Search Results

Search found 14956 results on 599 pages for 'mysql dba'.

Page 356/599 | < Previous Page | 352 353 354 355 356 357 358 359 360 361 362 363  | Next Page >

  • Preventing spam bots on site?

    - by Mike
    We're having an issue on one of our fairly large websites with spam bots. It appears the bots are creating user accounts and then posting journal entries which lead to various spam links. It appears they are bypassing our captcha somehow -- either it's been cracked or they're using another method to create accounts. We're looking to do email activation for the accounts, but we're about a week away from implementing such changes (due to busy schedules). However, I don't feel like this will be enough if they're using an SQL exploit somewhere on the site and doing the whole cross site scripting thing. So my question to you: If they are using some kind of XSS exploit, how can I find it? I'm securing statements where I can but, again, its a fairly large site and it'd take me awhile to actively clean up SQL statements to prevent XSS. Can you recommend anything to help our situation?

    Read the article

  • How to get rank based on SUM's?

    - by Kenan
    I have comments table where everything is stored, and i have to SUM everything and add BEST ANSWER*10. I need rank for whole list, and how to show rank for specified user/ID. Here is the SQL: SELECT m.member_id AS member_id, (SUM(c.vote_value) + SUM(c.best)*10) AS total FROM comments c LEFT JOIN members m ON c.author_id = m.member_id GROUP BY c.author_id ORDER BY total DESC LIMIT {$sql_start}, 20

    Read the article

  • what is the question for the query?

    - by Kevinniceguy
    Sorry...I mean what question will be for this query? SELECT SUM(price) FROM Room r, Hotel h WHERE r.hotelNo = h.hotelNo and hotelName = 'Paris Hilton' and roomNo NOT IN (SELECT roomNo FROM Booking b, Hotel h WHERE (dateFrom <= CURRENT_DATE AND dateTo >= CURRENT_DATE) AND b.hotelNo = h.hotelNo AND hotelName = 'Paris Hilton');

    Read the article

  • Calculate time from timezones in php

    - by Ramya
    Hai I have the system with employees having different timezones in their profile. I would like to show the date according to their timezones specified. The GMT time zone values are placed in the database. could you guys help me

    Read the article

  • select query from mysql_num_rows

    - by Andi Nugroho
    i want create multiple search where statement $where_search is a multiple condition from post form. but stil error when iam using this code ".where_search." in where condition with mysql_num_rows for paging $tampil2 = mysql_query("SELECT * FROM bb where ".$where_search." and kd_kelompok='2' and kd_komoditi='11' and nm_sebutan IS NOT NULL " ); this is the complete code. $where_search = "kd_pok='2' and kd_komoditi='11' "; if (isset($_POST['lakpus'])) { if (empty($_POST['lakpus'])) { } else { if (empty($where_search)) { $where_search .= "lakpus = '$lakpus' "; } else { $where_search .= "AND lakpus = '$lakpus' "; } } } if (isset($_POST['kd_por'])) { $kd_por = $_POST['kd_por'] ; if (empty($_POST['kd_por'])) { } else { if (empty($where_search)) { $where_search .= "kd_por = '$kd_por' "; } else { $where_search .= "AND tab1.kd_por = '$kd_por' "; } } } $max=15; $tampil2 = mysql_query("SELECT * FROM bb where ".$where_search." and kd_kelompok='2' and kd_komoditi='11' and nm_sebutan IS NOT NULL " ); $jml = mysql_num_rows($tampil2); $jmlhal = ceil($jml/$max);

    Read the article

  • Extending URIs with 2 queries (i.e. 'viewauthorbooks.php?authorid=4' AND 'orderby=returndate") Possi

    - by Jess
    I have a link in my system as displayed above; 'viewauthorbooks.php?authorid=4' which works fine and generates a page displaying the books only associated with the particular author. However I am implementing another feature where the user can sort the columns (return date, book name etc) and I am using the ORDER BY SQL clause. I have this also working as required for other pages, which do not already have another query in the URI. But for this particular page there is already a paramter returned in the URL, and I am having difficulty in extending it. When the user clicks on the a table column title I'm getting an error, and the original author ID is being lost!! This is the URI link I am trying to use: <th><a href="viewauthorbooks.php?authorid=<?php echo $row['authorid']?>&orderby=returndate">Return Date</a></th> This is so that the data can be sorted in order of Return Date. When I run this; the author ID gets lost for some reason, also I want to know if I am using correct layout to have 2 parameters run in the address? Thanks.

