Notice: Undefined index: error

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hi all,

i am trying to connect to a simple database and i am given an error that is unable to connect, but i have no rights so no need of username and password, and the information is correct. please any help?

PHP

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<?php
$username="uname";
$password="pword";
$database="db";
$hostname="localhost";

mysql_connect($hostname) or die("Unable to connect to SQLSERVER");
mysql_select_db ($database)or die( "Unable to select database");

$query="SELECT * FROM table";
$result=mysql_query($query) or die("SQL error: ".mysql_error());

echo "result = " + $result;

$num=mysql_num_rows($result);

mysql_close();

error :: Unable to select database

Posted 12-Feb-12 21:11pm

adnama

Updated 12-Feb-12 22:23pm

v3

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Comments

CRDave1988 13-Feb-12 3:41am

use
mysql_connect($hostname) or die( mysql_error());
and see what is actual Error and tell us.

adnama 13-Feb-12 4:19am

You were right the error was not that one. It was actually
Notice: Undefined index: $event_id in C:\xampp\htdocs\dbConn\index.php on line 26

and on line 26 i have echo "event_id : {$row['$event_id']} ";

i am trying to display the data.

Dhol Gaurav 13-Feb-12 4:28am

Please try below code it might work. there are two ways to resolve this problem

1) echo "event_id : {$row['event_id']} " ;

2) $event_id = 0;
echo "event_id : {$row['$event_id']} " ;

2 solutions

Solution 1

Hello

You have to pass user name and password into mysql_connect like

PHP

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mysql_connect($hostname,$username,$password) or die("Unable to connect to SQLSERVER");

here you must have to pass user name and it can not be blank, in case of password you can pass blank like

PHP

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$password = ""
mysql_connect($hostname,$username,$password) or die("Unable to connect to SQLSERVER");

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Posted 12-Feb-12 21:51pm

Dhol Gaurav

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This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

adnama · Accepted Answer · 2012-02-13T00:41:00

Solution 2

I have solved this myself as it was simply removing the $.

from "event_id : {$row['$event_id']} ";
to
"event_id : {$row['event_id']} ";

thanks for your help

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Posted 13-Feb-12 0:41am

adnama

Notice: Undefined index: error

2 solutions

Solution 1

Solution 2

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Notice: Undefined index: error

2 solutions

Solution 1

Solution 2

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