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PHP - Display Single Blog Post From Db In Php Mvc
Hello everyone. I'm new here and to php mvc. Been trying to hone my skills with mvc, mysql, classes and functions. But I am stuck. Can display all blog posts from DB but not a single post. The Readmore link only leads to post 1 (or the next published post from the db). Please I need your help. My code is below. Thanx in advance. BlogsModel.php (PDO db): public static function getBlog($blogsID) { try { $db = static::getDB(); $stmt = $db->query('SELECT * FROM blogs WHERE blogsStatus=1 AND blogsID=blogsID '); $result = $stmt->fetch(PDO::FETCH_ASSOC); return $result; } catch (PDOException $e) { echo $e->getMessage(); } } Blogs.php (Controller): public function blogAction() { $blog = BlogsModel::getBlog($blogsID = 'blogsID'); $user = BlogsModel::getBlog($blogsUserID = 'blogsUserID'); $data = [ 'blog' => $blog, 'user' => $user ]; View::renderTemplate("Services/Blogs/blog.php", $data); } index.php (front controller): $router->add('blogs/{id:[\w-]+}', ['controller' => 'Services\Blogs', 'action' => 'show']); ReadMore link: <a id="readMore" href="/blogs/{{ blog.blogsAlias }}" >ReadMore</a>
Similar TutorialsHi, I have a blog page I am working on and I am trying to make some enhancements that I need some help with. In MySQL I have a table with id, title and post. In my page display I want to enable the user to click on the title and display a post on it's own so the URL would look like www.mysite.com/blog/?id=1 or something like that. Here is the PHP function to display the post from MySQL. all on one page. function GetBlogPosts($inId=null, $inTagId =null) { if (!empty($inId)) { $query = mysql_query("SELECT * FROM blog_posts WHERE id = " . $inId . " ORDER BY id DESC LIMIT 2"); } else if (!empty($inTagId)) { $query = mysql_query("SELECT blog_posts.* FROM blog_post_tags LEFT JOIN (blog_posts) ON (blog_post_tags.postID = blog_posts.id) WHERE blog_post_tags.tagID =" . $tagID . " ORDER BY blog_posts.id DESC"); } else { $query = mysql_query("SELECT * FROM blog_posts ORDER BY id DESC"); } $postArray = array(); while ($row = mysql_fetch_assoc($query)) { $myPost = new BlogPost($row["id"], $row['title'], $row['post'], $row['postfull'], $row["author_id"], $row['date_posted']); array_push($postArray, $myPost); } return $postArray; } hey, i need some help if possible. I basicly have 2 tables, one has the blog posts and one has the blog comments. I'm trying to display the latest 5 blog posts (which works) but underneath each of the blog posts i need to display about 3 of the latest comments for that post but it isn't showing any. any ideas? here's my code: Code: [Select] <? session_start(); $myusername = $_SESSION['myusername']; $db_host = 'localhost'; $db_user = 'HIDDEN'; $db_pwd = 'HIDDEN'; $database = 'tjwebsol_dev'; $table = 'posts'; $table2 = 'blog_comments'; $tab = '&#38;#160;&#38;#160;&#38;#160;&#38;#160;&#38;#160;&#38;#160;&#38;#160;&#38;#160;&#38;#160;&#38;#160;&#38;#160;&#38;#160;'; if (!mysql_connect($db_host, $db_user, $db_pwd)) die("Can't connect to database"); if (!mysql_select_db($database)) die("Can't select database"); // sending query $result = mysql_query("SELECT * FROM {$table} ORDER BY blog_id DESC LIMIT 5"); if (!$result) { die("Query to show fields from table failed"); } $sql = mysql_query("SELECT * FROM {$table2} ORDER BY comment_id ASC LIMIT 5 "); $fields_num = mysql_num_fields($result); $fields_num2 = mysql_num_fields($sql); while($row2 = mysql_fetch_assoc($sql)) { $blog_comments = $row2['comment_content']; $date_commented = $row2['date_commented']; $time_commented = $row2['time_commented']; $blog_username2 = $row2['username']; $blogid = $row2['blogid']; $comment_id = $row2['comment_id']; } while($row = mysql_fetch_assoc($result)) { $blog_title = $row['title']; $blog_content = $row['content']; $blog_username = $row['username']; $blog_date = $row['date_posted']; $blog_time = $row['time_posted']; $blog_id = $row['blog_id']; echo "<hr>"; echo "<div class='post'>"; echo "<h2 class='title'><a href='#'>$blog_title</h2></a>"; echo "<p class='byline'>posted by <b>$blog_username</b> on <b>$blog_date</b> ($blog_time)</p>"; echo "<br>"; echo "<div class='entry'>$blog_content</div><br>"; if ($blogid == $blog_id) { echo "<b>User Posted Comments:</b><br><br>"; echo "$tab<b>$comment_id</b>. $blog_comments"; echo "<br><br>"; echo "Commented posted by <b>$blog_username2</b> on <b>$date_commented ($time_commented)</b>"; echo "<br><br><br>"; } if (!isset($_SESSION['logged']) || $_SESSION['logged'] !