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PHP - How To Create A Blog (not Use Mysql Or Any Other Databases)?
I want to create a text file based blog ( not using MySQL or other databases).
I have tried to google for tutorials but I couldn't find anything. is there anyone who can help me and give me some tips? is there any tutorials about this because I found nothing? Any good books? This is an assignment that I have to do and I am using this book: Spring into PHP 5 by Addison Wesley. I am a newbie in php and really want to do this because this is my last task and I really want to learn. So I'm grateful if anyone can help me. Similar TutorialsI would like some tips on how to create a monthly archive for my weblog. I am not using MySQL or any databases. Everything is saved on file. any tips or tutorials? Thanks hi all, I am trying to link 2 mysql tables and display some information from each of them. I have a list of all the possible items for sale in table1 and I am trying to count the number of rows in the other table2 where the items exist. E.g. 'table1' manufacturer model man1 item1 man1 item2 man1 item3 'table2' id model 1 item3 2 item3 3 item2 And the result would show: item1(0) item2(1) item3(2) It would list all the items from table1 and show next to it how many rows are related to that item from table2. I have inserted a quote where I have tried many times to enter something similar to that show in the note below - but I can not get it to work - it just shows the total number of models in table1 for a given manufacturer. The php I have made so far is: <?php case 'manufacturer': $query = " SELECT * FROM table1"; $query .= " WHERE manufacturer = '".$data."' "; $query .= " ORDER BY model "; $result = mysqli_query($cxn,$query); $returnData[''] = "Select a Model..."; while($row = mysqli_fetch_assoc($result)){ // I THINK I NEED TO INSERT SOMETHING LIKE $query2 = "SELECT * FROM table2 WHERE model = table1.model"; $k=$row['model']; $k2=$row2['model']; $counter[$k]+=1; $returnData[$k]=$k; } foreach($counter as $k => $row) { $returnData[$k] .= " ($row)"; } break; ?> is it possible to have two open connections to two different mysql dbs at the same time? when i tried it, only the one on the bottom of the list was active. my config file looks like this: //---------------------------------------------// $dbname = 'xxx'; # Database Name $dbuser = 'xxx'; # Database Username $dbpass = 'xxx'; # Database Password $dbhost = 'xxx'; # Database Host $conn2 = mysql_connect($dbhost,$dbuser,$dbpass) or die ("Could not connect to $dbname: ".mysql_error()); mysql_select_db($dbname) or die ("Could not access the database: ".mysql_error()); $dbname5 = 'yyy'; # Database Name $dbuser5 = 'yyy'; # Database Username $dbpass5 = 'yyy'; # Database Password $dbhost5 = 'yyy'; # Database Host $conn = mysql_connect($dbhost5,$dbuser5,$dbpass5) or die ("Could not connect to $dbname5: ".mysql_error()); mysql_select_db($dbname5) or die ("Could not access the database: ".mysql_error()); //--------------------------------------// so i want to be able to do mysql_query($query,$conn2) when i need to access xxx db, but it doesn't seem to work that way. am i doing something incorrectly? any help would be greatly appreciated. Okay so you may have noticed an earlier post from myself that problem have been fixed, Some background info, The website is currently on my test server so not connected to the internet. I have only been learning PHP for the last 3-4 days so I am fairly new to most things so please keep your explanations simple, The finished code here is being designed for use on large scale websites with a lot of users around 100,000+ per 24 hour period and therefor will need to be secure. Yet I have no idea how to do that, The script once finished will also be freely distributed and contributors will be fairly credited. I have some new problems that I need help fixing, So I am creating a PHP Blog connected to my MYSQL Database, Currently there is no admin backend that's a job for later but I do have the following files. index.php - This displays all blog posts for all my categorys news.php - This displays each blog post on there own config.php - This contains my server connect code I also have the following fields in my database articleid - Unique ID number of each blog post category - The category the blog post is in title - The title of each blog post body - The body of each post author - The name of the author date - date of posting But now I need to add some comments so can anyone please help out, The comments should only appear below the article on the 'news.php' page so I was hoping someone could write me a script to post and display the comments, (No login system needed I got that covered surprisingly.) if you can also give me a .sql file to add the needed tables to my database that would be greatly appreciated, Lastly would it be possible for someone to secure the script so it is invulnerable to SQL Injection attacks and any other form of attack that someone could pull off on the site. So bring on the code, Index.php Code: [Select] <?php include("config/config.php"); $data = mysql_query("SELECT * FROM blog ORDER by date ASC") or die(mysql_error()); while($row = mysql_fetch_array($data)) { echo "<table class='main'> <tr> <td> <a href='/news.php?articleid=" . $row['articleid'] . "' class='article_title'>" . $row['title'] . "</a> <p>" . $row['introduction'] . "</p></td><tr><td ALIGN='RIGHT' class='small'> Posted by:" . $row['author'] . ", on " . $row['date'] . ",</td></tr></table>"; } ?