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Search found 1913 results on 77 pages for 'die in sente'.

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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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  • 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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  • php and mysql listing databases and looping through results

    - by Jacksta
    Beginner help needed :) I am doign an example form a php book which lists tables in databases. I am getting an error on line 36: $db_list .= "$table_list"; <?php //connect to database $connection = mysql_connect("localhost", "admin_cantsayno", "cantsayno") or die(mysql_error()); //list databases $dbs = @mysql_list_dbs($connection) or die(mysql_error()); //start first bullet list $db_list = "<ul>"; $db_num = 0; //loop through results of functions while ($db_num < mysql_num_rows($dbs)) { //get database names and make each a list point $db_names[$db_num] = mysql_tablename($dbs, $db_num); $db_list .= "<li>$db_names[$db_num]"; //get table names and make another list $tables = @mysql_list_tables($db_names[$db_num]) or die(mysql_error()); $table_list = "<ul>"; $table_num = 0; //loop through results of function while ($table_num < mysql_num_rows($tables)){ //get table names and make each bullet point $table_names[$table_num] = mysql_tablename($tables, $table_num); $table_list .= "<li>$table_names[$table_num]"; $table_num++; } //close inner bullet list and increment number to continue $table_list .= "</ul>" $db_list .= "$table_list"; $db_num++; } //close outer bullet list $db_list .= "</ul>"; ?> <html> <head> <title>MySQL Tables</title> </head> <body> <p><strong>Data bases and tables on local host</strong></p> <? echo "$db_list"; ?> </body>

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  • searching for multiple columns mysql and php

    - by addi
    i'm trying to search for multiple columns using this code: <?php // Connection Database $search = $_POST ['Search']; mysql_connect("xxxxxx", "xxxxxx", "xxxxx") or die ("Error Connecting to Database"); mysql_select_db("xxxxx") or die('Error'); $data = mysql_query("SELECT CourseName, CourseDescription, CourseLeader FROM course MATCH (CourseName, CourseDescription, CourseLeader) AGAINST ('". $search ."') or die('Error'); Print "<table border cellpadding=3>"; while($info = mysql_fetch_array( $data )) { Print "<tr>"; Print "<th>Course Name:</th> <td>".$info['CourseName'] . "</td> "; Print "<th>Course Description:</th><td>".$info['CourseDescription'] . "</td> "; Print "<th>Course Leader:</th><td>".$info['CourseLeader'] . " </td></tr>"; } Print "</table>"; ?> i'm getting the following error: Parse error: syntax error, unexpected T_STRING in /home/a7105766/public_html/website/scripts/coursesearchdb.php on line 30 what am I doing wrong?? cheers

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  • database connection OK,result not appear

    - by klox
    hi..all.for now i'm already connected to database but the result not appear at Tuner range is"+res+"this is my code: var str=data[0]; var matches=str.match(/[EE|EJU].*D/i); $.ajax({ type:"post", url:"process1.php", data:"tversion="+matches+"&action=tunermatches", cache:false, async:false, success: function(res){ $('#value').replacewith("<div id='value'><h6>Tuner range is"+res+".</h6></div>"); } }); }); and this is my process file: //connect to database $dbc=mysql_connect(_SRV,_ACCID,_PWD) or die(_ERROR15.": ".mysql_error()); $db=mysql_select_db("qdbase",$dbc) or die(_ERROR17.": ".mysql_error()); switch(postVar('action')) { case 'tunermatches' : tunermatches(postVar('tversion')); break; function tunermatches($tversion)){ $Tuner=mysql_real_escape_string($tversion); $sql= "SELECT remark FROM settingdata WHERE itemname='Tuner_range' AND itemdata='".$Tunermatches."'"; $res=mysql_query($sql) or die (_ERROR26.":".mysql_error()); $dat=mysql_fetch_array($res,MYSQL_NUM); if($dat[0]>0) { echo $dat[0]; } mysql_close($dbc); }

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  • Most efficient way to write over file after reading