== true) { // not logged in, move to login page echo "<b><a href='login.php'>Login</a></b> to post your comments!"; } echo "<br><br><br>"; include('blognewcomment.php'); echo ""; echo "</div>"; } mysql_free_result($result); ?> PHP script return 20 UL LIST values like, < ul >
A < /ul > How to display UL LIST into row wise 5 columns like
A B C D Hello, I made myself a simple blog system, but in my case "simple" becomes difficult when it comes to writing an article. Have you ever seen that movie about how facebook was created (I think it's called "Social network" or something like that)? As a newbie web developer I noticed how "Mark" was writting his blog post. I think it was an textarea, and when he was writing article, he wrote it with html tags, like <p>. So, I have now the same problem: when I'm writing article, I have to write it with all div's and p's and a's and img's and so on. I want to avoid that. Few ideas did pop to my head: use preg_replace for some elements, and set up some javascript WYSIWYG editor (but i really don't want it, and can't explain why, I just don't like it). I have a couple of css classes for images, both of them resizes images, one resizes it a lot, other not so much, both make them smaller. One class is for image to be on the left, so the text would wap that image from the right. After paragraph I must clear floats with css class. Other image class just takes full width available and if image is bigger that that available width - it resizes image. So no need to clear floats. How can I make my life easier when I'm writing article to my blog? Hi, I am trying to get a facebook like button displayed under each of my blog post, but so far I am just getting three like buttons displayed above the blog post instead of below each one. Below is the function code to display my blog post and also where I cam currently placing the FB like button code. <?php include 'blogpost.php'; $connection = mysql_connect('localhost', 'username', 'pw') or die ("<p class='error'>Sorry, we were unable to connect to the database server.</p>"); $database = "DBName"; mysql_select_db($database, $connection) or die ("<p class='error'>Sorry, we were unable to connect to the database.</p>"); function GetBlogPosts($inId=null, $inTagId =null) { if (!empty($inId)) { $query = mysql_query("SELECT * FROM blog_posts WHERE id = " . $inId . " ORDER BY id DESC LIMIT 2"); } else if (!empty($inTagId)) { $query = mysql_query("SELECT blog_posts.* FROM blog_post_tags LEFT JOIN (blog_posts) ON (blog_post_tags.postID = blog_posts.id) WHERE blog_post_tags.tagID =" . $tagID . " ORDER BY blog_posts.id DESC"); } else { $query = mysql_query("SELECT * FROM blog_posts ORDER BY id DESC"); } $postArray = array(); while ($row = mysql_fetch_assoc($query)) { $myPost = new BlogPost($row["id"], $row['title'], $row['post'], $row['postfull'], $row["author_id"], $row['date_posted']); array_push($postArray, $myPost); echo '<script src="http://connect.facebook.net/en_US/all.js#xfbml=1"></script><fb:like show_faces="true" width="450"></fb:like>';//this is where I am currently implementing the FB Like button } return $postArray; } ?> Hello! I am a 17 year old beginner Php student and I am currently creating a page for which I would like to create a blog management admin interface. I went very well, I can already add the categories to the database, through the admin interface or I can list them on the website. However, there would be a problem with creating a Blog post. Because I use the same create () function as for adding a category, but unfortunately the Blog Post does not add to the database. I don't get any php error code just when I try to list the post id after the create function. Instead of a value of 1, I get a 0. I enclose the code of the blog post and its control, as well as the data of my database!
