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  • how to specify a BIGINT in a ruby scaffold?

    - by webdestroya
    I am trying to create a model in ruby that uses a BIGINT datatype (as opposed to the INT done by :integer). I have search all over Google, but all I seem to find is "run an SQL statement to alter the table to a BIGINT" - This seems a bit hack-ish to me, so I wanted to know if there was a way to specify a bigint in the ruby system like :big_int or something Any ideas?

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  • Advanced count and join in Rails

    - by trobrock
    I am try to find the top n number of categories as they relate to articles, there is a habtm relationship set up between the two. This is the SQL I want to execute, but am unsure of how to do this with ActiveRecord, aside from using the find_by_sql method. is there any way of doing this with ActiveRecord methods: SELECT "categories".id, "categories".name, count("articles".id) as counter FROM "categories" JOIN "articles_categories" ON "articles_categories".category_id = "categories".id JOIN "articles" ON "articles".id = "articles_categories".article_id GROUP BY "categories".id ORDER BY counter DESC LIMIT 5;

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  • 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

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  • Fiscal year, quarters, student table, and faculty table... How do I relate these?!

    - by yuudachi
    I have a student and faculty table. The primary key for student is studendID (SID) and faculty's primary key is facultyID, naturally. Student has an advisor column and a requested advisor column, which are foreign key to faculty. That's simple enough, right? However, now I have to throw in dates. I want to be able to view who their advisor was for a certain quarter (such as 2009 Winter) and who they had requested. The result will be a table like this: Year | Term | SID | Current | Requested ------------------------------------------------ 2009 | Winter | 860123456 | 1 | NULL 2009 | Winter | 860445566 | 3 | NULL 2009 | Winter | 860369147 | 5 | 1 And then if I feel like it, I could also go ahead and view a different year and a different term. I am not sure how these new table(s) will look like. Will there be a year table with three columns that are Fall, Spring and Winter? And what will the Fall, Spring, Winter table have? I am new to the art of tables, so this is baffling me... Also, I feel I should clarify how the site works so far now. Admin can approve student requests, and what happens is that the student's current advisor gets overwritten with their request. However, I think I should not do that anymore, right?

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  • 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

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  • 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.

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  • 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?

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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 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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  • 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)'>"; }

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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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  • 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.

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  • 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.

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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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  • How can one fetch partial objects in NHibernate?

    - by mark
    Dear ladies and sirs. I have an object O with 2 fields - A and B. How can I fetch O from the database so that only the field A is fetched? Of course, my real application has objects with many more fields, but two fields are enough to understand the principal. I am using NHibernate 2.1. Thanks.

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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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  • 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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  • Why is the ( ) mandatory in the SQL statement select * from gifts INNER JOIN sentgifts using (giftID

    - by Jian Lin
    Why is the ( ) mandatory in the SQL statement select * from gifts INNER JOIN sentgifts using (giftID); ? The ( ) usually is for specifying grouping of something. But in this case, are we supposed to be able to use 2 or more field names... in the example above, it can be all clear that it is 1 field, is it just that the parser is not made to bypass the ( ) when it is all clear? (such as in the language Ruby).

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  • Use Zip to Pre-Populate City/State Form with jQuery AJAX

    - by Paul
    I'm running into a problem that I can solve fine by just submitting a form and calling a db to retrieve/echo the information, but AJAX seems to be a bit different for doing this (and is what I need). Earlier in a form process I ask for the zip code like so: <input type="text" maxlength="5" size="5" id="zip" /> Then I have a button to continue, but this button just runs a javascript function that shows the rest of the form. When the rest of the form shows, I want to pre-populate the City input with their city, and pre-populate the State dropdown with their state. I figured I would have to find a way to set city/state to variables, and echo the variables into the form. But I can't figure out how to get/set those variables with AJAX as opposed to a form submit. Here's how I did it without ajax: $zip = mysql_real_escape_string($_POST['zip']); $q = " SELECT city FROM citystatezip WHERE zip = $zip"; $r = mysql_query($q); $row = mysql_fetch_assoc($r); $city = $row['city']; Can anybody help me out with using AJAX to set these variables? Thanks!

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  • Calculate the retrieved rows in database Visual C#

    - by Tanya Lertwichaiworawit
    I am new in Visual C# and would want to know how to calculate the retrieved data from a database. Using the above GUI, when "Calculate" is click, the program will display the number of students in textBox1, and the average GPA of all students in textBox2. Here is my database table "Students": I was able to display the number of students but I'm still confused to how I can calculate the average GPA Here's my code: private void button1_Click(object sender, EventArgs e) { string connection = @"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=C:\Database1.accdb"; OleDbConnection connect = new OleDbConnection(connection); string sql = "SELECT * FROM Students"; connect.Open(); OleDbCommand command = new OleDbCommand(sql, connect); DataSet data = new DataSet(); OleDbDataAdapter adapter = new OleDbDataAdapter(command); adapter.Fill(data, "Students"); textBox1.Text = data.Tables["Students"].Rows.Count.ToString(); double gpa; for (int i = 0; i < data.Tables["Students"].Rows.Count; i++) { gpa = Convert.ToDouble(data.Tables["Students"].Rows[i][2]); } connect.Close(); }

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