql_error());

i have tried up every settings but they are all helpless, let me know what i am doing wrong ? and what is to be done to fix it up ?

You need to remove the `'root','');` semi-colon after closing the round bracket as you are ending your statement there and than you are using OR operator — Mr. Alien, Apr 23 '13 at 07:13
Try to use PDO or Mysqli the mysql functions ar deprecated from php version 5.0. Read more: http://stackoverflow.com/questions/548986/mysql-vs-mysqli-in-php — S.Visser, Apr 23 '13 at 07:15

score 2 · Answer 1 · answered Apr 23 '13 at 07:12

2

Remove semicolon

$objConnect = mysql_connect('localhost','root',''); or  die(mysql_error());

after mysql_connect function change it to this

$objConnect = mysql_connect('localhost','root','') or  die(mysql_error());

answered Apr 23 '13 at 07:12

chandresh_cool

score 2 · Accepted Answer · answered Apr 23 '13 at 07:12

2

Drop the ; behind the mysql_connect():

$objConnect = mysql_connect('localhost','root','') or die(mysql_error());

^{Besides, have a look at PDO and mysqli, as the mysql_x functions are deprecated!}

answered Apr 23 '13 at 07:12

Sirko

oh yea ! Thanks man. Crazy how can i even forget it.. ! – devilcrab Apr 23 '13 at 07:19

score 1 · Answer 3 · answered Apr 23 '13 at 07:12

1

$objConnect = mysql_connect('localhost','root',''); or  die(mysql_error());

should be

$objConnect = mysql_connect('localhost','root','') or  die(mysql_error());
                                                  ^ removed ;

answered Apr 23 '13 at 07:12

Class

score 1 · Answer 4 · answered Apr 23 '13 at 07:12

1

Remove ; from here

 'root',''); or  die(mysql_error());
           ^

answered Apr 23 '13 at 07:12

Yogesh Suthar

4 Answers4