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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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  • 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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  • 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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  • 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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  • how store date in myqsl database?

    - by Syom
    i have date in dd/mm/yyyy format. how can i store it in databse, if i fant to do some operations on them after? for example i must find out the rows, where date > something what type i must set to date field? thanks

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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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  • login/logout problem in PHP

    - by user356170
    i have a question. I have a website in which i am giving security like login id and password( as usual). No what i want is that, 1) I don't want to allow a single user to login in different machine at the same time. 2) For this i am using a column in database which is keeping the current status of user(i.e. loging/logout). I am allowing user to login only when has session has not closed and status is login. 3) So my problem is that when i am logging out manually. it is closing the session as well as updating the database with status "logout". 4) but when i am closing the window from Cross buttonat top right corner. it is closing the ssion but table data is still "login". so later on i can't be able to login into the same user. 5) So how could i solve this problem. Please help me!

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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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  • 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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  • 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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  • 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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  • 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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  • Can I do this in only one query ?

    - by Paté
    Merry christmas everyone, I Know my way around SQL but I'm having a hard time figuring this one out. First here are my tables (examples) User id name friend from //userid to //userid If user 1 is friend with user 10 then you a row with 1,10. User 1 cannot be friend with user 10 if user 10 is not friend with user 1 so you have 1,10 10,1 It may look weird but I need those two rows per relations. Now I'm trying to make a query to select the users that have the most mutual friend with a given user. For example User 1 is friend with user 10,9 and 7 and user 8 is friend with 10,9 and 7 too ,I want to suggest user 1 to invite him (like facebook). I want to get like the 10 first people with the most mutual friend. The output would be like User,NumOfMutualFriends I dont know if that can be done in a single query ? Thanks in advance for any help.

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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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