    Read the article

  • What's wrong with this SQL query?

    - by ThinkingInBits
    I have two tables: photographs, and photograph_tags. Photograph_tags contains a column called photograph_id (id in photographs). You can have many tags for one photograph. I have a photograph row related to three tags: boy, stream, and water. However, running the following query returns 0 rows SELECT p.* FROM photographs p, photograph_tags c WHERE c.photograph_id = p.id AND (c.value IN ('dog', 'water', 'stream')) GROUP BY p.id HAVING COUNT( p.id )=3 Is something wrong with this query?

    Read the article

  • getting notice like undefined index

    - by user2533308
    $result = mysql_query("SELECT * FROM customers WHERE loginid='$_POST[login]' AND accpassword='$_POST[password]'"); if(mysql_num_rows($result) == 1) { while($recarr = mysql_fetch_array($result)) { $_SESSION[customerid] = $recarr[customerid]; $_SESSION[ifsccode] = $recarr[ifsccode]; $_SESSION[customername] = $recarr[firstname]. " ". $recarr[lastname]; $_SESSION[loginid] = $recarr[loginid]; $_SESSION[accstatus] = $recarr[accstatus]; $_SESSION[accopendate] = $recarr[accopendate]; $_SESSION[lastlogin] = $recarr[lastlogin]; } $_SESSION["loginid"] =$_POST["login"]; header("Location: accountalerts.php"); } else { $logininfo = "Invalid Username or password entered"; } Notice: Undefined index:login and Notice: Undefined index:password try to help me out getting error message in second line

    Read the article

  • sql query question / count

    - by scheibenkleister
    Hi, I have houses that belongs to streets. A user can buy several houses. How do I find out, if the user owns an entire street? street table with columns (id/name) house table with columns (id/street_id [foreign key] owner table with columns (id/house_id/user_id) [join table with foreign keys] So far, I'm using count which returns the result: select count(*), street_id from owner left join house on owner.house_id = house.id group by street_id where user_id = 1 count(*) | street_id 3 | 1 2 | 2 A more general count: select count(*) from house group by street_id returns: count(*) | street_id 3 | 1 3 | 2 How can I find out, that user 1 owns the entire street 1 but not street 2? Thanks.

    Read the article

  • How to get Joomla users data into a json array

    - by sami
    $sql = "SELECT * FROM `jos_users` LIMIT 0, 30 "; $response = array(); $posts = array(); $result=mysql_query($sql); while($row=mysql_fetch_array($result)) { $id=$row['id']; $id=$row['name']; $posts[] = array('id'=> $title, 'name'=> $name); } $response['jos_users'] = $posts; $fp = fopen('results.json', 'w'); fwrite($fp, json_encode($response)); fclose($fp); I want to fetch the user id and name to the json file.i thought id did wrong code.can anyone correct it ?

