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  • How to display thumbnails with category by category

    - by shin
    http://pastie.org/962429 I have the above SQL. There are three categories, webdesign, logos and prints. I want to display them as following html. http://pastie.org/962432 Basically pulling thumbnail image and name from database category by category. I am not really sure how to do it. I will appreciate any inputs or help. Thanks in advance.

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  • Recalculate Counter Cache of 120k Records [Rails / ActiveRecord]

    - by Sebastian
    The following situation: I have a poi model, which has many pictures (1:n). I want to recalculate the counter_cache column, because the values are inconsistent. I've tried to iterate within ruby over each record, but this takes much too long and quits sometimes with some "segmentation fault" bugs. So i wonder, if its possible to do this with a raw sql query?

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  • Create fulltext index on a VIEW

    - by kylex
    Is it possible to create a full text index on a VIEW? If so, given two columns column1 and column2 on a VIEW, what is the SQL to get this done? The reason I'd like to do this is I have two very large tables, where I need to do a FULLTEXT search of a single column on each table and combine the results. The results need to be ordered as a single unit. Suggestions? EDIT: This was my attempt at creating a UNION and ordering by each statements scoring. (SELECT a_name AS name, MATCH(a_name) AGAINST('$keyword') as ascore FROM a WHERE MATCH a_name AGAINST('$keyword')) UNION (SELECT s_name AS name,MATCH(s_name) AGAINST('$keyword') as sscore FROM s WHERE MATCH s_name AGAINST('$keyword')) ORDER BY (ascore + sscore) ASC sscore was not recognized.

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  • SQL to CodeIgniter Array Missing Data Issue

    - by SamD
    $query = $this->db->query("SELECT t1.numberofbets, t1.profit, t2.seven_profit, t3.28profit, user.user_id, username, password, email, balance, user.date_added, activation_code, activated FROM user LEFT JOIN (SELECT user_id, SUM(amount_won) AS profit, count(tip_id) AS numberofbets FROM tip GROUP BY user_id) as t1 ON user.user_id = t1.user_id LEFT JOIN (SELECT user_id, SUM(amount_won) AS seven_profit FROM tip WHERE date_settled > '$seven_daystime' GROUP BY user_id) as t2 ON user.user_id = t2.user_id LEFT JOIN (SELECT user_id, SUM(amount_won) AS 28profit FROM tip WHERE date_settled > '$twoeight_daystime' GROUP BY user_id) as t3 ON user.user_id = t3.user_id where activated = 1 GROUP BY user.user_id ORDER BY user.date_added DESC"); return $query->result_array(); The query works fine running it in phpMyAdmin and returns complete results (in image attached). However, printing the array in CodeIgniter, it has no value for one field ,seven_profit, where it is there in the SQL query ran in phpMyAdmin, just the discrepancy in this one field, from sql to php array... I just can’t see why, when printing the array, that one field, which should have value of 26, contains nothing? Any ideas? I changed the field name from starting with a number in attempt to fix it, but no difference. I know this is complex and looks horrible, any help or just people coming across something similar would be great to know about, thanks. Sam

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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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  • WordPress on other parts of my site

    - by SHiNKiROU
    I have a WordPress installation on my site, and I want to display WP posts on other parts of my site (that is outside the WP installation). How do I do that with PHP? I tried to search this type of question on Stack Overflow, Google and WP official site but I didn't find anything.

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  • uploading image & getting back from database

