Chibi Posted September 25, 2019 Share Posted September 25, 2019 (edited) I'm creating an edit page that has a many to many relationship between the banners table and the frames table. The problem I'm having is that I'm getting duplicates of the frames. I hope I've explained everything clearly. So for example I have Quote Frame 1 Frame 1 Frame 2 Frame 2 Frame 3 Instead of Quote Frame 1 Frame 2 Frame 3 This is my tables banners table Quote id | title 1 | Home banner frames table Quote id | title | description | image 1 | Frame 1 | Frame 1 | frame.jpg 2 | Frame 2 | Frame 2 | frame2.jpg 3 | Frame 3 | Frame 3 | frame3.jpg 4 | Frame 4 | Frame 4 | frame4.jpg 5 | Frame 5 | Frame 5 | frame5.jpg banner_frame table Quote id | banner_id | frame_id 1 | 1 | 1 2 | 2 | 1 2 | 2 | 5 Here is my code $query = "SELECT frames.id as frameId, frames.title as frameTitle, banners.id as banner_id "; $query .= "FROM frames "; $query .= "LEFT JOIN banner_frame ON banner_frame.frame_id = frames.id "; $query .= "LEFT JOIN banners ON banners.id = banner_frame.banner_id"; $banner_frame = mysqli_query($conn, $query); confirmQuery($banner_frame); while($row = mysqli_fetch_assoc($banner_frame)) { $frame_id = $row['frameId']; $frame_title = $row['frameTitle']; $checked = ''; if ($row['banner_id'] == $banner_id) { $checked = 'checked'; } echo "<div class='form-check'>"; echo "<input type='checkbox' name='frames[]' id='frame_checkbox' value='$frame_id' $checked> "; echo "<label for='frame_checkbox'>"; echo $frame_title; echo "</label>"; echo "</div>"; } Edited September 25, 2019 by Chibi Quote Link to comment https://forums.phpfreaks.com/topic/309276-duplicate-frame-titles/ Share on other sites More sharing options...
Barand Posted September 25, 2019 Share Posted September 25, 2019 With the provided data your query gives mysql> SELECT frames.id as frameId, frames.title as frameTitle, banners.id as banner_id -> FROM frames -> LEFT JOIN banner_frame ON banner_frame.frame_id = frames.id -> LEFT JOIN banners ON banners.id = banner_frame.banner_id -> ORDER BY frames.id; +---------+------------+-----------+ | frameId | frameTitle | banner_id | +---------+------------+-----------+ | 1 | Frame 1 | 1 | | 1 | Frame 1 | NULL | | 2 | Frame 2 | NULL | | 3 | Frame 3 | NULL | | 4 | Frame 4 | NULL | | 5 | Frame 5 | NULL | +---------+------------+-----------+ Beacause you have "... FROM frames LEFT JOIN ..." you get every frames record even if there is no matching data in the other tables. You get 2 rows for Frame 1 because there are 2 matching records for frame 1 in "banner_frame" table 57 minutes ago, Chibi said: id | banner_id | frame_id 1 | 1 | 1 2 | 2 | 1 2 | 2 | 5 So the question is "What is the purpose of the query?" Quote Link to comment https://forums.phpfreaks.com/topic/309276-duplicate-frame-titles/#findComment-1569948 Share on other sites More sharing options...
Chibi Posted September 25, 2019 Author Share Posted September 25, 2019 The purpose of the query is to list all the frames from the frames table and then only put a check next to those that is linked to the banner table through the pivot table. Quote Link to comment https://forums.phpfreaks.com/topic/309276-duplicate-frame-titles/#findComment-1569949 Share on other sites More sharing options...
