Search Results

Search found 14310 results on 573 pages for 'mysql sock'.

Page 348/573 | < Previous Page | 344 345 346 347 348 349 350 351 352 353 354 355  | Next Page >

  • 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

  • Better way to do SELECT with GROUP BY

    - by Luca Romagnoli
    Hi i've wrote a query that works: SELECT `comments`.* FROM `comments` RIGHT JOIN (SELECT MAX( id ) AS id, core_id, topic_id FROM comments GROUP BY core_id, topic_id order by id desc) comm ON comm.id = comments.id LIMIT 10 I want know if it is possible (and how) to rewrite it to get better performance. Thanks

    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

  • Query broke down and left me stranded in the woods

    - by user1290323
    I am trying to execute a query that deletes all files from the images table that do not exist in the filters tables. I am skipping 3,500 of the latest files in the database as to sort of "Trim" the table back to 3,500 + "X" amount of records in the filters table. The filters table holds markers for the file, as well as the file id used in the images table. The code will run on a cron job. My Code: $sql = mysql_query("SELECT * FROM `images` ORDER BY `id` DESC") or die(mysql_error()); while($row = mysql_fetch_array($sql)){ $id = $row['id']; $file = $row['url']; $getId = mysql_query("SELECT `id` FROM `filter` WHERE `img_id` = '".$id."'") or die(mysql_error()); if(mysql_num_rows($getId) == 0){ $IdQue[] = $id; $FileQue[] = $file; } } for($i=3500; $i<$x; $i++){ mysql_query("DELETE FROM `images` WHERE id='".$IdQue[$i]."' LIMIT 1") or die("line 18".mysql_error()); unlink($FileQue[$i]) or die("file Not deleted"); } echo ($i-3500)." files deleted."; Output: 0 files deleted. Database contents: images table: 10,000 rows filters table: 63 rows Amount of rows in filters table that contain an images table id: 63 Execution time of php script: 4 seconds +/- 0.5 second Relevant DB structure TABLE: images id url etc... TABLE: filter id img_id (CONTAINS ID FROM images table) etc...

    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

  • Compare structures of two databases?

    - by streetparade
    Hello, I wanted to ask whether it is possible to compare the complete database structure of two huge databases. We have two databases, the one is a development database, the other a production database. I've sometimes forgotten to make changes in to the production database, before we released some parts of our code, which results that the production database doesn't have the same structure, so if we release something we got some errors. Is there a way to compare the two, or synchronize?

    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

  • How do you sort php and sql arrays?

    - by Jon
    How can I sort this array by city or by id in descending order? if ($num > 0 ) { $i=0; while ($i < $num) { $city = mysql_result($result,$i,"city"); $state = mysql_result($result,$i,"state"); $id = mysql_result($result,$i,"id"); echo "$city"; echo "$state"; ++$i; } } else { echo "No results."; } ?>

    Read the article

  • update myqsl table

    - by Simon
    how can i write the query, to update the table videos, and set the value of field name to 'something' where the average is max(), or UPDATE the table, where average has the second value by size!!! i think the query must look like this!!! UPDATE videos SET name = 'something' WHERE average IN (SELECT `average` FROM `videos` ORDER BY `average` DESC LIMIT 1) but it doesn't work!!!

    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

  • 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

  • can you make an sql query for this situation?

    - by saurav
    i have a table as below. name and 10 cities in which he lived during his lifetime. name , city1 , city2 , city3 ,city4 , city5 ,city6 , city7 , city8 , city9 city10 suppose for a particular name i want to fetch other names in table matching with maximum number of cities. for example if i want to fetch other people who have lived in three or more cities lived by this person.

    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

  • 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

  • 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

  • Procedure in converting int to decimal data type?

    - by Fedor
    I have an int(11) column which is used to store money. I read some of the answers on SO and it seems I just need to update it to be a decimal (19,4) data type. Are there any gotchas I should know about before I actually do the converting? My application is in PHP/Zend and I'm not using an ORM so I doubt I would need to update any sort of class to consistently identify the data type.