    - by Ryan McClure
    I'm reading in some data from a file, manipulating it, and then overwriting it to the same file. Until now, I've been doing it like so: open (my $inFile, $file) or die "Could not open $file: $!"; $retString .= join ('', <$inFile>); ... close ($inFile); open (my $outFile, $file) or die "Could not open $file: $!"; print $outFile, $retString; close ($inFile); However I realized I can just use the truncate function and open the file for read/write: open (my $inFile, '+<', $file) or die "Could not open $file: $!"; $retString .= join ('', <$inFile>); ... truncate $inFile, 0; print $inFile $retString; close ($inFile); I don't see any examples of this anywhere. It seems to work well, but am I doing it correctly? Is there a better way to do this?

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  • authorise user from mysql database

    - by Jacksta
    I suck at php, and cant find the error here. The script gets 2 variables "username" and "password" from a html from then check them against a MySQL databse. When I run this I get the follow error "Query was empty" <? if ((!$_POST[username]) || (!$_POST[password])) { header("Location: show_login.html"); exit; } $db_name = "testDB"; $table_name = "auth_users"; $connection = @mysql_connect("localhost", "admin", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $slq = "SELECT * FROM $table_name WHERE username ='$_POST[username]' AND password = password('$_POST[password]')"; $result = @mysql_query($sql, $connection) or die(mysql_error()); $num = mysql_num_rows($result); if ($num != 0) { $msg = "<p>Congratulations, you're authorised!</p>"; } else { header("Location: show_login.html"); exit; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Secret Area</title> </head> <body> <? echo "$msg"; ?> </body> </html>

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  • php + MySQL editing table data.