    Read the article

  • Having a problem displaying data from last inserted data

    - by Gideon
    I'm designing a staff rota planner....have three tables Staff (Staff details), Event (Event details), and Job (JobId, JobDate, EventId (fk), StaffId (fk)). I need to display the last inserted job detail with the staff name. I've been at it for couple of hours and getting nowhere. Thanks for the help in advance. My code is the following: $eventId = $_POST['eventid']; $selectBox = $_POST['selectbox']; $timePeriod = $_POST['time']; $selectedDate = $_POST['date']; $count = count($selectBox); //constructing the staff selection if (empty($selectBox)) { echo "<p>You didn't select any member of staff to be assigned."; echo "<p><input type='button' value='Go Back' onClick='history.go(-1)'>"; } else { echo "<p> You selected ".$count. " staff for this show."; for ($i=0;$i<$count;$i++) { $selectId = $selectBox[$i]; //insert the details into the Job table in the database $insertJob = "INSERT INTO Job (JobDate, TimePeriod, EventId, StaffId) VALUES ('".$selectedDate."', '".$timePeriod."', ".$eventId.", ".$selectId.")"; $exeinsertJob = mysql_query($insertJob) or die (mysql_error()); } } //display the inserted job details $insertedlist = "SELECT Job.JobId, Staff.LastName, Staff.FirstName, Job.JobDate, Job.TimePeriod FROM Staff, Job WHERE Job.StaffId = Staff.StaffId AND Job.EventId = $eventId AND Job.JobDate = ".$selectedDate; $exeinsertlist = mysql_query($insertedlist) or die (mysql_error()); if ($exeinsertlist) { echo "<p><table cellspacing='1' cellpadding='3'>"; echo "<tr><th colspan=5> ".$eventname."</th></tr>"; echo "<tr><th>Job Id</th><th>Last Name</th> <th>First Name </th><th>Date</th><th>Hours</th></tr>"; while ($joblistarray = mysql_fetch_array($exeinsertlist)) { echo "<tr><td align=center>".$joblistarray['JobId']." </td><td align=center>".$joblistarray['LastName']."</td><td align=center>".$joblistarray['FirstName']." </td><td align=center>".$joblistarray['JobDate']." </td><td align=center>".$joblistarray['TimePeriod']."</td></tr>"; } echo "</table>"; echo "<h3><a href=AssignStaff.php>Add More Staff?</a></h3>"; } else { echo "The Job list can not be displayed at this time. Try again."; echo "<p><input type='button' value='Go Back' onClick='history.go(-1)'>"; }

    Read the article

  • PHP Login, Store Session Variables.

    - by Andreas Carlbom
    Yo. I'm trying to make a simple login system in PHP and my problem is this: I don't really understand sessions. Now, when I log a user in, I run session_register("user"); but I don't really understand what I'm up to. Does that session variable contain any identifiable information, so that I for example can get it out via $_SESSION["user"] or will I have to store the username in a separate variable? Thanks.

    Read the article

  • PHP PDO - Num Rows

    - by Ian
    PDO apparently has no means to count the number of rows returned from a select query (mysqli has the num_rows variable). Is there a way to do this, short of using count($results->fetchAll()) ?

