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• One on One table relation - is it harmful to keep relation in both tables?

  - by EBAGHAKI

  I have 2 tables that their rows have one on one relation.. For you to understand the situation, suppose there is one table with user informations and there is another table that contains a very specific informations and each user can only link to one these specific kind of informations ( suppose second table as characters ) And that character can only assign to the user who grabs it, Is it against the rules of designing clean databases to hold the relation key in both tables? User Table: user_id, name, age, character_id Character Table: character_id, shape, user_id I have to do it for performance, how do you think about it?

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• Easy way to compute how close an auto_increment is to its maximum value?

  - by David M

  So yesterday we had a table that has an auto_increment PK for a smallint that reached its maximum. We had to alter the table on an emergency basis, which is definitely not how we like to roll. Is there an easy way to report on how close each auto_increment field that we use is to its maximum? The best way I can think of is to do a SHOW CREATE TABLE statement, parse out the size of the auto-incremented column, then compare that to the AUTO_INCREMENT value for the table. On the other hand, given that the schema doesn't change very often, should I store information about the columns' maximum values and get the current AUTO_INCREMENT with SHOW TABLE STATUS?

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• Is this a secure way to structure a mysql_query in PHP

  - by Supernovah

  I have tried and tried to achieve an SQL injection by making custom queries to the server outside of firefox. Inside the php, all variables are passed into the query in a string like this. Note, by this stage, $_POST has not been touched. mysql_query('INSERT INTO users (password, username) VALUES(' . sha1($_POST['password']) . ',' . $_POST['username'] . ')); Is that a secure way to make a change?

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• TV Guide script - getting current date programmes to show

  - by whitstone86

  This is part of my TV guide script: //Connect to the database mysql_connect("localhost","root","PASSWORD"); //Select DB mysql_select_db("mytvguide"); //Select only results for today and future $result = mysql_query("SELECT programme, channel, episode, airdate, expiration, setreminder FROM mediumonair where airdate >= now()"); The episodes show up, so there are no issues there. However, it's getting the database to find data that's the issue. If I add a record for a programme that airs today this should show: Medium showing on TV4 8:30pm "Episode" Set Reminder Medium showing on TV4 May 18th - 6:25pm "Episode 2" Set Reminder Medium showing on TV4 May 18th - 10:25pm "Episode 3" Set Reminder Medium showing on TV4 May 19th - 7:30pm "Episode 3" Set Reminder Medium showing on TV4 May 20th - 1:25am "Episode 3" Set Reminder Medium showing on TV4 May 20th - 6:25pm "Episode 4" Set Reminder but this shows instead: Medium showing on TV4 May 18th - 6:25pm "Episode 2" Set Reminder Medium showing on TV4 May 18th - 10:25pm "Episode 3" Set Reminder Medium showing on TV4 May 19th - 7:30pm "Episode 3" Set Reminder Medium showing on TV4 May 20th - 1:25am "Episode 3" Set Reminder Medium showing on TV4 May 20th - 6:25pm "Episode 4" Set Reminder I almost have the SQL working; just not sure what the right code is here, to avoid the second mistake showing - as the record (which indicates a show currently airing) does not seem to work at present. Please can anyone help me with this? Thanks

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• Tracking Votes and only allowing 1 vote per member