    - by Jacksta
    This question is relating to 2 php scripts. The first script is called pick_modcontact.php where I choose a contact (from a contact book like phone book), then posts to the script show_modcontact.php When I click the submit button on the form on pick.modcontact.php. As a result of submitting the form I am then taken to show_modcontact.php. As the variables are not present the user is directed back to pick_modcontact.php I can not work out how to correct the code so that it will show the results of the script show_modcontact.php This script shows all contacts in a database which is an "address book" this part works fine. please see below. Name:pick_modcontact.php if ($_SESSION['valid'] != "yes") { header( "Location: contact_menu.php"); exit; } $db_name = "testDB"; $table_name = "my_contacts"; $connection = @mysql_connect("localhost", "admin", "user") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT id, f_name, l_name FROM $table_name ORDER BY f_name"; $result = @mysql_query($sql, $connection) or die(mysql_error()); $num = @mysql_num_rows($result); if ($num < 1) { $display_block = "<p><em>Sorry No Results!</em></p>"; } else { while ($row = mysql_fetch_array($result)) { $id = $row['id']; $f_name = $row['f_name']; $l_name = $row['l_name']; $option_block .= "<option value\"$id\">$f_name, $l_name</option>"; } $display_block = "<form method=\"POST\" action=\"show_modcontact.php\"> <p><strong>Contact:</strong> <select name=\"id\">$option_block</select> <input type=\"submit\" name=\"submit\" value=\"Select This Contact\"></p> </form>"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Modify A Contact</title> </head> <body> <h1>My Contact Management System</h1> <h2><em>Modify a Contact</em></h2> <p>Select a contact from the list below, to modify the contact's record.</p> <? echo "$display_block"; ?> <br> <p><a href="contact_menu.php">Return to Main Menu</a></p> </body> </html> This script is for modifying the contact: named show_modcontact.php <?php if (!$_POST['id']) { header( "Location: pick_modcontact.php"); exit; } else { session_start(); } if ($_SESSION['valid'] != "yes") { header( "Location: pick_modcontact.php"); exit; } $db_name = "testDB"; $table_name = "my_contacts"; $connection = @mysql_connect("localhost", "admin", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT f_name, l_name, address1, address2, address3, postcode, prim_tel, sec_tel, email, birthday FROM $table_name WHERE id = '" . $_POST['id'] . "'"; $result = @mysql_query($sql, $connection) or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $f_name = $row['f_name']; $l_name = $row['l_name']; $address1 = $row['address1']; $address2 = $row['address2']; $address3 = $row['address3']; $country = $row['country']; $prim_tel = $row['prim_tel']; $sec_tel = $row['sec_tel']; $email = $row['email']; $birthday = $row['birthday']; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Modify A Contact</title> </head> <body> <form action="do_modcontact.php" method="post"> <input type="text" name="id" value="<? echo $_POST['id']; ?>" /> <table cellpadding="5" cellspacing="3"> <tr> <th>Name & Address Information</th> <th> Other Contact / Personal Information</th> </tr> <tr> <td align="top"> <p><strong>First Name:</strong><br /> <input type="text" name="f_name" value="<? echo "$f_name"; ?>" size="35" maxlength="75" /></p> <p><strong>Last Name:</strong><br /> <input type="text" name="l_name" value="<? echo "$l_name"; ?>" size="35" maxlength="75" /></p> <p><strong>Address1:</strong><br /> <input type="text" name="f_name" value="<? echo "$address1"; ?>" size="35" maxlength="75" /></p> <p><strong>Address2:</strong><br /> <input type="text" name="f_name" value="<? echo "$address2"; ?>" size="35" maxlength="75" /></p> <p><strong>Address3:</strong><br /> <input type="text" name="f_name" value="<? echo "$address3"; ?>" size="35" maxlength="75" /> </p> <p><strong>Postcode:</strong><br /> <input type="text" name="f_name" value="<? echo "$postcode"; ?>" size="35" maxlength="75" /></p> <p><strong>Country:</strong><br /> <input type="text" name="f_name" value="<? echo "$country"; ?>" size="35" maxlength="75" /> </p> <p><strong>First Name:</strong><br /> <input type="text" name="f_name" value="<? echo "$f_name"; ?>" size="35" maxlength="75" /></p> </td> <td align="top"> <p><strong>Prim Tel:</strong><br /> <input type="text" name="f_name" value="<? echo "$prim_tel"; ?>" size="35" maxlength="75" /></p> <p><strong>Sec Tel:</strong><br /> <input type="text" name="f_name" value="<? echo "$sec_tel"; ?>" size="35" maxlength="75" /></p> <p><strong>Email:</strong><br /> <input type="text" name="f_name" value="<? echo "$email;" ?>" size="35" maxlength="75" /> </p> <p><strong>Birthday:</strong><br /> <input type="text" name="f_name" value="<? echo "$birthday"; ?>" size="35" maxlength="75" /> </p> </td> </tr> <tr> <td align="center"> <p><input type="submit" name="submit" value="Update Contact" /></p> <br /> <p><a href="contact_menu.php">Retuen To Menu</a></p> </td> </tr> </table> </form> </body> </html> note for site admin, I am re posting this question with the hope of someone else reading over it. older questions seem to go dead after a while.

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  • Error with MySQL Query

    - by Ken
    Okay, I must be an idiot, because this is my 3rd question for today. Here's my code: date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "********"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db("`users`, $con) or die(mysql_error()"); $query = ("INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password'))"); mysql_query('$query') or die(mysql_error()); mysql_close($con); echo("Thank you for registering!"); I always get the error returned as: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '$query' at line 1. Help a newbie. I'm about to stab my monitor.

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  • I can't insert data into my database

    - by Ken
    I don't know why, but my data doesn't go into my database 'users' with the table 'data'. <html> <body> <?php date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "g00dfor@boy"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db(`users`, $con) or die(mysql_error()); mysql_query(INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password')) or die(mysql_error()); mysql_close($con) echo("Thank you for registering!"); ?> </body> </html> All i get is a blank page.

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  • Why doesn't this PHP execute?