    Read the article

  • Unnecessary Error Message Being Displayed

    - by ThatMacLad
    I've set up a form to update my blog and it was working fine up until about this morning. It keeps on turning up with an Invalid Entry ID error on the edit post page when I click the update button despite the fact that it updates the homepage. All help is seriously appreciated. <html> <head> <title>Ultan's Blog | New Post</title> <link rel="stylesheet" href="css/editpost.css" type="text/css" /> </head> <body> <div class="new-form"> <div class="header"> </div> <div class="form-bg"> <?php mysql_connect ('localhost', 'root', 'root') ; mysql_select_db ('tmlblog'); if (isset($_POST['update'])) { $id = htmlspecialchars(strip_tags($_POST['id'])); $month = htmlspecialchars(strip_tags($_POST['month'])); $date = htmlspecialchars(strip_tags($_POST['date'])); $year = htmlspecialchars(strip_tags($_POST['year'])); $time = htmlspecialchars(strip_tags($_POST['time'])); $entry = $_POST['entry']; $title = htmlspecialchars(strip_tags($_POST['title'])); if (isset($_POST['password'])) $password = htmlspecialchars(strip_tags($_POST['password'])); else $password = ""; $entry = nl2br($entry); if (!get_magic_quotes_gpc()) { $title = addslashes($title); $entry = addslashes($entry); } $timestamp = strtotime ($month . " " . $date . " " . $year . " " . $time); $result = mysql_query("UPDATE php_blog SET timestamp='$timestamp', title='$title', entry='$entry', password='$password' WHERE id='$id' LIMIT 1") or print ("Can't update entry.<br />" . mysql_error()); header("Location: post.php?id=" . $id); } if (isset($_POST['delete'])) { $id = (int)$_POST['id']; $result = mysql_query("DELETE FROM php_blog WHERE id='$id'") or print ("Can't delete entry.<br />" . mysql_error()); if ($result != false) { print "The entry has been successfully deleted from the database."; exit; } } if (!isset($_GET['id']) || empty($_GET['id']) || !is_numeric($_GET['id'])) { die("Invalid entry ID."); } else { $id = (int)$_GET['id']; } $result = mysql_query ("SELECT * FROM php_blog WHERE id='$id'") or print ("Can't select entry.<br />" . $sql . "<br />" . mysql_error()); while ($row = mysql_fetch_array($result)) { $old_timestamp = $row['timestamp']; $old_title = stripslashes($row['title']); $old_entry = stripslashes($row['entry']); $old_password = $row['password']; $old_title = str_replace('"','\'',$old_title); $old_entry = str_replace('<br />', '', $old_entry); $old_month = date("F",$old_timestamp); $old_date = date("d",$old_timestamp); $old_year = date("Y",$old_timestamp); $old_time = date("H:i",$old_timestamp); } ?> <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <p><input type="hidden" name="id" value="<?php echo $id; ?>" /> <strong><label for="month">Date (month, day, year):</label></strong> <select name="month" id="month"> <option value="<?php echo $old_month; ?>"><?php echo $old_month; ?></option> <option value="January">January</option> <option value="February">February</option> <option value="March">March</option> <option value="April">April</option> <option value="May">May</option> <option value="June">June</option> <option value="July">July</option> <option value="August">August</option> <option value="September">September</option> <option value="October">October</option> <option value="November">November</option> <option value="December">December</option> </select> <input type="text" name="date" id="date" size="2" value="<?php echo $old_date; ?>" /> <select name="year" id="year"> <option value="<?php echo $old_year; ?>"><?php echo $old_year; ?></option> <option value="2004">2004</option> <option value="2005">2005</option> <option value="2006">2006</option> <option value="2007">2007</option> <option value="2008">2008</option> <option value="2009">2009</option> <option value="2010">2010</option> </select> <strong><label for="time">Time:</label></strong> <input type="text" name="time" id="time" size="5" value="<?php echo $old_time; ?>" /></p> <p><strong><label for="title">Title:</label></strong> <input type="text" name="title" id="title" value="<?php echo $old_title; ?>" size="40" /> </p> <p><strong><label for="password">Password protect?</label></strong> <input type="checkbox" name="password" id="password" value="1"<?php if($old_password == 1) echo " checked=\"checked\""; ?> /></p> <p><textarea cols="80" rows="20" name="entry" id="entry"><?php echo $old_entry; ?></textarea></p> <p><input type="submit" name="update" id="update" value="Update"></p> </form> <p><strong>Be absolutely sure that this is the post that you wish to remove from the blog!</strong><br /> </p> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <input type="hidden" name="id" id="id" value="<?php echo $id; ?>" /> <input type="submit" name="delete" id="delete" value="Delete" /> </form> </div> </div> </div> <div class="bottom"></div> </body> </html>

    Read the article

  • Group / User based security. Table / SQL question

    - by Brett
    Hi, I'm setting up a group / user based security system. I have 4 tables as follows: user groups group_user_mappings acl where acl is the mapping between an item_id and either a group or a user. The way I've done the acl table, I have 3 columns of note (actually 4th one as an auto-id, but that is irrelevant) col 1 item_id (item to access) col 3 user_id (user that is allowed to access) col 3 group_id (group that is allowed to access) So for example item1, peter, , item2, , group1 item3, jane, , so either the acl will give access to a user or a group. Any one line in the ACL table with either have an item - user mapping, or an item group. If I want to have a query that returns all objects a user has access to, I think I need to have a SQL query with a UNION, because I need 2 separate queries that join like.. item - acl - group - user AND item - acl - user This I guess will work OK. Is this how its normally done? Am I doing this the right way? Seems a little messy. I was thinking I could get around it by creating a single user group for each person, so I only ever deal with groups in my SQL, but this seems a little messy as well..

