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Search found 15 results on 1 pages for 'user309381'.

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• How to find level of employee position using RECURSIVE COMMON TABLE EXPRESSION

  - by user309381

  ;with Ranked(Empid,Mngrid,Empnm,RN,level) As ( select Empid,Mngrid ,Empnm ,row_number() over (order by Empid)AS RN , 0 as level from dbo.EmpMngr ), AnchorRanked(Empid,Mngrid,Empnm,RN,level) AS(select Empid,Mngrid,Empnm,RN ,level from Ranked ), RecurRanked(Empid,Mngrid,Empnm,RN,level) AS(select Empid,Mngrid,Empnm,RN,level from AnchorRanked Union All select Ranked.Empid,Ranked.Mngrid,Ranked.Empnm,Ranked.RN,Ranked.level + 1 from Ranked inner join RecurRanked on Ranked.Empid = RecurRanked.Empid AND Ranked.RN = RecurRanked.RN+1) select Empid,Empnm,level from RecurRanked

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• MySQL Connection Error in PHP

  - by user309381

  I have set the password for root and grant all privileges for root. Why does it say it is denied? ****mysql_query() [function.mysql-query]: Access denied for user 'SYSTEM'@'localhost' (using password: NO) in C:\wamp\www\photo_gallery\includes\database.php on line 56 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\wamp\www\photo_gallery\includes\database.php on line 56 The Query has problemAccess denied for user 'SYSTEM'@'localhost' (using password: NO) Code as follows: <?php include("DB_Info.php"); class MySQLDatabase { public $connection; function _construct() { $this->open_connection(); } public function open_connection() { /* $DB_SERVER = "localhost"; $DB_USER = "root"; $DB_PASS = ""; $DB_NAME = "photo_gallery";*/ $this->connection = mysql_connect($DBSERVER,$DBUSER,$DBPASS); if(!$this->connection) { die("Database Connection Failed" . mysql_error()); } else { $db_select = mysql_select_db($DBNAME,$this->connection); if(!$db_select) { die("Database Selection Failed" . mysql_error()); } } } function mysql_prep($value) { if (get_magic_quotes_gpc()) { $value = stripslashes($value); } // Quote if not a number if (!is_numeric($value)) { $value = "'" . mysql_real_escape_string($value) . "'"; } return $value; } public function close_connection() { if(isset($this->connection)) { mysql_close($this->connection); unset($this->connection); } } public function query($sql) { //$sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //$found_user = mysql_fetch_assoc($result); //echo $found_user; return $found_user; } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); ?>

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• i have problem with include file

  - by user309381

  //this is intializer.php defined('DS')? null :define('DS',DIRECTORY_SEPARATOR); defined('SITE_ROOT')? null : define('SITE_ROOT',DS.'C:',DS.'wamp',DS.'www',DS.'photo_gallery'); defined('LIB_PATH')?null:define('LIB_PATH',SITE_ROOT.DS.'includes'); require_once(LIB_PATH.DS.'datainfo.php'); require_once(LIB_PATH.DS.'function.php'); require_once(LIB_PATH.DS.'session.php'); require_once(LIB_PATH.DS.'database.php'); require_once(LIB_PATH.DS.'user.php'); //this is other file where i call php file // ERROR Use of undefined constant LIB_PATH - assumed 'LIB_PATH' in //C:\wamp\www\photo_gallery\includes\database.php on //Notice: Use of undefined constant DS - assumed 'DS' in //C:\wamp\www\photo_gallery\includes\database.php on include(LIB_PATH.DS."database.php") ?

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• Is it possible to pass the value in php file using Ajax.Request?

  - by user309381

  function reload(form) { var val = $('seltab').getValue(); new Ajax.Request('Website.php?$cat = val', { method:'post', onSuccess: function(transport){ ..... code in php: echo "<select id = seltab onchange='reload(this.form)'>"; $querysel = "SELECT title_id,author FROM authors NATURAL JOIN books"; $result1 = mysql_query($querysel) ; while($rowID = mysql_fetch_assoc($result1)) {

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• i am getting error like mysql_connect() acces denied for system@localhost(using password NO)

  - by user309381

  class MySQLDatabase { public $connection; function _construct() { $this->open_connection();} public function open_connection() {$this->connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if(!$this->connection){die("Database Connection Failed" . mysql_error());} else{$db_select = mysql_select_db(DB_NAME,$this->connection); if(!$db_select){die("Database Selection Failed" . mysql_error()); } }} public function close_connection({ if(isset($this->connection)){ mysql_close($this->connection); unset($this->connection);}} public function query(/*$sql*/){ $sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //return $result;while( $found_user = mysql_fetch_assoc($result)) { echo $found_user ['username']; } } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); $database->open_connection(); $database->query(); $database->close_connection(); ?>

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• How to retrieve the value from Select html element using JS prototype in php?