    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

  • 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

  • how to have defined connection within function for pdo communication with DB

    - by Scarface
    hey guys I just started trying to convert my query structure to PDO and I have come across a weird problem. When I call a pdo query connection within a function and the connection is included outside the function, the connection becomes undefined. Anyone know what I am doing wrong here? I was just playing with it, my example is below. include("includes/connection.php"); function query(){ $user='user'; $id='100'; $sql = 'SELECT * FROM users'; $stmt = $conn->prepare($sql); $result=$stmt->execute(array($user, $id)); // now iterate over the result as if we obtained // the $stmt in a call to PDO::query() while($r = $stmt->fetch(PDO::FETCH_ASSOC)) { echo "$r[username] $r[id] \n"; } } query();

    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

  • Table not Echoing out if another Table has a Zero value

    - by John
    Hello, The table below with mysql_query($sqlStr3) (the one with the word "Joined" in its row) does not echo if the result associated with mysql_query($sqlStr1) has a value of zero. This happens even if mysql_query($sqlStr3) returns a result. In other words, if a given loginid has an entry in the table "login", but not one in the table "submission", then the table associated with mysql_query($sqlStr3) does not echo. I don't understand why the "submission" table would have any effect on mysql_query($sqlStr3), since the $sqlStr3 only deals with another table, called "login", as seen below. Any ideas why this is happening? Thanks in advance, John W. <?php echo '<div class="profilename">User Profile for </div>'; echo '<div class="profilename2">'.$profile.'</div>'; $tzFrom = new DateTimeZone('America/New_York'); $tzTo = new DateTimeZone('America/Phoenix'); $profile = mysql_real_escape_string($_GET['profile']); $sqlStr = "SELECT l.username, l.loginid, s.loginid, s.submissionid, s.title, s.url, s.datesubmitted, s.displayurl FROM submission AS s INNER JOIN login AS l ON s.loginid = l.loginid WHERE l.username = '$profile' ORDER BY s.datesubmitted DESC"; $result = mysql_query($sqlStr); $arr = array(); echo "<table class=\"samplesrec1\">"; while ($row = mysql_fetch_array($result)) { $dt = new DateTime($row["datesubmitted"], $tzFrom); $dt->setTimezone($tzTo); echo '<tr>'; echo '<td class="sitename3">'.$dt->format('F j, Y &\nb\sp &\nb\sp g:i a').'</a></td>'; echo '<td class="sitename1"><a href="http://www.'.$row["url"].'">'.$row["title"].'</a></td>'; echo '</tr>'; } echo "</table>"; $sqlStr1 = "SELECT l.username, l.loginid, s.loginid, s.submissionid, s.title, s.url, s.datesubmitted, s.displayurl, l.created, count(s.submissionid) countSubmissions FROM submission AS s INNER JOIN login AS l ON s.loginid = l.loginid WHERE l.username = '$profile'"; $result1 = mysql_query($sqlStr1); $arr1 = array(); echo "<table class=\"samplesrec2\">"; while ($row1 = mysql_fetch_array($result1)) { echo '<tr>'; echo '<td class="sitename5">Submissions: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$row1["countSubmissions"].'</td>'; echo '</tr>'; } echo "</table>"; $sqlStr2 = "SELECT l.username, l.loginid, c.loginid, c.commentid, c.submissionid, c.comment, c.datecommented, l.created, count(c.commentid) countComments FROM comment AS c INNER JOIN login AS l ON c.loginid = l.loginid WHERE l.username = '$profile'"; $result2 = mysql_query($sqlStr2); $arr2 = array(); echo "<table class=\"samplesrec3\">"; while ($row2 = mysql_fetch_array($result2)) { echo '<tr>'; echo '<td class="sitename5">Comments: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$row2["countComments"].'</td>'; echo '</tr>'; } echo "</table>"; $tzFrom3 = new DateTimeZone('America/New_York'); $tzTo3 = new DateTimeZone('America/Phoenix'); $sqlStr3 = "SELECT created, username FROM login WHERE username = '$profile'"; $result3 = mysql_query($sqlStr3); $arr3 = array(); echo "<table class=\"samplesrec4\">"; while ($row3 = mysql_fetch_array($result3)) { $dt3 = new DateTime($row3["created"], $tzFrom3); $dt3->setTimezone($tzTo3); echo '<tr>'; echo '<td class="sitename5">Joined: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$dt->format('F j, Y').'</td>'; echo '</tr>'; } echo "</table>"; ?> </body> </html>

    Read the article

  • Connecting to 3rd party databse in Joomla!?

    - by Michael
    I need to connect to another database in Joomla! that's on another server. This is for a plugin and I need to pull some data from a table. Now what I don't want is to use this database to run Joomla!, I already have Joomla! installed and running on its own database on its server but I want to connect to another database (ON TOP of the current one) to pull some data, then disconnect from that 3rd party database - all while keeping the original Joomla database connection in tact.

    Read the article

< Previous Page | 344 345 346 347 348 349 350 351 352 353 354 355  | Next Page >