  - by MikeAdams

  What I'm trying to do is count the votes when someone votes on a "page". I think I lost myself trying to figure out how to track when a member votes or not. I can't seem to get the code to tell when a member has voted. //Generate code ID $useXID = intval($_GET['id']); $useXrank = $_GET['rank']; //if($useXrank!=null && $useXID!=null) { $rankcheck = mysql_query('SELECT member_id,code_id FROM code_votes WHERE member_id="'.$_MEMBERINFO_ID.'" AND WHERE code_id="'.$useXID.'"'); if(!mysql_fetch_array($rankcheck) && $useXrank=="up"){ $rankset = mysql_query('SELECT * FROM code_votes WHERE member_id="'.$_MEMBERINFO_ID.'"'); $ranksetfetch = mysql_fetch_array($rankset); $rankit = htmlentities($ranksetfetch['ranking']); $rankit+="1"; mysql_query("INSERT INTO code_votes (member_id,code_id) VALUES ('$_MEMBERINFO_ID','$useXID')") or die(mysql_error()); mysql_query("UPDATE code SET ranking = '".$rankit."' WHERE ID = '".$useXID."'"); } elseif(!mysql_fetch_array($rankcheck) && $useXrank=="down"){ $rankset = mysql_query('SELECT * FROM code_votes WHERE member_id="'.$_MEMBERINFO_ID.'"'); $ranksetfetch = mysql_fetch_array($rankset); $rankit = htmlentities($ranksetfetch['ranking']); $rankit-="1"; mysql_query("INSERT INTO code_votes (member_id,code_id) VALUES ('$_MEMBERINFO_ID','$useXID')") or die(mysql_error()); mysql_query("UPDATE code SET ranking = '".$rankit."' WHERE ID = '".$useXID."'"); } // hide vote links since already voted elseif(mysql_fetch_array($rankcheck)){$voted="true";} //}

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• If we export the table into another DB then does its all triggers also gets exported along with it?

  - by OM The Eternity

  If we export the table into another DB then does its all triggers also gets exported along with it?

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• Should I change $_REQUEST to $_POST

  - by Scarface

  Hey guys quick question, I have a checkbox system where a list of items can be checked and deleted on the click of a button. I currently use request and it does the job but I was wondering if $_REQUEST was some sort of security risk or improper. If anyone has any advice I would appreciate it. Should I change to $_POST? If so, what is the best way to go about it? foreach ($_REQUEST as $key=>$value) { if (substr($key,0,3)==="img") { $id = substr($key,3); if(isset($_REQUEST['Delete'])) { $sql = 'SELECT file_name,username FROM images WHERE id=?'; $stmt = $conn->prepare($sql); $result=$stmt->execute(array($id)); while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) { $image=$row['file_name']; $user=$row['username']; $myFile = "$user/images/$image"; unlink($myFile); } <input id=\"img".$id."\" name=\"img".$id."\" type=\"checkbox\">

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• PHP: How to get the days of the week?

  - by fwaokda

  I'm wanting to store items in my database with a DATE value for a certain day. I don't know how to get the current Monday, or Tuesday, etc. from the current week. Here is my current database setup. menuentry id int(10) PK menu_item_id int(10) FK day_of_week date message varchar(255) So I have a class setup that holds all the info then I was going to do something like this... foreach ( $menuEntryArray as $item ) { if ( $item->getDate() == [DONT KNOW WHAT TO PUT HERE] ) { // code to display menu_item information } } So I'm just unsure what to put in "[DONT KNOW WHAT TO PUT HERE]" to compare to see if the date is specified for this week's Monday, or Tuesday, etc. The foreach above runs for each day of the week - so it'll look like this... Monday Item 1 Item 2 Item 3 Tuesday Item 1 Wednesday Item 1 Item 2 ... Thanks!

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• Stopping users posting more than once

  - by user342391

  Before posting my form I am checking the database to see if there are any previous posts from the user. If there are previous posts then the script will kick back a message saying you have already posted. The problem is that what I am trying to achieve isn't working it all goes wrong after my else statement. It is also probable that there is an sql injection vulnerability too. Can you help??4 <?php include '../login/dbc.php'; page_protect(); $customerid = $_SESSION['user_id']; $checkid = "SELECT customerid FROM content WHERE customerid = $customerid"; if ($checkid = $customerid) {echo 'You cannot post any more entries, you have already created one';} else $sql="INSERT INTO content (customerid, weburl, title, description) VALUES ('$_POST[customerid]','$_POST[webaddress]','$_POST[pagetitle]','$_POST[pagedescription]')"; if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } echo "1 record added"; ?>

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• SQL with codition on calculated value

  - by user619893

  I have a table with products, their amount and their price. I need to select all entries where the average price per article is between a range. My query so far: SELECT productid,AVG(SUM(price)/SUM(amount)) AS avg FROM stock WHERE avg=$from AND avg<=$to GROUP BY productid If do this, it tells me avg doesnt exist. Also i obviously need to group by because the sum and average need to be per wine

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• What SQL query should I perform to get the result set expected?