    - by cam
    I copied the code from this site exactly: http://davidwalsh.name/web-service-php-mysql-xml-json as follows, /* require the user as the parameter */ if(isset($_GET['user']) && intval($_GET['user'])) { /* soak in the passed variable or set our own */ $number_of_posts = isset($_GET['num']) ? intval($_GET['num']) : 10; //10 is the default $format = strtolower($_GET['format']) == 'json' ? 'json' : 'xml'; //xml is the default $user_id = intval($_GET['user']); //no default /* connect to the db */ $link = mysql_connect('localhost','username','password') or die('Cannot connect to the DB'); mysql_select_db('db_name',$link) or die('Cannot select the DB'); /* grab the posts from the db */ $query = "SELECT post_title, guid FROM wp_posts WHERE post_author = $user_id AND post_status = 'publish' ORDER BY ID DESC LIMIT $number_of_posts"; $result = mysql_query($query,$link) or die('Errant query: '.$query); /* create one master array of the records */ $posts = array(); if(mysql_num_rows($result)) { while($post = mysql_fetch_assoc($result)) { $posts[] = array('post'=>$post); } } /* output in necessary format */ if($format == 'json') { header('Content-type: application/json'); echo json_encode(array('posts'=>$posts)); } else { header('Content-type: text/xml'); echo '<posts>'; foreach($posts as $index => $post) { if(is_array($post)) { foreach($post as $key => $value) { echo '<',$key,'>'; if(is_array($value)) { foreach($value as $tag => $val) { echo '<',$tag,'>',htmlentities($val),'</',$tag,'>'; } } echo '</',$key,'>'; } } } echo '</posts>'; } /* disconnect from the db */ @mysql_close($link); } And the php doesn't execute, it just displays as plain text. What's the dealio? The host supports PHP, I use it to run a Wordpress blog and other things.

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  • Javascript function with PHP throwing a "Illegally Formed XML Syntax" error

    - by Joe
    I'm trying to learn some javascript and i'm having trouble figuring out why my code is incorrect (i'm sure i'm doing something wrong lol), but anyways I am trying to create a login page so that when the form is submitted javascript will call a function that checks if the login is in a mysql database and then checks the validity of the password for the user if they exist. however I am getting an error (Illegally Formed XML Syntax) i cannot resolve. I'm really confused, mostly because netbeans is saying it is a xml syntax error and i'm not using xml. here is the code in question: function validateLogin(login){ login.addEventListener("input", function() { $value = login.value; if (<?php //connect to mysql mysql_connect(host, user, pass) or die(mysql_error()); echo("<script type='text/javascript'>"); echo("alert('MYSQL Connected.');"); echo("</script>"); //select db mysql_select_db() or die(mysql_error()); echo("<script type='text/javascript'>"); echo("alert('MYSQL Database Selected.');"); echo("</script>"); //query $result = mysql_query("SELECT * FROM logins") or die(mysql_error()); //check results against given login while($row = mysql_fetch_array($result)){ if($row[login] == $value){ echo("true"); exit(0); } } echo("false"); exit(0); ?>) { login.setCustomValidity("Invalid Login. Please Click 'Register' Below.") } else { login.setCustomValidity("") } }); } the code is in an external js file and the error throws on the last line. Also from reading i understand best practices is to not mix js and php so how would i got about separating them but maintaining the functionality i need? thanks!

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  • HELP A NEWB (AGAIN) PLZ

    - by Ken
    Okay, I must be an idiot, because this is my 3rd question for today. Here's my code: date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "********"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db("`users`, $con) or die(mysql_error()"); $query = ("INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password'))"); mysql_query('$query') or die(mysql_error()); mysql_close($con); echo("Thank you for registering!"); I always get the error returned as: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '$query' at line 1. Help a newbie. I'm about to stab my monitor.