    Read the article

  • SQL conditional row insert

    - by Pablo
    Is it possible to insert a new row if a condition is meet? For example, i have this table with no primary key nor uniqueness +----------+--------+ | image_id | tag_id | +----------+--------+ | 39 | 8 | | 8 | 39 | | 5 | 11 | +----------+--------+ I would like to insert a row if a combination of image_id and tag_id doesn't exists for example; INSERT ..... WHERE image_id!=39 AND tag_id!=8

    Read the article

  • Multitenant shared user account?

    - by jpartogi
    Dear all, Based on your experience, which is the route to go for a multi-tenant user login? One user login per account. Which means if there is one user that has access to multiple account, there will be redundancy of record in the database One user login for all account that she has privileges to. Which means one user record has access to multiple account if she has privileges to that account. From your experience, which one is better and why? I was thinking to choose the latter, but I don't know whether it will cause security issue or less flexibility. Thank you for sharing your experience.

    Read the article

  • Set primary key with two integers

    - by user299196
    I have a table with primary key (ColumnA, ColumnB). I want to make a function or procedure that when passed two integers will insert a row into the table but make sure the largest integer always goes into ColumnA and the smaller one into ColumnB. So if we have SetKeysWithTheseNumbers(17, 19) would return |-----------------| |ColumnA | ColumnB| |-----------------| |19 | 17 | |-----------------| SetKeysWithTheseNumbers(19, 17) would return the same thing |-----------------| |ColumnA | ColumnB| |-----------------| |19 | 17 | |-----------------|

    Read the article

  • declaring constraint to consider prog logic

    - by shantanuo
    I can open a trip only once but can close it multiple times. I can not declare the Trip_no + status as primary key since there can be multiple entries while closing the trip. Is there any way that will assure me that a trip number is opened only once? For e.g. there should not be the second row with "Open" status for trip No. 3 since it is already there in the following table. Trip No | Status 1 Open 1 Close 1 Close 2 Open 2 Close 3 Open 3 Close 3 Close 3 Close 3 Close

    Read the article

  • Can I join two tables whereby the joined table is sorted by a certain column?

    - by Ferdy
    I'm not much of a database guru so I need some help on a query I'm working on. In my photo community project I want to richly visualize tags by not only showing the tag name and counter (# of images inside them), I also want to show a thumb of the most popular image inside the tag (most karma). The table setup is as follow: Image table holds basic image metadata, important is the karma field Imagefile table holds multiple entries per image, one for each format Tag table holds tag definitions Tag_map table maps tags to images In my usual trial and error query authoring I have come this far: SELECT * FROM (SELECT tag.name, tag.id, COUNT(tag_map.tag_id) as cnt FROM tag INNER JOIN tag_map ON (tag.id = tag_map.tag_id) INNER JOIN image ON tag_map.image_id = image.id INNER JOIN imagefile on image.id = imagefile.image_id WHERE imagefile.type = 'smallthumb' GROUP BY tag.name ORDER BY cnt DESC) as T1 WHERE cnt > 0 ORDER BY cnt DESC [column clause of inner query snipped for the sake of simplicity] This query gives me somewhat what I need. The outer query makes sure that only tags are returned for which there is at least 1 image. The inner query returns the tag details, such as its name, count (# of images) and the thumb. In addition, I can sort the inner query as I want (by most images, alphabetically, most recent, etc) So far so good. The problem however is that this query does not match the most popular image (most karma) of the tag, it seems to always take the most recent one in the tag. How can I make sure that the most popular image is matched with the tag?

    Read the article

< Previous Page | 352 353 354 355 356 357 358 359 360 361 362 363  | Next Page >