  - by texai

  What SQL query should I perform to get the result set expected, giving the first element of the chain (2) as input data, or any of them ? table name: changes +----+---------------+---------------+ | id | new_record_id | old_record_id | +----+---------------+---------------+ | 1| 4| 2| | -- non relevant data -- | | 6| 7| 4| | -- non relevant data -- | | 11| 13| 7| | 12| 14| 13| | -- non relevant data -- | | 31| 20| 14| +----+---------------+---------------+ Result set expected: +--+ | 2| | 4| | 7| |13| |14| |20| +--+ I know I should consider change my data model, but: What if I couldn't? Thank you in advance!

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• select distinct over specific columns

  - by Midhat

  A query in a system I maintain returns QID AID DATA 1 2 x 1 2 y 5 6 t As per a new requirement, I do not want the (QID, AID)=(1,2) pair to be repeated. We also dont care what value is selected from "data" column. either x or y will do. What I have done is to enclose the original query like this SELECT * FROM (<original query text>) Results group by QID,AID Is there a better way to go about this? The original query uses multiple joins and unions and what not, So I would prefer not to touch it unless its absolutely necesary

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• Php random row help...

  - by Skillman

  I've created some code that will return a random row, (well, all the rows in a random order) But i'm assuming its VERY uneffiecent and is gonna be a problem in a big database... Anyone know of a better way? Here is my current code: $count3 = 1; $count4 = 1; //Civilian stuff... $query = ("SELECT * FROM `*Table Name*` ORDER BY `Id` ASC"); $result = mysql_query($query); while($row = mysql_fetch_array($result)) { $count = $count + 1; $civilianid = $row['Id']; $arrayofids[$count] = $civilianid; //echo $arrayofids[$count]; } while($alldone != true) { $randomnum = (rand()%$count) + 1; //echo $randomnum . "<br>"; //echo $arrayofids[$randomnum] . "<br>"; $currentuserid = $arrayofids[$randomnum]; $count3 += 1; while($count4 < $count3) { $count4 += 1; $currentarrayid = $listdone[$count4]; //echo "<b>" . $currentarrayid . ":" . $currentuserid . "</b> "; if ($currentarrayid == $currentuserid){ $found = true; //echo " '" .$found. "' "; } } if ($found == true) { //Reset array/variables... $count4 = 1; $found = false; } else { $listdone[$count3] = $currentuserid; //echo "<u>" . $count3 .";". $listdone[$count3] . "</u> "; $query = ("SELECT * FROM `*Tablesname*` WHERE Id = '$currentuserid'"); $result = mysql_query($query); $row = mysql_fetch_array($result); $username = $row['Username']; echo $username . "<br>"; $count4 = 1; $amountdone += 1; if ($amountdone == $count) { //$count $alldone = true; } } } Basically it will loop until its gets an id (randomly) that hasnt been chosen yet. -So the last username could take hours :P Is this 'bad' code? :P :(

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• why this query result return wrong id ,name

  - by mohamed

  select id,name,type,MAX(vote) as maxvote from channelsvote group by type

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• Is it expensive to hold on to PreparedStatements? (Java & JDBC)

  - by sbook

  I'm trying to figure out if it's efficient for me to cache all of my statements when I create my database connection or if I should only create those that are most used and create the others if/when they're needed.. It seems foolish to create all of the statements in all of the client threads. Any feedback would be greatly appreciated.

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• MOD_REWRITE HELP!

