[ danijela21 @ 06.07.2004. 01:51 ] @
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] |