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  • Matching process , issue with query

    - by Blerta Blerta
    i have this code which helps me match two different tables.. now, each of this tables, has a epos_id and a rbpos_id ! I have another table which has pairs of rbpos_id and epos_id, something like: id | epos_id | rbpos_id 1 a3566 465jd 2 hkiyb rbposi When i join this other table, i need to check this condition, i mean, the matching should be done, only and if, the epos_id and rbpos_id of the join i'm doing, have the same id,i mean, belong to the same row.. Here is my current query... Thanks! SELECT retailer.date, retailer.time, retailer.location, retailer.user_id,imovo.mobile_number ". "FROM retailer LEFT JOIN imovo ". " ON addtime(retailer.time, '0:0:50')>imovo.time AND retailer.time <imovo.time AND retailer.date=imovo.date

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  • What is the meanning of 'idx_categories_desc_categories_name' in osCommerce

    - by Sumant
    while working on osCommerce-3 i got the table structure for category & categories_description as CREATE TABLE IF NOT EXISTS `osc_categories` ( `categories_id` int(10) unsigned NOT NULL AUTO_INCREMENT, `categories_image` varchar(255) DEFAULT NULL, `parent_id` int(10) unsigned DEFAULT NULL, `sort_order` int(11) DEFAULT NULL, `date_added` datetime DEFAULT NULL, `last_modified` datetime DEFAULT NULL, PRIMARY KEY (`categories_id`), KEY `idx_categories_parent_id` (`parent_id`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=5 ; CREATE TABLE IF NOT EXISTS `osc_categories_description` ( `categories_id` int(10) unsigned NOT NULL, `language_id` int(10) unsigned NOT NULL, `categories_name` varchar(255) NOT NULL, PRIMARY KEY (`categories_id`,`language_id`), KEY `idx_categories_desc_categories_id` (`categories_id`), KEY `idx_categories_desc_language_id` (`language_id`), KEY `idx_categories_desc_categories_name` (`categories_name`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8; here i am not getting the meanning of indexing "idx_categories_desc_categories_id", "idx_categories_desc_language_id", "idx_categories_desc_categories_name" What is the use of this indexing.What does it mean?

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  • 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

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  • PHP - How to get, and display the biggest values from a database?

    - by Dodi300
    Hello. Can anyone tell me how to get and display the biggest values from a database? I have multiple values in my database with the heading "gmd", but how would I get only the first 3 biggest ones to be displayed? How would I do it in this example: $query = "SELECT gmd FROM account"; $result = mysql_query($query); while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { } Thanks.

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  • 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.

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  • 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

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  • how to link table to table

    - by Niño Seymour L. Rodriguez
    I am a comsci student and I'm taking up database now. I got a problem in or should I say I dont know how to link table to table. It is not like you'll just use a foreign key and connect it to the primary key. The outcome should be like this: In the table Course there are three fields namely "course_id", "Description" and "subjects". When you click the name field Subject, a table named Subject should appear. Can you help me with this? hope you understnd my grammar, hehe..im not good in english......it will be a big help if you can answer it.........thank you po..............

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  • Javascript Confirm Delete in One PHP File (on href)

    - by gamerzfuse
    <p><span class="linky"><a href="deletephone.php?id=' . $row['id'] . '">Delete Phone</a></span></p><br /> I have the above code that I am using to link to a delete script. I want to somehow incorporate Javascript with a simple onclick confirmation. This way if they choose OK, I can run the code to delete the item from the database, but if they choose Cancel then I can cancel the operation and do nothing. I have tried a whole variety of functions with changing the window.location to the delete file, and trying to cancel the href= if they choose Cancel, but it always goes to the link regardless of what the user clicks. I would like to be able to keep the delete functions inside the same PHP file if possible, but this is not necessary at all. Thanks in advance! ASIDE: If there is a simple PHP way to check IF the alert was confirmed or denied, that could work also. Any way to check what the user chooses and then run my simple delete PHP command.

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  • to take values of checkbox in table attributes

    - by mwj
    i have a database patient with 3-4 tables n each table has about 8 attributes.... i have a table medical history which has attribute additional info ... under which i have 5 checkboxes.... all the values entered are taken up except the chekbox values..... plz help

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  • 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?

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  • User Getting Logged Out After Making First Comment