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  • Error loading font file using Imager::Font module in perl

    - by user211808
    use strict; use Imager; use Imager::Font; my $img = Imager->new(); my $file = "D:\\table.png"; $img->open(file=>$file) or die $img->errstr(); # Create smaller version my $thumb = $img->scale(scalefactor=>1.2); my $black = Imager::Color->new( 0, 0, 0 ); my $format; # Autostretch individual channels $thumb->filter(type=>'autolevels'); my $font_filename = "D:\\courbd.ttf"; my $font = Imager::Font->new(file=>$font_filename) or die "Cannot load $font_filename: ", Imager->errstr; for $format ( qw( png gif jpg tiff ppm ) ) { # Check if given format is supported if ($Imager::formats{$format}) { $file.="_low.$format"; print "Storing image as: $file\n"; $thumb->string(x => 50, y => 70, font =>$font, string => "Hello, World!", color => 'red', size => 30, aa => 1); $thumb->write(file=>$file) or die $thumb->errstr; } }

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  • How to loop through an array return from the Query of Mysql

    - by Jerry
    This might be easy for you guys but i could't get it. I have a php class that query the database and return the query result. I assign the result to an array and wants to use it on my main.php script. I have tried to use echo $var[0] or echo $var[1] but the output are 'array' instead of my value. Anyone can help me about this issue? Thanks a lot! My php class <?php class teamQuery { function teamQuery(){ } function getAllTeam(){ $connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if (!$connection) { die("Database connection failed: " . mysql_error()); } $db_select = mysql_select_db(DB_NAME,$connection); if (!$db_select) { die("Database selection failed: " . mysql_error()); } $teamQuery=mysql_query("SELECT * FROM team", $connection); if (!$teamQuery){ die("database has errors: ".mysql_error()); } $ret = array(); while($row=mysql_fetch_array($teamQuery)){ $ret[]=$row; } mysql_free_result($teamQuery); return $ret; } } ?> My php on the main.php $getTeam=new teamQuery(); $team=$getTeam->getAllTeam(); //echo $team[0] or team[1] output 'array' string! // while($team){ // do something } can't work either // How to loop through the values?? Thanks!

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  • How to output multiple rows from an SQL query using the mysqli object

    - by Jonathan
    Assuming that the mysqli object is already instantiatied (and connected) with the global variable $mysql, here is the code I am trying to work with. class Listing { private $mysql; function getListingInfo($l_id = "", $category = "", $subcategory = "", $username = "", $status = "active") { $condition = "`status` = '$status'"; if (!empty($l_id)) $condition .= "AND `L_ID` = '$l_id'"; if (!empty($category)) $condition .= "AND `category` = '$category'"; if (!empty($subcategory)) $condition .= "AND `subcategory` = '$subcategory'"; if (!empty($username)) $condition .= "AND `username` = '$username'"; $result = $this->mysql->query("SELECT * FROM listing WHERE $condition") or die('Error fetching values'); $this->listing = $result->fetch_array() or die('could not create object'); foreach ($this->listing as $key => $value) : $info[$key] = stripslashes(html_entity_decode($value)); endforeach; return $info; } } there are several hundred listings in the db and when I call $result-fetch_array() it places in an array the first row in the db. however when I try to call the object, I can't seem to access more than the first row. for instance: $listing_row = new Listing; while ($listing = $listing_row-getListingInfo()) { echo $listing[0]; } this outputs an infinite loop of the same row in the db. Why does it not advance to the next row? if I move the code: $this->listing = $result->fetch_array() or die('could not create object'); foreach ($this->listing as $key => $value) : $info[$key] = stripslashes(html_entity_decode($value)); endforeach; if I move this outside the class, it works exactly as expected outputting a row at a time while looping through the while statement. Is there a way to write this so that I can keep the fetch_array() call in the class and still loop through the records?

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  • Is it best to make fewer calls to the database and output the results in an array?

