pokusavam da ubacim neke informacije iz formulara na starnici u bazu podataka medjutim dobijam sledecu poruku :
*******************************************************
Warning: mysql_connect(): Access denied for user: 'mkbobi_danijela@localhost' (Using password: YES) in /home/mkbobi/public_html/DBInsert.php on line 25
The query is:
INSERT into formular values ('dasd', 'dada', 'asdsa', 'sdada')
Warning: mysql_db_query(): supplied argument is not a valid MySQL-Link resource in /home/mkbobi/public_html/DBInsert.php on line 32
The query could not be executed!
Warning: mysql_close(): supplied argument is not a valid MySQL-Link resource in /home/mkbobi/public_html/DBInsert.php on line 38
**********************************************
ne razumem zasto ?
evo koda koji sam napisala
***********************************************
<HTML>
<HEAD><TITLE>Inserting Data into a Database</TITLE></HEAD>
<BODY>
<?php
/* This page receives and handles the data generated by "DBform.html". */
import_request_variables("gP");
$Array["FirstName"] = trim ($Array["FirstName"]);
$Array["LastName"] = trim ($Array["LastName"]);
$Array["Email"] = trim ($Array["Email"]);
$Array["Comments"] = trim ($Array["Comments"]);
// Set the variables for the database access:
$Host = "localhost";
$User = "mkbobi_danijela";
$Password = "******";
$DBName = "mkbobi_mydb";
$TableName = "formular";
$Link = mysql_connect ($Host, $User, $Password);
$Query = "INSERT into $TableName values ('$Array[FirstName]', '$Array[LastName]', '$Array[Email]', '$Array[Comments]')";
print ("The query is:<BR>$Query<P>\n");
if (mysql_db_query ($DBName, $Query, $Link)) {
print ("The query was successfully executed!<BR>\n");
} else {
print ("The query could not be executed!<BR>\n");
}
mysql_close ($Link);
?>
</BODY>
</HTML>
****************************************************
unapred hvala puno :)
[Ovu poruku je menjao Goran Rakić dana 07.07.2004. u 19:02 GMT]