> Comments:I have removed all the junk out and left the basic script, news.php Code: [Select] <?php include("config/config.php"); $data = mysql_query("SELECT * FROM blog WHERE articleid = {$_GET['articleid']} ORDER by date ASC") or die(mysql_error()); while($row = mysql_fetch_array($data)) { echo "<table class='main'> <tr> <td> <a href='/news.php?articleid=" . $row['articleid'] . "' class='article_title'>" . $row['title'] . "</a> <p>" . $row['introduction'] . "</p></td><tr><td ALIGN='RIGHT' class='small'> Posted by:" . $row['author'] . ", on " . $row['date'] . ",</td></tr></table>"; } ?>Comments:So here's the complex bit if someone could please add the comments below the article here, Thanks again, Blaze, (Really bad PHP Programmer) This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=323879.0 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); ?> Hello guys, i want to create array from mysql what is the script for this...? value like $data = array("1" => .0032, "2" => .0028, "3" => .0021, "4" => .0033, "5" => .0034, "6" => .0031, "7" => .0036, "8" => .0027, "9" => .0024, "10" => .0021, "11" => .0026, "12" => .0024, "13" => .0036, "14" => .0028, "15" => .0025); What I mean by dynamic is say I created a function to generate a random code. I put it in a variable named $key. I make a form with a hidden input type with the name and value of $key. When the form posts, it generates the table name to whatever $key is. Because I tried something like this & its not querying. Heres my code: <?php $dbc = mysqli_connect('localhost', 'cpacrop_todo', 'pass', 'cpacrop_todo') or die('Failed to connect to database!'); function make_key($num_chars) { if ((is_numeric($num_chars)) && ($num_chars > 0) && (! is_null($num_chars))) { $key = ''; $accepted_chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890'; //seed srand(((int)((double)microtime()*1000003)) ); for ($i=0; $i<=$num_chars; $i++) { $random_number = rand(0, (strlen($accepted_chars) -1)); $key .= $accepted_chars[$random_number] ; } return $key; } } $key = make_key(6); ?> <form action="" method="post"> <input type="hidden" name="<?php echo "$key"; ?>" /> <input type="submit" name="submit" value="Submit" /> </form> <?php if ($_POST['submit'] == "Submit") { $query = "CREATE TABLE `$key` ( `id` INT AUTO_INCREMENT, `title` VARCHAR(35), `description` VARCHAR(365) );"; mysqli_query($dbc,$query) or die('Unable to query database.'); echo "Success"; } ?> If this is possible, what am I doing wrong? Thanks! i have two tables, employee details table(empno, emp_name , branch) and attendance table(date, Ot, attendence). I want to create a relation between two tables Hey yall! I'm working on a new site idea and I've run across a problem that I know is simple enough but I'm stumped. It's in the signup form What I want to do is insert the new user into the 'Login' table and get the user id that was just created and use that to create a table with the user id in the name. Here is what I have: // now we insert user into 'Login' table mysql_real_escape_string($insert = "INSERT INTO `Login` (`UID`, `pass`, `HR`, `mail`, `FullName`) VALUES ('{$_POST['username']}', '{$_POST['pass']}', '{$_POST['pass2']}', '{$_POST['e-mail']}', '{$_POST['FullName']}')"); mysql_query($insert) or die( 'Query string: ' . $insert . '<br />Produced an error: ' . mysql_error() . '<br />' ); $error="Thank you, you have been registered."; setcookie('Errors', $error, time()+20); // Get user ID mysql_real_escape_string($checkID = "SELECT * FROM Login WHERE `mail` = '{$_POST['e-mail']}'"); while ($checkIDdata = mysql_fetch_assoc($checkID)) { $userID = $checkIDdata; // now we create table 'Transactions" for the user mysql_real_escape_string($create = "CREATE TABLE `financewatsonn`.`Transactions_{$userID}` ( `ID` INT( 20 ) NOT NULL AUTO_INCREMENT COMMENT 'Transaction ID', `name` VARCHAR( 50 ) NOT NULL COMMENT 'Name/Location', `amount` VARCHAR( 50 ) NOT NULL COMMENT 'Amount', `date` VARCHAR( 50 ) NOT NULL COMMENT 'Date', `category` VARCHAR( 50 ) DEFAULT NULL COMMENT 'Category', `delete` INT( 1 ) NOT NULL DEFAULT '0', UNIQUE KEY `ID` ( `ID` ) ) ENGINE = MYISAM DEFAULT CHARSET = utf8 COMMENT = 'User ID {$userID}'"); mysql_query($create) or die( 'Query string: ' . $create . '<br />Produced an error: ' . mysql_error() . '<br />' ); } I know in 'Get user ID' that I need to get the ID but I'm not sure how to get that information. i get the error Quote Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource at 166 which is the while line under Get user ID Hey guys, Currently Im using the code: while($row=mysql_fetch_array($nt)){ echo" $row[user_login],$row[user_email],$row[CallHistoryTimes],$row[CallHistoryInfo],$row[CallHistoryLastOp]$row[CallHistoryLastDate],<br>"; } To list a mysql array; what I am trying to do is to use the mysql row ID and turn the entire string above into a list. So for example, the database would get the ID information & create a link of edit.php/?id=3 If someone could help me on this it would be mega appreciated. Thanks I have a mysql table with the structure of Code: [Select] ID Menu_Name Parent_ID 1 Finance NULL 2 Business NULL 3 Investment 1 4 Trading 2 How can I create a html <ul><li> list based on the parent? Hi to everybody I need ur help because I’m trying to make a script in php to write the same data but with different date in MySQL depending on the splitting ($data06).... like a schedule...