    - by Anup Prakash
    Putting a set of code which is pushing image to database and fetching back from database: <!-- <?php error_reporting(0); // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","root","")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("test"); $q = <<<CREATE create table image ( pid int primary key not null auto_increment, title text, imgdata longblob, friend text) CREATE; @mysql_query($q); // Insert any new image into database if (isset($_POST['submit'])) { move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($instr) < 149000) { $image_query="insert into image (title, imgdata,friend) values (\"". $_REQUEST['title']. "\", \"". $image. "\",'".$_REQUEST['friend']."')"; mysql_query ($image_query) or die("query error"); } else { $errmsg = "Too large!"; } $resultbytes=''; // Find out about latest image $query = "select * from image where pid=1"; $result = @mysql_query("$query"); $resultrow = @mysql_fetch_assoc($result); $gotten = @mysql_query("select * from image order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; $resultbytes = $row[imgdata]; $friend=$row[friend]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../wellimg/ctco.jpg","rb"); $bytes = fread($instr,filesize("../wellimg/ctco.jpg")); } if ($resultbytes!='') { echo $resultbytes; } } ?> <html> <head> <title>Upload an image to a database</title> </head> <body bgcolor="#FFFF66"> <form enctype="multipart/form-data" name="file_upload" method="post"> <center> <div id="image" align="center"> <h2>Heres the latest picture</h2> <font color=red><?php echo $errmsg; ?></font> <b><?php echo $title ?></center> </div> <hr> <h2>Please upload a new picture and title</h2> <table align="center"> <tr> <td>Select image to upload: </td> <td><input type="file" name="imagefile"></td> </tr> <tr> <td>Enter the title for picture: </td> <td><input type="text" name="title"></td> </tr> <tr> <td>Enter your friend's name:</td> <td><input type="text" name="friend"></td> </tr> <tr> <td><input type="submit" name="submit" value="submit"></td> <td></td> </tr> </table> </form> </body> </html> --> Above set of code has one problem. The problem is whenever i pressing the "submit" button. It is just displaying the image on a page. But it is leaving all the html codes. even any new line message after the // Printing image on browser echo $resultbytes; //************************// So, for this i put this set of code in html tag: This is other sample code: <!-- <?php error_reporting(0); // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","root","")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("test"); $q = <<<CREATE create table image ( pid int primary key not null auto_increment, title text, imgdata longblob, friend text) CREATE; @mysql_query($q); // Insert any new image into database if (isset($_POST['submit'])) { move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($instr) < 149000) { $image_query="insert into image (title, imgdata,friend) values (\"". $_REQUEST['title']. "\", \"". $image. "\",'".$_REQUEST['friend']."')"; mysql_query ($image_query) or die("query error"); } else { $errmsg = "Too large!"; } $resultbytes=''; // Find out about latest image $query = "select * from image where pid=1"; $result = @mysql_query("$query"); $resultrow = @mysql_fetch_assoc($result); $gotten = @mysql_query("select * from image order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; $resultbytes = $row[imgdata]; $friend=$row[friend]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../wellimg/ctco.jpg","rb"); $bytes = fread($instr,filesize("../wellimg/ctco.jpg")); } } ?> <html> <head> <title>Upload an image to a database</title> </head> <body bgcolor="#FFFF66"> <form enctype="multipart/form-data" name="file_upload" method="post"> <center> <div id="image" align="center"> <h2>Heres the latest picture</h2> <?php if ($resultbytes!='') { // Printing image on browser echo $resultbytes; } ?> <font color=red><?php echo $errmsg; ?></font> <b><?php echo $title ?></center> </div> <hr> <h2>Please upload a new picture and title</h2> <table align="center"> <tr> <td>Select image to upload: </td> <td><input type="file" name="imagefile"></td> </tr> <tr> <td>Enter the title for picture: </td> <td><input type="text" name="title"></td> </tr> <tr> <td>Enter your friend's name:</td> <td><input type="text" name="friend"></td> </tr> <tr> <td><input type="submit" name="submit" value="submit"></td> <td></td> </tr> </table> </form> </body> </html> --> ** But in this It is showing the image in format of special charaters and digits. 1) So, Please help me to print the image with some HTML code. So that i can print it in my form to display the image. 2) Is there any way to convert the database image into real image, so that i can store it into my hard-disk and call it from tag? Please help me.

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  • Need help with SQL Query

    - by StackOverflowNewbie
    Say I have 2 tables: Person - Id - Name PersonAttribute - Id - PersonId - Name - Value Further, let's say that each person had 2 attributes (say, gender and age). A sample record would be like this: Person->Id = 1 Person->Name = 'John Doe' PersonAttribute->Id = 1 PersonAttribute->PersonId = 1 PersonAttribute->Name = 'Gender' PersonAttribute->Value = 'Male' PersonAttribute->Id = 2 PersonAttribute->PersonId = 1 PersonAttribute->Name = 'Age' PersonAttribute->Value = '30' Question: how do I query this such that I get a result like this: 'John Doe', 'Male', '30'

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  • Data Modeling Help - Do I add another table, change existing table's usage, or something else?

    - by StackOverflowNewbie
    Assume I have the following tables and relationships: Person - Id (PK) - Name A Person can have 0 or more pets: Pet - Id (PK) - PersonId (FK) - Name A person can have 0 or more attributes (e.g. age, height, weight): PersonAttribute _ Id (PK) - PersonId (FK) - Name - Value PROBLEM: I need to represent pet attributes, too. As it turns out, these pet attributes are, in most cases, identical to the attributes of a person (e.g. a pet can have an age, height, and weight too). How do I represent pet attributes? Do I create a PetAttribute table? PetAttribute Id (PK) PetId (FK) Name Value Do I change PersonAttribute to GenericAttribute and have 2 foreign keys in it - one connecting to Person, the other connecting to Pet? GenericAttribute Id (PK) PersonId (FK) PetId (FK) Name Value NOTE: if PersonId is set, then PetId is not set. If PetId is set, PersonId is not set. Do something else?

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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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  • Check for a unique value within a count, but get all results

    - by pedalpete
    I'm trying to create a single query which, similar to stack overflow, will give me the number of votes, but also make sure that the currently viewing user can't upvote again if they've already upvoted. my query currently looks like SELECT cid, text, COUNT(votes.parentid) FROM comments LEFT JOIN votes ON comments.cid=votes.parentid AND votes.type=3 WHERE comments.type=0 AND comments.parentid='$commentParentid' GROUP BY comments.cid But I'm completely stumpted on how to add the check to see if the userid is in the votes table. The other option is to add a seperate query where SELECT COUNT(*) FROM votes WHERE userid='$userid' AND parentid='$commentParentid' AND type=3 I'm just realizing I'm so lost with this that I don't even really know what tags to provide.