Blog Posting form admin html and php code (ujblogposzt.php) ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
<?php require_once('templates/head.php') ?> <!-- includes database and user controll -->
<?php include('app/controllers/posztok.php');?> <!-- Add new blog post controll file -->
<!-- Main content -->
<div class="card-footer">
</div>
<!-- bs-custom-file-input --> <?php require_once('templates/footer.php') ?>
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Blogpost controll code (posztok.php)
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
<?php
if (isset($_POST['add-post'])) {
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- I see everything okay in these codes, because I add the category to the database in the same way when adding a category my function called dd is at the end of the line of code, it's just for checking, it shows results. my create() function seen like this:
function create($table, $data)
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- The strange thing is that if I add 1-2 items to the database statically without php code, it is detected and displayed in the blog editing menu, but I can't add a post to the database via the admin interface itself, that's why I wrote here ! :) From testing, I also print the end result of the create function at the end of the control code, which should return an id, but instead of id 1 I get a value of 0 and the post doesn't added to the database after pressing the post button['add-post'] and unfortunately I can't figure out why it is ?! I also attach the pictures of the problem!
The pictures of my problem can be viewed he https://drive.google.com/drive/folders/1WTVNv4LlbJiosWvBZoIRzjiUDkvhMgYk?usp=sharing
Thanks to anyone in advance !!!
I have a problem about Blog Post Categorization Script. Can anyone review it ? Instead of showing how much posts that category have it just repeats. Code: [Select] <? $sqlCategories = "SELECT * FROM blog_entry, categories where blog_entry.CategoryID=categories.CategoryID and blog_entry.UserID=".$_REQUEST['UserID']." order by Category asc"; $resultCategories = mysql_query($sqlCategories, $conn); if (@mysql_num_rows($resultCategories)!=0){ $strcategory=""; while($row_categories = @mysql_fetch_array($resultCategories)) { $categoryduplicate=0; $arrcat = explode(",",$strcategory); for($i=0;$i<strlen($arrcat);$i++) { if($arrcat[$i]==$row_categories['CategoryID']) { $categoryduplicate=1; } } if($categoryduplicate==0) { if($strcategory=="") { $strcategory=$row_categories['CategoryID']; } else { $strcategory=$strcategory.",".$row_categories['CategoryID']; } $sqlCategories2 = "SELECT * FROM blog_entry where CategoryID=".$row_categories['CategoryID']." and UserID=".$_REQUEST['UserID']; $resultCategories2 = mysql_query($sqlCategories2, $conn); if($row_categories['CategoryID'] == $_REQUEST['CategoryID']) { echo " <Tr> <td class='content'><a href='postss.php?CategoryID=".$row_categories['CategoryID']."&UserID=".$_REQUEST['UserID']."&Category=".$row_categories['Category']."' class='in_sel'>".$row_categories['Category']."</a> [".@mysql_num_rows($resultCategories2)."]</td> </Tr> "; } else { echo " <Tr> <td class='content'><a href='posts.php?CategoryID=".$row_categories['CategoryID']."&UserID=".$_REQUEST['UserID']."&Category=".$row_categories['Category']."' class='in_sel'>".$row_categories['Category']."</a> [".@mysql_num_rows($resultCategories2)."]</td> </Tr> "; } } } } ?> Hi all - I'm setting up a custom PHP blog. It pulls the data from a MySQL database which includes HTML tags (<p><div><span> etc...). I would like to display only up to 50 words per post on the blog page, which users can then read and click a link to then see the entire post. I've developed some code which does this, however, it seems to be stripping my HTML tags... Very sad! Would be very very grateful if one (or more) of you kind lot would have a look at my code and let me know if there is a easier (I'm all for easy) and proper way of implementing this so that it works without stripping my HTML tags. Cheers!!! Code: [Select] // Counts number of blog words in the content $blog_content_words = str_word_count($blog["content"], 1); // Sets the blog_words variable to 0 $blog_words = 0; // Prints title on page as a permalink to a post echo '<h1><a href="blog.php?id='.