    - by Jonathan
    I'm trying to create a more succinct way to make hundreds of db calls. Instead of writing the whole query out every time I wanted to output a single field, I tried to port the code into a class that did all the query work. This is the class I have so far: class Listing { /* Connect to the database */ private $mysql; function __construct() { $this->mysql = new mysqli(DB_LOC, DB_USER, DB_PASS, DB) or die('Could not connect'); } function getListingInfo($l_id = "", $category = "", $subcategory = "", $username = "", $status = "active") { $condition = "`status` = '$status'"; if (!empty($l_id)) $condition .= "AND `L_ID` = '$l_id'"; if (!empty($category)) $condition .= "AND `category` = '$category'"; if (!empty($subcategory)) $condition .= "AND `subcategory` = '$subcategory'"; if (!empty($username)) $condition .= "AND `username` = '$username'"; $result = $this->mysql->query("SELECT * FROM listing WHERE $condition") or die('Error fetching values'); $info = $result->fetch_object() or die('Could not create object'); return $info; } } This makes it easy to access any info I want from a single row. $listing = new Listing; echo $listing->getListingInfo('','Books')->title; This outputs the title of the first listing in the category "Books". But if I want to output the price of that listing, I have to make another call to getListingInfo(). This makes another query on the db and again returns only the first row. This is much more succinct than writing the entire query each time, but I feel like I may be calling the db too often. Is there a better way to output the data from my class and still be succinct in accessing it (maybe outputting all the rows to an array and returning the array)? If yes, How?

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  • Storing data from database [mysql_num_rows]

    - by user1717305
    So I have this code to pass items from database to my order table. When I'm echoing the session. The session variable contains something so there's no problem with that. But when I echo those variables under numrows, it only shows nothing. Is there something wrong? <?php error_reporting(E_ALL ^ E_NOTICE); session_start(); require("connect.php"); $UserID = $_SESSION['CustNum']; $UserN = $_SESSION['UserName']; $ProdGTotal = $_SESSION['ProdGTotal']; $queryord = mysql_query("SELECT * FROM customer WHERE UserName = '$UserN'"); $numrows = mysql_num_rows($queryord); if(numrows == 1){ $row = mysql_fetch_assoc($queryord)or die ('Unable to run query:'.mysql_error()); // fetch associated: get function from a query for a database $dbstreet = $row['Street']; $dhousenum = $row['HouseNum']; $dbcnum = $row['CelNum']; $dbarea = $row['Area']; $dbbuilding = $row['Building']; $dbcity = $row['City']; $dbpnum = $row['PhoneNum']; $dbfname = $row['FName']; $dblname = $row['LName']; } else die(mysql_error()); $query4=mysql_query("INSERT INTO orderdetails VALUES ('', '$UserID', Now(), '$dbhousenum', '$dbstreet', '$dbarea', '$dbbuilding', '$dbcity', '$dbfname', '$dblname', '$dbcnum', '$dbpnum', '$ProdGTotal')",$connect); if ($query4){ header("location:index.php"); } else die(mysql_error()); ?>

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  • Server 2012 AD-DS Setup Fails (Microsoft.Directory.Services.Deployment.DeepTasks.DeepTasks not found)

    - by Daniel Steiner
    Good Morning everyone, I am currently trying to promote my 2012 Server to a Domain Controller but when I am at the first step in the setup I get the Error Message (German, Original Message): [Bereitstellungskonfiguration] Fehler bei der Bestimmung, ob der Zielserver bereits ein Domänencontroller ist: Der Typ [Microsoft.Directory.Services.Deployment.DeepTasks.DeepTasks] wurde nicht gefunden: Vergewissern Sie sich, dass die Assembly, die diesen Typ enthält, geladen ist. (Translated to English): Error while determining, if the Targetserver already is a Domain Controller: The Type [Microsoft.Directory.Services.Deployment.DeepTasks.DeepTasks] was not found: Make sure, that the assembly, that contains this type, is loaded. Thus I can neither Configure the AD-DS nor deinstall them via Server Manager. Any Help how to fix that problem would be greatly appricieated.