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  • how to specify a BIGINT in a rails 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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  • 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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  • problem in counting children category

    - by moustafa
    I have this table: fourn_category (id , sub) I am using this code to count: function CountSub($id){ $root = array($id); $query = mysql_query("SELECT id FROM fourn_category WHERE sub = '$id'"); while( $row = mysql_fetch_array( $query, MYSQL_ASSOC ) ){ array_push($root,$row['id']); CountSub($row['id']); } return implode(",",$root); } It returns the category id as 1,2,3,4,5 to using it to count the sub by IN() But the problem is that it counts this: category 1 category 2 category 3 category 4 category 5 Category 1 has 1 child not 4. Why? How can I get all children's trees?

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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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  • 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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  • How to add SQL elements to an array in PHP

    - by DanLeaningphp
    So this question is probably pretty basic. I am wanting to create an array from selected elements from a SQL table. I am currently using: $rcount = mysql_num_rows($result); for ($j = 0; $j <= $rcount; $j++) { $row = mysql_fetch_row($result); $patients = array($row[0] => $row[2]); } I would like this to return an array like this: $patients = (bob=>1, sam=>2, john=>3, etc...) Unfortunately, in its current form, this code is either copying nothing to the array or only copying the last element.

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  • Correlate GROUP BY and LEFT JOIN on multiple criteria to show latest record?

    - by Sunbird
    In a simple stock management database, quantity of new stock is added and shipped until quantity reaches zero. Each stock movement is assigned a reference, only the latest reference is used. In the example provided, the latest references are never shown, the stock ID's 1,4 should have references charlie, foxtrot respectively, but instead show alpha, delta. How can a GROUP BY and LEFT JOIN on multiple criteria be correlated to show the latest record? http://sqlfiddle.com/#!2/6bf37/107 CREATE TABLE stock ( id tinyint PRIMARY KEY, quantity int, parent_id tinyint ); CREATE TABLE stock_reference ( id tinyint PRIMARY KEY, stock_id tinyint, stock_reference_type_id tinyint, reference varchar(50) ); CREATE TABLE stock_reference_type ( id tinyint PRIMARY KEY, name varchar(50) ); INSERT INTO stock VALUES (1, 10, 1), (2, -5, 1), (3, -5, 1), (4, 20, 4), (5, -10, 4), (6, -5, 4); INSERT INTO stock_reference VALUES (1, 1, 1, 'Alpha'), (2, 2, 1, 'Beta'), (3, 3, 1, 'Charlie'), (4, 4, 1, 'Delta'), (5, 5, 1, 'Echo'), (6, 6, 1, 'Foxtrot'); INSERT INTO stock_reference_type VALUES (1, 'Customer Reference'); SELECT stock.id, SUM(stock.quantity) as quantity, customer.reference FROM stock LEFT JOIN stock_reference AS customer ON stock.id = customer.stock_id AND stock_reference_type_id = 1 GROUP BY stock.parent_id

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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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  • Php mysqli and stored-procedure

    - by Mneva skoko
    I have a stored procedure: Create procedure news(in dt datetime,in title varchar(10),in desc varchar(200)) Begin Insert into news values (dt,title,desc); End Now my php: $db = new mysqli("","","",""); $dt = $_POST['date']; $ttl = $_POST['title']; $desc = $_POST['descrip']; $sql = $db-query("CALL news('$dt','$ttl','$desc')"); if($sql) { echo "data sent"; }else{ echo "data not sent"; } I'm new with php please help thank you

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  • NSPredicate cause update editing to return NSFetchedResultsChangeDelete not NSFetchedResultsChangeUp

    - by Matthew Weiss
    I have predicate inside of - (NSFetchedResultsController *)fetchedResultsController in a standard way starting from the CoreDataBook example. NSPredicate *predicate = [NSPredicate predicateWithFormat:@"state=%@ && date = %@ && date < %@", @"1",fromDate,toDate]; [fetchRequest setPredicate:predicate]; This works fine however when editing an item, it returns with NSFetchedResultsChangeDelete not Update. When the main view returns, it is missing the item. If I restart the simulator the delete was not saved and the correct editing result is shown the the predicate working correctly. case NSFetchedResultsChangeDelete: [tableView deleteRowsAtIndexPaths:[NSArray arrayWithObject:indexPath] withRowAnimation:UITableViewRowAnimationFade]; break; I can confirm the behavior by commenting out the two predicate lines ONLY and then all works as it should correctly returning with the full set after editing and calling NSFetchedResultsChangeUpdate instead of NSFetchedResultsChangeDelete. I have read http://matteocaldari.it/2009/11/multiple-contexts-controllers-delegates-and-coredata-bug who reports similar behavior but I have not found a work around to my problem. I can

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