  - by user309381

  <script type="text/javascript" src="prototype.js"></script> <script> function reload(form){ var val = $("seltab");alert(val); }</script> echo "<form method = post name = f1 action = '' >"; echo "<select id = seltab onchange = 'reload(this.form)'>"; $querysel = "SELECT title_id,author FROM authors NATURAL JOIN books"; $result1 = mysql_query($querysel) ; while($rowID = mysql_fetch_assoc($result1)) { $TitleID = $rowID['title_id']; $author = $rowID['author']; print "<option value =$TitleID>$author\n"; print "</option>"; } print "</select>";

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• Trouble using Prototype JS

  - by user309381

  I have three files named like: AjxDm.php db_test website.php What I am trying to do is AjxDm.php in this file I click and I am using Ajax.Request and it directed me to the db_test.php. In this file I am retrieving a value from the database and I assign the value into a dropdown. Again when I try to select one of the options in the dropdown it is direceting me to the Website.php using Ajax.Request with the dropdown selected value and retrieve data from database and display. The problem is that when I call from AjxDm.php to Db_test it works but when I select option in db_test.php it doesn't find a JavaScript function which direct me to the website.php. What can I do to make it work?

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• Php 5 and mysql connecting gives error ...code and error in there check out and plz..plz help....

  - by user309381

  **mysql_query() [function.mysql-query]: Access denied for user 'SYSTEM'@'localhost' (using password: NO) in C:\wamp\www\photo_gallery\includes\database.php on line 56 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\wamp\www\photo_gallery\includes\database.php on line 56 The Query has problemAccess denied for user 'SYSTEM'@'localhost' (using password: NO) i have set the password for root and grant privilege all for root.Why it soes show like SYSTEM@localhost i dont have SYSTEM .** class MySQLDatabase { public $connection; function _construct() { $this-open_connection(); } public function open_connection() { /* $DB_SERVER = "localhost"; $DB_USER = "root"; $DB_PASS = ""; $DB_NAME = "photo_gallery";*/ $this-connection = mysql_connect($DBSERVER,$DBUSER,$DBPASS); if(!$this-connection) { die("Database Connection Failed" . mysql_error()); } else { $db_select = mysql_select_db($DBNAME,$this-connection); if(!$db_select) { die("Database Selection Failed" . mysql_error()); } } } function mysql_prep($value) { if (get_magic_quotes_gpc()) { $value = stripslashes($value); } // Quote if not a number if (!is_numeric($value)) { $value = "'" . mysql_real_escape_string($value) . "'"; } return $value; } public function close_connection() { if(isset($this->connection)) { mysql_close($this->connection); unset($this->connection); } } public function query($sql) { //$sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //$found_user = mysql_fetch_assoc($result); //echo $found_user; return $found_user; } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); ?

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• please help me to find out where i am doing mistake in this code? i wnat retieve the value that i am

  - by user309381

  function reload(form) { var val = $('seltab').getValue(); new Ajax.Request('Website.php?cat=' +escape(val), { method:'get', onSuccess: function(transport){ var response = transport.responseText ; $("MyDivDB").innerHTML = transport.responseText ; alert("Success! \n\n" + response); }, onFailure: function(){ alert('Something went wrong...') } }); } </script> </head> title author pages $con = mysql_connect($dbhostname,$dbuserid,$dbpassword); if(!$con) { die ("connection failed".mysql_error()); } $db = mysql_select_db($dbname,$con); if(!$db) { die("Database is not selected".mysql_error()); } $query ="SELECT * FROM books NATURAL JOIN authors" ; $result = mysql_query($query); if(!$query) { die("Database is not query".mysql_error()); } while($row = mysql_fetch_array($result,MYSQL_ASSOC)) { $title = $row["title"]; $author = $row["author"]; $page = $row["pages"]; echo "<tr>"; echo "<td>$title</td>"; echo "<td>$author</td>"; echo "<td>$page</td>"; echo "</tr>"; } print "</table>"; echo "<select id = seltab onchange = 'reload(this.form)'>"; $querysel = "SELECT title_id,author FROM authors NATURAL JOIN books"; $result1 = mysql_query($querysel) ; while($rowID = mysql_fetch_assoc($result1)) { $TitleID = $rowID['title_id']; $author = $rowID['author']; print "<option value = $author>$author\n"; print "</option>"; } print "</select>"; ? Wbsite.php

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• how to pass the value in php file using Ajax.Request() function.The way i do use it's not working Wh

  - by user309381

  <html><head> function reload(form) { var val = $('seltab').getValue(); new Ajax.Request('Website.php?$cat = val', { method:'post', onSuccess: function(transport) { ..... </head> Code in PHP: echo "<select id = seltab onchange = 'reload(this.form)'>"; $querysel = "SELECT title_id,author FROM authors NATURAL JOIN books"; $result1 = mysql_query($querysel) ; while($rowID = mysql_fetch_assoc($result1)) { .... }

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• i dont understand error while connecting php and mysql? user denied ? plz help me out to solve. ?