Barand Posted September 25, 2019 Share Posted September 25, 2019 Something like this? mysql> SELECT f.id as frameId -> , f.title as frameTitle -> , COUNT(DISTINCT bf.banner_id) as banner_tot -> FROM frames f -> LEFT JOIN banner_frame bf ON bf.frame_id = f.id -> GROUP BY f.id; +---------+------------+------------+ | frameId | frameTitle | banner_tot | +---------+------------+------------+ | 1 | Frame 1 | 2 | | 2 | Frame 2 | 0 | | 3 | Frame 3 | 0 | | 4 | Frame 4 | 0 | | 5 | Frame 5 | 1 | +---------+------------+------------+ Quote Link to comment https://forums.phpfreaks.com/topic/309276-duplicate-frame-titles/#findComment-1569950 Share on other sites More sharing options...
Chibi Posted September 25, 2019 Author Share Posted September 25, 2019 No that didn't work. The way it's looking now is Banner 1 has Frame 3, Frame 4 and Frame 5 checked, instead of Frame 1 Banner 2 has Frame 1 checked instead of Frame 1 and Frame 2 Then I created Banner 3 and that has nothing checked instead of Frame 3 and Frame 4 Here is my whole code. Maybe the issue is with my other code and not the sql statement <?php //GETS THE BANNER FROM THE b_id FROM THE URL if(isset($_GET['b_id'])) { $banner_id = $_GET['b_id']; } $query = "SELECT * FROM banners WHERE id = $banner_id"; $select_banner_by_id = mysqli_query($conn, $query); while($row = mysqli_fetch_assoc($select_banner_by_id)) { $banner_id = $row['id']; $banner_title = $row['title']; } //THIS UPDATES THE BANNER if(isset($_POST['edit_banner'])) { $banner_title = $_POST['banner_title']; $query = "UPDATE banners SET "; $query .= "title = '{$banner_title}' "; $query .= "WHERE id = $banner_id "; $update_banner = mysqli_query($conn, $query); confirmQuery($update_banner); //THIS REDIRECTS THE PAGE AFTER SAVING header("Location: banners.php"); } ?> <form action="" method="POST"> <div class="form-group"> <label for="banner_title">Title</label> <input type="text" class="form-control" name="banner_title" value="<?php echo $banner_title ?>"> </div> <!-- BEGIN FRAMES --> <?php $query = " SELECT f.id as frameId, f.title as frameTitle, COUNT(DISTINCT bf.banner_id) as banner_table "; $query .= "FROM frames f "; $query .= "LEFT JOIN banner_frame bf ON bf.frame_id = f.id "; $query .= "GROUP BY f.id"; $banner_frame = mysqli_query($conn, $query); while($row = mysqli_fetch_assoc($banner_frame)) { $frame_id = $row['frameId']; $frame_title = $row['frameTitle']; $checked = ''; if ($row['banner_table'] == $banner_id) { $checked = 'checked'; } echo "<div class='form-check'>"; echo "<input type='checkbox' name='frames[]' id='frame_checkbox' value='$frame_id' $checked> "; echo "<label for='frame_checkbox'>"; echo $frame_title; echo "</label>"; echo "</div>"; } ?> <div class="form-group"> <button type="submit" class="btn btn-dark btn-lg btn-block" name="edit_banner">Save</button> </div> </form> Quote Link to comment https://forums.phpfreaks.com/topic/309276-duplicate-frame-titles/#findComment-1569951 Share on other sites More sharing options...
Barand Posted September 25, 2019 Share Posted September 25, 2019 4 hours ago, Chibi said: The purpose of the query is to list all the frames from the frames table and then only put a check next to those that is linked to the banner table through the pivot table. I used COUNT(DISTINCT bf.banner_id) as banner_table so that it would be non-zero if the frame were linked to any banners. You are trying to match that that count value against a $banner_id (???). Does that mean you want to show as checked those frames that are link to s specific banner and not just any banner? If so the query will be different. Quote Link to comment https://forums.phpfreaks.com/topic/309276-duplicate-frame-titles/#findComment-1569953 Share on other sites More sharing options...
Chibi Posted September 25, 2019 Author Share Posted September 25, 2019 Quote Does that mean you want to show as checked those frames that are link to s specific banner and not just any banner? Yep that is what I was trying to get to in my question. Sorry it wasn't clear enough. Quote Link to comment https://forums.phpfreaks.com/topic/309276-duplicate-frame-titles/#findComment-1569954 Share on other sites More sharing options...