    - by John
    Hello, I am using a login system that works well. I am also using a comment system. The comment function does not show up unless the user is logged in (as shown in commentformonoff.php below). When a user makes a comment, the info is passed from the function "show_commentbox" to the file comments2a.php. Then, the info is passed to the file comments2.php. When the site is first pulled up on a browser, after logging in and making a comment, the user is logged out. After logging in a second time during the same browser session, the user is no longer logged out after making a comment. How can I keep the user logged in after making the first comment? Thanks in advance, John Commentformonoff.php: <?php if (!isLoggedIn()) { if (isset($_POST['cmdlogin'])) { if (checkLogin($_POST['username'], $_POST['password'])) { show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl); } else { echo "<div class='logintocomment'>Login to comment</div>"; } } else { echo "<div class='logintocomment'>Login to comment</div>"; } } else { show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl); } ?> Function "show_commentbox": function show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl) { echo '<form action="http://www...com/.../comments/comments2a.php" method="post"> <input type="hidden" value="'.$_SESSION['loginid'].'" name="uid"> <input type="hidden" value="'.$_SESSION['username'].'" name="u"> <input type="hidden" value="'.$submissionid.'" name="submissionid"> <input type="hidden" value="'.stripslashes($submission).'" name="submission"> <input type="hidden" value="'.$url.'" name="url"> <input type="hidden" value="'.$submittor.'" name="submittor"> <input type="hidden" value="'.$submissiondate.'" name="submissiondate"> <input type="hidden" value="'.$countcomments.'" name="countcomments"> <input type="hidden" value="'.$dispurl.'" name="dispurl"> <label class="addacomment" for="title">Add a comment:</label> <textarea class="checkMax" name="comment" type="comment" id="comment" maxlength="1000"></textarea> <div class="commentsubbutton"><input name="submit" type="submit" value="Submit"></div> </form> '; } Included in comments2a.php: $uid = mysql_real_escape_string($_POST['uid']); $u = mysql_real_escape_string($_POST['u']); $query = sprintf("INSERT INTO comment VALUES (NULL, %d, %d, '%s', NULL)", $uid, $subid, $comment); mysql_query($query) or die(mysql_error()); $lastcommentid = mysql_insert_id(); header("Location: comments2.php?submission=".$submission."&submissionid=".$submissionid."&url=".$url."&submissiondate=".$submissiondate."&comment=".$comment."&subid=".$subid."&uid=".$uid."&u=".$u."&submittor=".$submittor."&countcomments=".$countcomments."&dispurl=".$dispurl."#comment-$lastcommentid"); exit(); Included in comments2.php: if($_SERVER['REQUEST_METHOD'] == "POST"){header('Location: http://www...com/.../comments/comments2.php?submission='.$submission.'&submissionid='.$submissionid.'&url='.$url.'&submissiondate='.$submissiondate.'&submittor='.$submittor.'&countcomments='.$countcomments.'&dispurl='.$dispurl.'');} $uid = mysql_real_escape_string($_GET['uid']); $u = mysql_real_escape_string($_GET['u']);

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  • 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>

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  • 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>

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  • 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.

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  • How do I select distinct rows where a column may have a number of the same values but all their 2nd

    - by Martin Rose
    I have a table in the form: test_name| test_result | test1 | pass | test2 | fail | test1 | pass | test1 | pass | test2 | pass | test1 | pass | test3 | pass | test3 | fail | test3 | pass | As you can see all test1's pass while test2's and test3's have both passes and fails. Is there a SQL statement that I can use to return the distinct names of the tests that only pass? E.g. test1

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  • 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();

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  • Jquery/Javascript gmail style stuff for message inbox, such as select all message using checkbox etc

    - by Psychonetics
    I am enjoying the fact that I'm here building a private message inbox for my website after building a full user signup/login and activation system when a few months ago I thought I wouldn't have enough patience to learn this stuff. Anyway to my question. I am currently building the private message inbox for my users and wondering if there are any jquery/javascript stuff I can use to make my inbox more like the gmail inbox. E.G. Gmail allows you to select all read messages or unread or starred or unstarred or none of the messages using a checkbox. I would like to add this kind of feature to my website and I'm sure the easiest way to achieve this would be using a jquery/javascript script. I would appreciate if someone could provide some links or info to where I can find several of these types of scripts to use with my inbox page. Thanks EDIT: Would also like to note that I would like the checkbox to be in a dropdown just like gmails.

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  • 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?

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  • 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.

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  • How to get the value of a field in PHP?

    - by user272899
    I need to get the value of a field; I think I am along the right lines but not quite sure this is the proper code. The "Delete Movie" button is where I am trying to get the value of that row like so: value="'.$row['id'].'" Can you help? <?php //connect to database mysql_connect($mysql_hostname,$mysql_user,$mysql_password); @mysql_select_db($mysql_database) or die("<b>Unable to connect to specified database</b>"); //query databae $query = "select * from movielist"; $result=mysql_query($query) or die('Error, insert query failed'); $row=0; $numrows=mysql_num_rows($result); echo "<table border=1>"; echo "<tr> <td>ID</td> <td>Type</td> <td>Title</td> <td>Description</td> <td>Imdb URL</td> <td>Year</td> <td>Genre</td> <td>Actions</td> </tr>"; while($row<$numrows) { $id=mysql_result($result,$row,"id"); $type=mysql_result($result,$row,"type"); $title=mysql_result($result,$row,"title"); $description=mysql_result($result,$row,"description"); $imdburl=mysql_result($result,$row,"imdburl"); $year=mysql_result($result,$row,"year"); $genre=mysql_result($result,$row,"genre"); ?> <tr> <td><?php echo $id; ?></td> <td><?php echo $type; ?></td> <td><?php echo $title; ?></td> <td><?php echo $description; ?></td> <td><?php echo $imdburl; ?></td> <td><?php echo $year; ?></td> <td><?php echo $genre; ?></td> <td> <!-- Delete Movie Button --> <form style="display: inline;" action="delete/" method="post" onsubmit="return movie_delete()"> <input type="hidden" name="moviedeleteid" value="'.$row['id'].'"> <button type="submit" class="tooltip table-button ui-state-default ui-corner-all" title="Delete trunk"><span class="ui-icon ui-icon-trash"></span></button> </form> </td> </tr> <?php $row++; } echo "</table>"; ?>

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  • Rails advanced queries with join and sum calculation

    - by Dustin Brewer
    I have two models: companies and expenses. Companies have many expenses and expenses belong to companies. My expense model has an 'amount' column. I was wondering if there is a way to perform a find based on a date range and the amount column of the expenses. Something like top 3 companies by total expense amounts over a 7 day period. I've tried for the better part of the day to get this to work, I've attempted joins, chaining named scopes, raw sql, etc. and I'm not having any luck. Thanks for the help.

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