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  • Laptop Akku defekt? [closed]

    - by venturebeats
    Ich habe ein großes Problem und zwar startet mein Laptop akku: akku Dell Inspiron 9400 Gut, dann machte ich ihn aber sofort "aus", ich klappte ihn eben zu, was heißt Energie sparen, er ist nicht heruntergefahren! Eine halbe Stunde später wollte mein Vater noch ran, er war dann auch ca. 5min und dann fuhr der Laptop wie gewöhnlich, wwenn er kein Akku mehr hat, heerunter. Dann wollten wir den Laptop wieder aufladen, doch es brannte nicht einmal die rote LED, was soviel heißt wie der Akku wird geladen. Und der Laptop (übrigens erst 7 Monate alt) kann man nicht startet! Nun meine Frage, hat das was mit dem Aufladekabel zu tun( dort brennt immer noch die grüne Lampe, was soviel heißt, wie dass aufjedenfall von der Seckdose Strom kommt) oder ist etwas am Laptop, bzw. ist er kaputt?

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  • Mapi session exceeds maximum count of type objtMessage

    - by wullxz
    one client (it's allways the same client) has often problems with mapi sessions killed by the exchange server. The Application Eventlog on the exchange logs eventid 9646 with source MSExchangeIS: Die MAPI-Sitzung '/o=xx/ou=Exchange Administrative Group (FYDIBOHF23SPDLT)/cn=Recipients/cn=xxxx' hat die maximal zulässige Anzahl von 250 Objekten vom Typ 'objtMessage' überschritten. The client has no eventlogs logged about this error. I looked for installed Outlook Add-Ins and found the default add-ins from microsoft, an adobe pdf add-in (which I deactivated because it's not needed) and an "Octopus" plugin from telekom. Octopus is a CTI-application that connects to Outlook. My guess is, that Octopus (or its add-in) causes this error because this client has over 1100 contacts. My question is: how can I find out, which application/add-in causes this problem? Edit: I already looked at eventid.net but nothing helped. Edit2: Exchange-Cache-Mode is not used nor are there any shared folders / mailboxes open.

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  • Bandwidth Limit User

    - by user45611
    Hello, i'm saxtor i would like to know how to limit users bandwidth for 10gb per day however i dont want to limit them by ipaddress because if they where to go to an internet cafe the users at the cafe will be restricted with that quota, i need to log them via sockets, example the user request to download a file from http://localhost with there username and password, when they download the file sql will update there bandwidth they used, i have a script here but its not working my buffer doesnt work that rate when a user uses multiple connections thanks for the help!. /** * @author saxtor if you can improve this code email me @saxtorinc.com * @copyright 2010 / /* * CREATE TABLE IF NOT EXISTS max_traffic ( id int(255) NOT NULL AUTO_INCREMENT, limit int(255) NOT NULL, PRIMARY KEY (id) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=0 ; */ //SQL Connection [this is hackable for testing] date_default_timezone_set("America/Guyana"); mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("Quota") or die(mysql_error()); function quota($id) { $result = mysql_query("SELECT `limit` FROM max_traffic WHERE id='$id' ") or die(error_log(mysql_error()));; $row = mysql_fetch_array($result); return $row[0]; } function update_quota($id,$value) { $result = mysql_query("UPDATE `max_traffic` SET `limit`='$value' WHERE id='$id'") or die(mysql_error()); return $value; } if ( quota(1) != 0) $limit = quota(1); else $limit = 0; $multipart = false; //was a part of the file requested? (partial download) $range = $_SERVER["HTTP_RANGE"]; if ($range) { //pass client Range header to rapidshare // _insert($range); $cookie .= "\r\nRange: $range"; $multipart = true; header("X-UR-RANGE-Range: $range"); } $url = 'http://127.0.0.1/puppy.iso'; $filename = basename($url); //octet-stream + attachment = client always stores file header('Content-type: application/octet-stream'); header('Content-Disposition: attachment; filename="'.$filename.'"'); //always included so clients know this script supports resuming header("Accept-Ranges: bytes"); //awful hack to pass rapidshare the premium cookie $user_agent = ini_get("user_agent"); ini_set("user_agent", $user_agent . "\r\nCookie: enc=$cookie"); $httphandle = fopen($url, "r"); $headers = stream_get_meta_data($httphandle); $size = $headers["wrapper_data"][6]; $sizer = explode(' ',$size); $size = $sizer[1]; //let's check the return header of rapidshare for range / length indicators //we'll just pass these to the client foreach ($headers["wrapper_data"] as $header) { $header = trim($header); if (substr(strtolower($header), 0, strlen("content-range")) == "content-range") { // _insert($range); header($header); header("X-RS-RANGE-" . $header); $multipart = true; //content-range indicates partial download } elseif (substr(strtolower($header), 0, strlen("Content-Length")) == "content-length") { // _insert($range); header($header); header("X-RS-CL-" . $header); } } if ($multipart) header('HTTP/1.1 206 Partial Content'); flush(); $speed = 4128; $packet = 1; //this is private dont touch. $bufsize = 128; //this is private dont touch/ $bandwidth = 0; //this is private dont touch. while (!(connection_aborted() || connection_status() == 1) && $size > 0) { while (!feof($httphandle) && $size > 0) { if ($limit <= 0 ) $size = 0; if ( $size < $bufsize && $size != 0 && $limit != 0) { echo fread($httphandle,$size); $bandwidth += $size; } else { if( $limit != 0) echo fread($httphandle,$bufsize); $bandwidth += $bufsize; } $size -= $bufsize; $limit -= $bufsize; flush(); if ($speed > 0 && ($bandwidth > $speed*$packet*103)) { usleep(100000); $packet++; //update_quota(1,$limit); } error_log(update_quota(1,$limit)); $limit = quota(1); //if( $size <= 0 ) // exit; } fclose($httphandle); } exit; ?