Barand Posted September 25, 2019 Share Posted September 25, 2019 In that case you need to specify the banner you are looking for in the LEFT JOIN's ON clause EG (looking for banner #2) SELECT DISTINCT f.id as frameId , f.title as frameTitle , bf.banner_id FROM frames f LEFT JOIN banner_frame bf ON bf.frame_id = f.id AND bf.banner_id = 2 ORDER BY f.id; +---------+------------+-----------+ | frameId | frameTitle | banner_id | +---------+------------+-----------+ | 1 | Frame 1 | 2 | | 2 | Frame 2 | NULL | | 3 | Frame 3 | NULL | | 4 | Frame 4 | NULL | | 5 | Frame 5 | 2 | +---------+------------+-----------+ 1 Quote Link to comment https://forums.phpfreaks.com/topic/309276-duplicate-frame-titles/#findComment-1569955 Share on other sites More sharing options...
Chibi Posted September 25, 2019 Author Share Posted September 25, 2019 That was it. THANK YOU SO MUCH. 😀 Quote Link to comment https://forums.phpfreaks.com/topic/309276-duplicate-frame-titles/#findComment-1569956 Share on other sites More sharing options...
Barand Posted September 25, 2019 Share Posted September 25, 2019 Another option is to display, for each frame, all the referenced banners. For example if you have banner_frames table thus +----+-----------+----------+ | id | banner_id | frame_id | +----+-----------+----------+ | 1 | 1 | 1 | | 2 | 2 | 1 | | 3 | 2 | 3 | | 4 | 3 | 3 | | 5 | 1 | 4 | | 6 | 2 | 4 | | 7 | 3 | 4 | +----+-----------+----------+ then the output would look like CODE: <?php include 'db_inc.php'; $db = pdoConnect('chibi'); // // Get the banner data // $res = $db->query("SELECT id , title FROM banners ORDER BY id "); $banner_data = $res->fetchAll(); $banner_count = count($banner_data); $banner_heads = '<th>' . join('</th><th>', array_column($banner_data,'title')) . '</th>'; $banner_template = array_fill_keys(array_column($banner_data, 'id'), ''); // // Get frame-banner data // store in array // $res = $db->query("SELECT f.id as fid , f.title , b.id as bid , CASE WHEN bf.banner_id IS NULL THEN '' ELSE 'Y' END as checked FROM frames f CROSS JOIN banners b LEFT JOIN banner_frame bf ON f.id = bf.frame_id AND b.id = bf.banner_id ORDER BY f.id, b.id "); $data = []; foreach ($res as $r) { if (!isset($data[$r['fid']])) { $data[$r['fid']] = [ 'title' => $r['title'], 'banners' => $banner_template ]; } $data[$r['fid']]['banners'][$r['bid']] = $r['checked']; } ?> <!DOCTYPE html > <html> <head> <meta http-equiv="content-language" content="en"> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <link rel="stylesheet" href="https://www.w3schools.com/w3css/4/w3.css"> <title>Banner Frames</title> </head> <body> <header class="w3-container w3-blue"> <h1>Banner Frames</h1> </header> <div class="w3-container w3-margin w3-padding"> <table class="w3-table w3-bordered"> <tr class="w3-brown"> <th rowspan="2">Frame ID</th> <th rowspan="2">Title</th> <th colspan="<?=$banner_count?>">Banners</th> </tr> <tr class="w3-brown"> <?=$banner_heads?> </tr> <?php // output the data array as table rows foreach ($data as $fid => $fdata) { echo "<tr><td>$fid</td><td>{$fdata['title']}</td>"; foreach ($fdata['banners'] as $bid => $chkd) { $checked = $chkd=='Y' ? 'checked' : ''; echo "<td><input type=\"checkbox\" name=\"banner[$fid][]\" value=\"$bid\" $checked></td>"; } echo "</tr>\n"; } ?> </table> </div> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/309276-duplicate-frame-titles/#findComment-1569958 Share on other sites More sharing options...
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