For example if $data06 = annual and $data03 = “2019/01/01” programm must create in MySQL : 2019/01/01 2020/01/01 2021/01/01 2022/01/01 2023/01/01
if $data06=“half year” will create 10 date , increasing 6 moths ...
But I have a problem at finish of code switch ($data06) { case 'annual': $numrate = 5; $aumdata = "+12 months"; break; case 'half year': $numrate = 10; $aumdata = "+6 month"; break;
default: echo "error"; $sql = "INSERT into $nometb ( name, scadenza ) values ( '$data01', '$data03' )"; if ($conna->query($sql) === TRUE) {} else {die('ERROR'. $conna->error); }
//IMPORT NEXT AND NEW DATE $newDate = date_create($data03); for ($mul = 2; $mul <= $numrate; ++$mul) {
$datanuova = date_create($data03); $datanuova->modify($aumdata); $datanuova->format('yy/m/d'); $newDate = $datanuova->format('yy/m/d');
$sql = "INSERT into $nometb ( name, scadenza ) values ( '$data01', '$newDate' )"; if ($conna->query($sql) === TRUE) {} else {die('ERROR'. $conna->error); }
//HERE THERE IS ERROR $data03 = $newDate;
if ($conna->query($sql) === TRUE) {} else {die('ERRORE NELL\'IMPORTAZIONE'. $conna->error); } }
} }
}
How I can resolve ?
many thanks Francesco How would I create a prepared mysql temp table using PHP? I currently write my prepared statements like the example below. How would I create a "TEMP" table? Thanks!
$connection8y = new mysqli("host", "xxx", "xxx", "db"); This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=321735.0 Hi guys, I am thinking of create an free email system by sending an messages using with a username. Would I have to store the messages via in mysql? If I can do that, then I just curious that how would a user delete their messages if their messages is display on php page? What method would I have to use? I've just learned the basics of MySQL last night and made a simple app (where a user enters a song name, song title, and rating (out of 5) and it displays the list of songs with the highest rated songs first). So I decided I wanted to make something similar yet slightly more complicated and I'm not sure if I'm on the right track. Basically it's a "flash card" game whe - The user gets to input their own questions and answers into a simple form, which would then go into an MySQL table with 2 columns, questions and answers - A new page where the user see's random question, must input an answer, if the answer matches goes to next question otherwise gives error Now I'm confused as to how to go about making this. I was thinking of making some session variables and then do something along the lines of Code: [Select] $display_query = "SELECT * FROM questions"; $display = mysqli_query($connection, $display_query); while ($row = mysqli_fetch_array($display)) { $_SESSION['question'] = $row[0]; }[/syntax] //and then make some PHP that might do something like this: if ($_SESSION['answer'] != $row[1]) { echo "error, try again" } else { // I would write something here to grab the next question } So am I just completely off the wall with this, how would I make such an app? How would I make it, for example, fetch new questions? Hello
I have a mysql table like this:
id | user | car 1 | 1 | fiat 500 2 | 2 | vw polo 3 | 2 | vw golf 4 | 3 | renault clio 5 | 2 | fiat panda 6 | 3 | seat ibiza From this table how can i get a query that the result be like: user 2 - 3 cars user 3 - 2 cars user 1 - 1 cars Thanks |
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