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  • Cannot connect remote mysql, Cpanel?

    - by BerkErarslan
    I use centos and cpanel I need to use remote mysql with php. I add the ip which I try connect to server to "Remote Database Access Hosts" on Cpanel. However, when I try to connect server with this code: <?php $link = mysql_connect("server_ip", "xxx", "xxx") or die(mysql_error()); $db = mysql_select_db("xxx", $link) or die (mysql_error()); print_r($db); I have error like this: Warning: mysql_connect() [function.mysql-connect]: Can't connect to MySQL server on 'xxxx' (10060) in C:\AppServ\www\test.php on line 3 Can't connect to MySQL server on '94.138.204.234' (10060) I also try to connect using "server_ip:3306" but it still doesn't work. How Can I solve this problem ?

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  • Need to set mailx variable to specify the From address

    - by user256817
    Running Oracle Linux 5.8 (which is just re-branded RedHat EL 5.8) I must change the From address. But we have scripts that use mailx which cannot be re-written to use any extra flags, so I'd like to use internal variables instead, which I see on the linux.die.net manpage on mailx is an alternative to the -r flag: -r address Sets the From address. Overrides any from variable specified in environment or startup files. Tilde escapes are disabled. The -r address options are passed to the mail transfer agent unless SMTP is used. This option exists for compatibility only; it is recommended to set the from variable directly instead. (Source: http://linux.die.net/man/1/mailx) How can we use these mailx variables? I tried adding this to /root/.mailrc, no go: set [email protected] I also added that to /etc/mail.rc with no gold. So I am turning to you, SuperUsers...

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  • root directory - www or public_html

    - by Phil Jackson
    Is the root directory where all files are kept (directly from accessing from FTP) always "www" or "public_html" depending on what OS? Or is it possible to rename this folder? And if so, what would be unique about this folder to be able to identify it? i.e. currently I just wrote this; my $root; my $ftp = Net::FTP->new($DB_ftpserver, Debug => 0) or die "Cannot connect to some.host.name: $@"; $ftp->login($DB_ftpuser, $DB_ftppass) or die "Cannot login ", $ftp->message; my @list = $ftp->dir; if( scalar @list != 0 ) { foreach( @list ){ if( $_ =~ m/www$/g ){ $root = "www"; last; }elsif( $_ =~ m/public_html$/g ){ $root = "public_html"; last; } } } but would not work if it has a different name. Any help much appreciated.

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