$blog["id"].'">'.$blog["title"].'</a></h1><p>'; // Loops while blog_word is under 50 while($blog_words < 50) { // Prints a word from blog_content_words array of blog post and adds a space afterwards echo $blog_content_words["$blog_words"]." "; // Adds 1 to the blog_words counter ++$blog_words; } // Adds a read more link to the post which links to full blog post echo '... <a href="blog.php?id='.$blog["id"].'">[read more]</a></p>'; I had a wordpress site setup for my maintenance company. I am not trying to setup a blog on the site. I have the blogroll showing up on this page: http://handymore.com/blog/ but when you click the individual articles the single post page shows a Error 404 message. Could anyone point me in the right direction as to how to fix this so that the individual post pages are found? Any guidance would be appreciated. This topic has been moved to PHP Applications. http://www.phpfreaks.com/forums/index.php?topic=348057.0 Hi I have a simple form and when the user submits, php is putting a \ before every single quote entered in the field. So for example, if a user enters O'Neill, once i do $lastName = $_POST["lastname"]; $lastName comes back as: O\'Neill is there some way I can turn this off? i have a while loop that builds up a table of records, each row has a button that needs to have an action to delete that specific record. I can think of the obvious way around this and add a form around each row and set the button to type="submit". However there is a problem i have the entire table of records wrapped within in a form object for other functionalities. Does anyone have a solution for this problem? hi i need help on posting multiple inputs in a single button...while using mysql_fetch_array here is my codes: <? $re6 = mysql_query('select username from users where course = "BSIT" and yearlevel = "FOURTH"'); ?> <br /> <h1>Post Grade</h1> <h1>IT, Fourth Year</h1> <br />Please fill the following form to send The Grade<br /> <? $n = 0; while($row = mysql_fetch_row($re6)) { echo'<form action="grade_post.php" method="post">'; echo'Recipient<span class="small">(Username)</span><input type="text" value="'.$row['username'].'" readonly="readonly" id="recip" name="recip[' . $n . ']" />'; echo'Subject<input type="text" value="'.htmlentities($otitle, ENT_QUOTES, 'UTF-8').'" id="title" name="title[' . $n . ']" />'; echo'<input type="hidden" value="FOURTH" id="year" name="year[' . $n . ']" />'; echo'<input type="hidden" value="FIRST" id="sem" name="sem[' . $n . ']" />'; echo'Grade<input type="text" id="message" name="message[' . $n . ']" ><br />'; ++$n; } ?> i get all my recipients in every input type, but when i tried to post it in my database not all of them are posted rather only one of them are posted in my database ...what i want to happen is that all of my recipients in every input type will be posted in my database with different ids'...help pls... This topic has been moved to PHP Applications. http://www.phpfreaks.com/forums/index.php?topic=318539.0 ==> MySQL version 5.0.91-community So once upon a time I was working on a CMS. I worked on it for 3 months, had to wipe my computer - but didn't worry about it because I backed up the files on a flash drive... which got smushed by some boxes when I moved <.< awesome -.- Anyway, a year later (ie: today), I've decided to start again from scratch. For me - for whatever reason - the simplest things always seem to give me the most trouble. I decided to start with a "user" table - naturally - and began working on the first page: user_view.php and added an include file that connects to the database (connectDB.php). Since I started today, everything is simple - no design elements yet - just good 'ol code. The View Users Page is simply a page that grabs all the rows from Table: users and prints the information to the screen, as if you were viewing a Members page on a forum. This Page works fine. No complaints, does everything I need it to, prints all the rows and cols from Table Users. Then I created the Edit User Page. When I did this a year ago, and I wanted to print out all the data from the table, I used the while loop. When I wanted to only grab certain variables from a table, I just created a new set of variables, as show below. Using the same method from a year ago, I started working on printing various data to the screen (such as "Welcome $username"). 