  - by shahinkian

  I want to use mode rewrite to display the following: mydomain.com/Florida/Tampa/ instead of mydomain.com/place.php?state=Florida&city=Tampa I've akready done this: (since I think it might make a difference!) mydomain.com/[name].html instead of mydomain.com/profile?user=[name] Here is the code! Options +FollowSymLinks Options +Indexes RewriteEngine On RewriteBase / RewriteCond %{SCRIPT_FILENAME}! !-f RewriteCond %{SCRIPT_FILENAME}! !-d RewriteRule (.*).html profile.php?user=$1 [QSA.L]

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• Adding to database. No repeat on refresh

  - by kevstarlive

  I have this code: Episode.php <?$feedback = new feedback; $articles = $feedback->fetch_all(); if (isset($_POST['name'], $_POST['post'])) { $cast = $_GET['id']; $name = $_POST['name']; $email = $_POST['email']; $post = nl2br ($_POST['post']); $ipaddress = $_SERVER['REMOTE_ADDR']; if (empty($name) or empty($post)) { $error = 'All Fields Are Required!'; }else{ $query = $pdo->prepare('INSERT INTO comments (cast, name, email, post, ipaddress) VALUES(?, ?, ?, ?, ?)'); $query->bindValue(1, $cast); $query->bindValue(2, $name); $query->bindValue(3, $email); $query->bindValue(4, $post); $query->bindValue(5, $ipaddress); $query->execute(); } }?> <div align="center"> <strong>Give us your feedback?</strong><br /><br /> <?php if (isset($error)) { ?> <small style="color:#aa0000;"><?php echo $error; ?></small><br /><br /> <?php } ?> <form action="episode.php?id=<?php echo $data['cast_id']; ?>" method="post" autocomplete="off" enctype="multipart/form-data"> <input type="text" name="name" placeholder="Name" /> / <input type="text" name="email" placeholder="Email" /><small style="color:#aa0000;">*</small><br /><br /> <textarea rows="10" cols="50" name="post" placeholder="Comment"></textarea><br /><br /> <input type="submit" onclick="myFunction()" value="Add Comment" /> <br /><br /> <small style="color:#aa0000;">* <b>Email will not be displayed publicly</b></small><br /> </form> </div> Include.php class feedback { public function fetch_all(){ global $pdo; $query = $pdo->prepare("SELECT * FROM comments"); $query->bindValue(1, $cast); $query->execute(); return $query->fetchAll(); } } This code updates to the database as it is suppose to. But after submission it reloads the current page as mentioned in the form action. But when I refresh the page to see the comment being added it asks to re submit. If I hit submit then the comment adds again. How can I stop this from happening? Maybe I could hide the comment box and display a thank you message but that would not stop a repeat entry. Please help. Thank you. Kev

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• Java: Do something on event in SQL Database?

  - by wretrOvian

  Hello I'm building an application with distributed parts. Meaning, while one part (writer) maybe inserting, updating information to a database, the other part (reader) is reading off and acting on that information. Now, i wish to trigger an action event in the reader and reload information from the DB whenever i insert something from the writer. Is there a simple way about this? Would this be a good idea? : // READER while(true) { connect(); // reload info from DB executeQuery("select * from foo"); disconnect(); }

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• Can't use method return value in write context; Not sure where to go from here

  - by Morgan Green

  This is my source for the variable. <?php if ($admin->get_permissions()=3) echo 'Welcome to the Admin Panel'; else echo 'Sorry, You do not have access to this page'; ?> And the code that I'm actually trying to call with the if statement is: public function get_permissions() { $username = $_SESSION['admin_login']; global $db; $info = $db->get_row("SELECT `permissions` FROM `user` WHERE `username` = '" . $db->escape($username) . "'"); if(is_object($info)) return $info->permissions; else return ''; } This should be a simple way to call my pages that the user is authorized for by using an else if statement. Or So I thought

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• Error during data INSERT in php

  - by nectar

  here my code- $sql = "INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status') VALUES "; for($i=0; $i0) { $sql .= ", "; } $sql .= "('$pin[$i]', '$ownerid', 'Free', '1')"; } $sql .= ";"; echo $sql; mysql_query($sql); if(mysql_affected_rows() 0) { echo "done"; } else { echo "Fail"; } output: ** INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status') VALUES ('13837927', 'admin', 'Free', '1'), ('59576082', 'admin', 'Free', '1'); Fail why it is not inserting values when $sql query is right?

