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  • PHP + MYSQLI: Variable parameter/result binding with prepared statements.

    - by Brian Warshaw
    In a project that I'm about to wrap up, I've written and implemented an object-relational mapping solution for PHP. Before the doubters and dreamers cry out "how on earth?", relax -- I haven't found a way to make late static binding work -- I'm just working around it in the best way that I possibly can. Anyway, I'm not currently using prepared statements for querying, because I couldn't come up with a way to pass a variable number of arguments to the bind_params() or bind_result() methods. Why do I need to support a variable number of arguments, you ask? Because the superclass of my models (think of my solution as a hacked-up PHP ActiveRecord wannabe) is where the querying is defined, and so the find() method, for example, doesn't know how many parameters it would need to bind. Now, I've already thought of building an argument list and passing a string to eval(), but I don't like that solution very much -- I'd rather just implement my own security checks and pass on statements. Does anyone have any suggestions (or success stories) about how to get this done? If you can help me solve this first problem, perhaps we can tackle binding the result set (something I suspect will be more difficult, or at least more resource-intensive if it involves an initial query to determine table structure).

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  • PHP coding a price comparaison tool

    - by Tristan
    Hello, it's the first time I developp such tool you all know (the possibility to compare articles according to price and/or options) Since I never did that i want to tell me what do you think of the way i see that : On the database we would have : offer / price / option 1 / option 2 / option 3 / IDseller / IDoffer best buy / 15$ / full FTP / web hosting / php.ini / 10 / 1 .../..../.... And the request made by the client : "SELECT * FROM offers WHERE price <= 20 AND option1 = fullFTP"; I don't know if it seems OK to you. Plus i was wondering, how to avoid multiples entries for the same seller. Imagine you have multiple offers with a price <= 20 with the option FullFTP for the same seller, i don't want him to be shown 5 times on the comparator. If you have any advices ;) Thanks

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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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  • Php INNER JOING jqGrid help

    - by yanike
    I'm trying to get INNER JOIN to work with JQGRID, but I can't get it working. I want the code to get the first_name and last_name from members using the "efrom" from messages that matches the "id" from members. $col = array(); $col["title"] = "From"; $col["name"] = "messages.efrom"; $col["width"] = "70"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "First Name"; $col["name"] = "members.first_name"; $col["width"] = "80"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Last Name"; $col["name"] = "members.last_name"; $col["width"] = "80"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Subject"; $col["name"] = "messages.esubject"; $col["width"] = "300"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Date"; $col["name"] = "messages.edatetime"; $col["width"] = "150"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $g = new jqgrid(); $grid["sortname"] = 'messages.edatetime'; $g->select_command = "SELECT messages.efrom, messages.esubject, messages.edatetime, members.first_name, members.last_name FROM messages INNER JOIN members ON messages.efrom = members.id";

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  • How to stop looking in a database after X rows are found?

    - by morningface
    I have a query to a database that returns a number X of results. I am looking to return a maximum of 10 results. Is there a way to do this without using LIMIT 0,9? I'll use LIMIT if I have to, but I'd rather use something else that will literally stop the searching, rather than look at all rows and then only return the top 10.

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  • PHP - Select from database the same query

    - by How to PHP
    I created a table that contains the name of the user and his job, and created PHP page that shows me all the users that works doctor, I entered doctor into a variable then I selected from the table where Jobs equal to $doctor, that is great, but I need it to get the same Jobs into a table in the page and the other same jobs into a table in the same page. this is my code that shows only the users works doctor in one table, <html> <h1>Doctors</h1> </html> <?php mysql_connect('localhost','root',''); mysql_select_db('data'); $doctor='doctor'; $query= mysql_query("SELECT * FROM `users` WHERE `job` = '$doctor'")or die(mysql_error()); while ($arr = mysql_fetch_array($query)) $name= $arr['name']; echo $name; } ?> That shows me doctors when I put doctor in a variable I want to show all same Jobs in a table. Is there is a way to do this? Thanks :)

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  • 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.

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  • 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...

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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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  • InnoDB or MyISAM - Why not both?

    - by Skoder
    Hey. I'm new to databases, and I've read various threads about which is better between InnoDB and MyISAM. It seems that the debates are to use or the other. Is it not possible to use both, depending on the table? What would be the disadvantages in doing this? As far as I can tell, the engine can be set during the CREATE TABLE command. Therefore, certain tables which are often read can be set to MyISAM, but tables that need transaction support can use InnoDB. I'm sure there must be a problem, otherwise this would be the ultimate answer :).

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  • counting twice in a query, once using restrictions

    - by Andrew Heath
    Given the following tables: Table1 [class] [child] math boy1 math boy2 math boy3 art boy1 Table2 [child] [glasses] boy1 yes boy2 yes boy3 no If I want to query for number of children per class, I'd do this: SELECT class, COUNT(child) FROM Table1 GROUP BY class and if I wanted to query for number of children per class wearing glasses, I'd do this: SELECT Table1.class, COUNT(table1.child) FROM Table1 LEFT JOIN Table2 ON Table1.child=Table2.child WHERE Table2.glasses='yes' GROUP BY Table1.class but what I really want to do is: SELECT class, COUNT(child), COUNT(child wearing glasses) and frankly I have no idea how to do that in only one query. help?

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  • Converting a certain SQL query into relational algebra

    - by Fumler
    Just doing an assignment for my database course and I just want to double check that I've correctly wrapped my head around relational algebra. The SQL query: SELECT dato, SUM(pris*antall) AS total FROM produkt, ordre WHERE ordre.varenr = produkt.varenr GROUP BY dato HAVING total >= 10000 The relational algebra: stotal >= 10000( ?R(dato, total)( sordre.varenr = produkt.varenr( datoISUM(pris*antall(produkt x ordre)))) Is this the correct way of doing it?

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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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  • 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');

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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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  • 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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  • 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?

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  • Favouriting things in a database - most efficient method of keeping track?

    - by a2h
    I'm working on a forum-like webapp where I'd like to allow users to favourite an item so that they can keep track of it, and also so that others can see how many times an item's been favourited. The problem is, I'm unsure on the best practices for databases, which includes this situation. I have two ideas in my head on how to do this: Add an extra column to the user table and store things like so: "|2|5|73|" Add an extra table with at least two columns, one for referencing an item, the other for referencing a user. I feel uncomfortable about going for the second method as it involves an extra table, and potentially more queries would be required. Perhaps these beliefs aren't an issue, as I have little understanding of databases beyond simply working with table layouts and basic queries.

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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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  • Jetty 6: Unknown Error 99

    - by Silvio Donnini
    The system I'm developing is comprised of a jetty server (v6.1.2rc4) and a php frontend that sends http requests to jetty via curl_exec. The server and the client are on the same machine. The requests I send can be both POSTs and GETs, I get the same error for either which is: Failed to connect to 127.0.0.1: Unknown error 99 This is rather cryptic. It seems that after the first problematic request, some of the following (unrelated) requests also get corrupted. It looks like jetty is simply refusing the connection, but I can't read more than that into the error message. I thought it was a problem with the server's configuration, so I tried changing jetty's maxIdleTimeMs, but without success. Any idea about what to do is welcome thanks, Silvio

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  • 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 ?

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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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  • 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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  • 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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