5 minutes of typing... refresh page... nada. So I stripped away just about every tag but the essentials and it still wouldn't work Code: [Select] <?php ... $userID = 3; $sql = "SELECT * FROM users WHERE userID='$userID'"; $result = mysql_query($sql); $rows = mysql_fetch_row($result); echo $rows['username']; ... ?> Normally, this would just grab the username from a row in Table users where the userID = 3 But instead - it does nothing. I just see a blank page. It's a simple 5 lines of code and it's like "im just gonna do w.e the fk i wanna do mmmk thxforplaying" PS: Keep in mind there is nothing wrong with the connection to the database, and as already stated I'm able to print ALL values from the Users Table using this same method: Code: [Select] <?php ... $user_sql = "SELECT * FROM users ORDER BY username ASC"; $user_result = mysql_query($user_sql); while ( $user_rows = mysql_fetch_array($user_result) ) ... ?> Hello. I'm new to pHp and I would like to know how to get my $date_posted to read as March 12, 2012, instead of 2012-12-03. Here is the code: Code: [Select] <?php $sql = " SELECT id, title, date_posted, summary FROM blog_posts ORDER BY date_posted ASC LIMIT 10 "; $result = mysql_query($sql); while($row = mysql_fetch_assoc($result)) { $id = $row['id']; $title = $row['title']; $date_posted = $row['date_posted']; $summary = $row['summary']; echo "<h3>$title</h3>\n"; echo "<p>$date_posted</p>\n"; echo "<p>$summary</p>\n"; echo "<p><a href=\"post.php?id=$id\" title=\"Read More\">Read More...</a></p>\n"; } ?> I have tried the date() function but it always updates with the current time & date so I'm a little confused on how I get this to work. This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=323879.0 Hi,
Webapp noob here.
I have an application that is going to be data-heavy, so I would like to avoid saving things to the database whenever possible. The application in a nutshell is an android application wired up to a php web app(with MySQL) where you can send requests to the phone and the phone will respond with text or image data with POST.
The normal way to do this would be for the phone to post the data to the database, then have the webapp display it. However, what im wondering is if it is possible for a user on the webapp to request information from the phone, and then when the phone replies using a POST to display the information temporarily in maybe a popup window instead of having to save it to the database.
Thanks!
Hi, I've been scratching my head for a while now about how to do this, I'm relatively new to php and mysql and perhaps foolishly taking on creating a user area for a website. I have everything else working, all of my register account functions and confirmations and all of the login scripts etc. I have created a profile page which returns various information to the user (this bit works fine) and I've got some nice show/hide toggles running with some javascript/css but my intention is to allow the user to change thier information (e-mail address, contact phone number and also whether they are subscribed to the e-mail list), it also displays any support tickets or messages. So after the long intro, here's what I'm struggling with... I have a form in a visibility toggled <div> which submits a 'change_email' script, so a user wants to change their e-mail, clicks on change, the <div> appears, they bang in the new e-mail and hit submit. My php script appears to work (because it doesn't throw up any errors), until you realise that actually it's not updated the record in the db... I'm using mysql_query("UPDATE users SET email='$new_email' WHERE username='$user'"); Do I need to setup variables for all of the information in the db (name, username, password, email, contno etc etc) and include them in the command to get it to work or should that just pick the correct record and then update it? If that is the case is there a way I can include 'blank' variables so I don't have to set them all up... e.g. mysql_query("UPDATE users SET user='',password='',email='$new_email', etc WHERE username='$user'"); Many thanks in anticipation |
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