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• How to send the my sql result to controller in codeigniter as a variable

  - by udaya

  Hi In my mode I am selecting a field as $query1 = $this->db->query("SELECT dPassword FROM tbl_login WHERE dEmailID='[email protected]'"); How to return dpassword as a variable to my controller I tried this way return dpassword;

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• Problem with sending "SetCookie" first in php code

  - by Camran

  According to this manual: http://us2.php.net/setcookie I have to set the cookie before anything else. Here is my cookie code: if (isset($_COOKIE['watched_ads'])){ $expir = time()+1728000; //20 days $ad_arr = unserialize($_COOKIE['watched_ads']); $arr_elem = count($ad_arr); if (in_array($ad_id, $ad_arr) == FALSE){ if ($arr_elem>10){ array_shift($ad_arr); } $ad_arr[]=$ad_id; setcookie('watched_ads', serialize($ad_arr), $expir, '/'); } } else { $expir = time()+1728000; //20 days $ad_arr[] = $ad_id; setcookie('watched_ads', serialize($ad_arr), $expir, '/'); } As you can see I am using variables in setting the cookie. The variables comes from a mysql_query and I have to do the query first. But then, if I do, I will get an error message: Cannot modify header information - headers already sent by ... The error points to the line where I set the cookie above. What should I do?a

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• Turning user ID into name (seperate tables) in PHP

  - by mobile

  I am currently trying to display the username of people who i am following, the problem is that during the following process, only the ID of me and the person i'm following is stored. I've got it to the point where the ID's are displayed but i'd like to show the names hyperlinked. $p_id is the profile ID. Here's what I've got: $following = mysql_query("SELECT `follower`, `followed` FROM user_follow WHERE follower=$p_id"); I am following: <?php while($apple = mysql_fetch_array($following)){ echo '<a href="'.$apple['followed'].'">+'.$apple['followed'].'</a> '; }?> The usernames are in a different table "users" under the field "username" - I need them to match up with the ID's that are currently displayed, and be displayed. Any help appreciated, thanks guys

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• is Payment table needed when you have an invoice table like this?

  - by EBAGHAKI

  this is my invoice table: Invoice Table: invoice_id creation_date due_date payment_date status enum('not paid','paid','expired') user_id total_price I wonder if it's Useful to have a payment table in order to record user payments for invoices. payment table can be like this: payment_id payment_date invoice_id price_paid status enum('successful', 'not successful')

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• How grouping and totaling are done into three tables using JOIN

  - by text

  Here are my tables respondents: field sample value respondentid : 1 age : 2 gender : male survey_questions: id : 1 question : Q1 answer : sample answer answers: respondentid : 1 question : Q1 answer : 1 --id of survey question I want to display all respondents who answered the certain survey, display all answers and total all the answer and group them according to the age bracket. I tried using this query: $sql = "SELECT res.Age, res.Gender, answer.id, answer.respondentid, SUM(CASE WHEN res.Gender='Male' THEN 1 else 0 END) AS males, SUM(CASE WHEN res.Gender='Female' THEN 1 else 0 END) AS females, CASE WHEN res.Age < 1 THEN 'age1' WHEN res.Age BETWEEN 1 AND 4 THEN 'age2' WHEN res.Age BETWEEN 4 AND 9 THEN 'age3' WHEN res.Age BETWEEN 10 AND 14 THEN 'age4' WHEN res.Age BETWEEN 15 AND 19 THEN 'age5' WHEN res.Age BETWEEN 20 AND 29 THEN 'age6' WHEN res.Age BETWEEN 30 AND 39 THEN 'age7' WHEN res.Age BETWEEN 40 AND 49 THEN 'age8' ELSE 'age9' END AS ageband FROM Respondents AS res INNER JOIN Answers as answer ON answer.respondentid=res.respondentid INNER JOIN Questions as question ON answer.Answer=question.id WHERE answer.Question='Q1' GROUP BY ageband ORDER BY res.Age ASC"; I was able to get the data but the listing of all answers are not present. What's wrong with my query. I want to produce something like this: ex: # of Respondents is 3 ages: 2,3 and 6 Question: what are your favorite subjects? Ages 1-4: subject 1: 1 subject 2: 2 subject 3: 2 total respondents for ages 1-4 : 2 Ages 5-10: subject 1: 1 subject 2: 1 subject 3: 0 total respondents for ages 5-10 : 1

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