  - by user309381

  class MySQLDatabase { public $connection; function _construct() { $this->open_connection();} public function open_connection() {$this->connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if(!$this->connection){die("Database Connection Failed" . mysql_error());} else{$db_select = mysql_select_db(DB_NAME,$this->connection); if(!$db_select){die("Database Selection Failed" . mysql_error()); } }} public function close_connection({ if(isset($this->connection)){ mysql_close($this->connection); unset($this->connection);}} public function query(/*$sql*/){ $sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //return $result;while( $found_user = mysql_fetch_assoc($result)) { echo $found_user ['username']; } } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); $database->open_connection(); $database->query(); $database->close_connection(); I am getting error like denied for user system@locahost(using password no).i have also other database but it runs fine and i dont also i have set the password after encountered the error what else can do to solve plz help ?

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• mysql_connect() acces denied for system@localhost(using password NO)

  - by user309381

  class MySQLDatabase { public $connection; function _construct() { $this-open_connection(); } public function open_connection() { $this-connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if(!$this-connection) { die("Database Connection Failed" . mysql_error()); } else { $db_select = mysql_select_db(DB_NAME,$this-connection); if(!$db_select) { die("Database Selection Failed" . mysql_error()); } } } function mysql_prep($value) { if (get_magic_quotes_gpc()) { $value = stripslashes($value); } // Quote if not a number if (!is_numeric($value)) { $value = "'" . mysql_real_escape_string($value) . "'"; } return $value; } public function close_connection() { if(isset($this->connection)) { mysql_close($this->connection); unset($this->connection); } } public function query(/*$sql*/) { $sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //return $result; while( $found_user = mysql_fetch_assoc($result)) { echo $found_user ['username']; } } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); $database-open_connection(); $database-query(); $database-close_connection(); ?

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• when i click on checkbox ,the image should be hiden though i dont make it happen somehow and i can g

  - by user309381

  function Psend() { new Ajax.Request('Handler.ashx', { method: 'get', onSuccess: function(transport) { var response = transport.responseText || "no response text"; //alert("Success! \n\n" + response); var obj = response.evalJSON(true); for (i = 0; i < 4; i++) { DeCheBX = $('MyDiv').insert(new Element('input', { 'type': 'checkbox', 'id': "Img" + obj[i].Nam, 'value': obj[i].IM, 'onClick': 'SayHi(this,i)' })); document.body.appendChild(DeCheBX); DeImg = $('MyDiv').insert(new Element('img', { 'id': "img" + obj[i].Nam, 'src': obj[i].IM })); document.body.appendChild(DeImg); SayHi = function(x,i) { try { if ($(x).checked == true) { img = "img" + obj[i].Nam; alert(img); $('img').hide(); } } catch (e) { alert("error"); } }; } }, onFailure: function() { alert('Something went wrong...') } }); }

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• Using $this when not in object context--i am using the latest version of php and mysql

  - by user309381

  This is user.php: include("databse.php");//retrieving successfully first name and lastname from databse file into user.php class user { public $first_name; public $last_name; public static function full_name() { if(isset($this->first_name) && isset($this->last_name)) { return $this->first_name . " " . $this->last_name; } else { return ""; } } } ?> Other php file, index.php: <?php include(databse.php); include(user.php); $record = user::find_by_id(1); $object = new user(); $object->id = $record['id']; $object->username = $record['username']; $object->password = $record['password']; $object->first_name = $record['first_name']; $object->last_name = $record['last_name']; // echo $object->full_name(); echo $object->id;// successfully print the id echo $object->username;//success fully print the username echo->$object->full_name();//**ERROR:Using $this when not in object context** ?>

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• how to hide the image? how can i do ?

  - by user309381

  function Psend() { new Ajax.Request('Handler.ashx', { method: 'get', onSuccess: function(transport) { var response = transport.responseText || "no response text"; //alert("Success! \n\n" + response); var obj = response.evalJSON(true); for (i = 0; i < 4; i++) { DeCheBX = $('MyDiv').insert(new Element('input', { 'type': 'checkbox', 'id': "img" + obj[i].Nam, 'value': obj[i].IM, 'onClick': 'SayHi(this)' })); DeImg = $('MyDiv').insert(new Element('img', { 'id': "img" + obj[i].Nam, 'src': obj[i].IM, 'style': 'display = inline', 'onClick': 'Say(this)' })); document.body.appendChild(DeCheBX); document.body.appendChild(DeImg); } }, onFailure: function() { alert('Something went wrong...') } }); SayHi = function(x) { if ($(x).checked == true) { // $('id').hide(); **$('img'+i).style.visibility = "hidden";**// doesnt work } };

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