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PHP - From Csv Into Mysql Database
How would I go about doing the following:
I have a csv like this Quote "Division","Section","Group","Product Code","Description","Description + Secondary Description" "Division 1","Section 1","Group 1","BMSLPL25","Test Name","Test Description" "Division 1","Section 1","Group 2","BMSLPL26","Test Name 2","Test Description 2" "Division 2","Section 2","Group 2","BMSLPL27","Test Name 3","Test Description 3" I have a database structured like this Quote Divisions --- id name parent_id Groups --- id name division_id Products --- id code description secondary_description Section is a sub division. What is the best way to get the information from CSV into this database? Should I have another table and store the CSV data as is and then query that to make the other tables. Any help much appreciated. Similar TutorialsAt the moment I am creating a search function for my website. The approach I have in mind is a pseudo-PHP database. To give an example: A HTML form will submit the results to a PHP file. HTML FORM - Colour: Black PHP RESULT PAGE - if ($_POST['color'] == 'Black') {readfile("./products/black/*.html");} HTML FORM - Price: <$50 PHP RESULT PAGE - if ($_POST['Price'] == '<$50') {readfile("./products/less50/*.html");} The problem here is if there is an item that is black and costs less than $50, then its going to be listed twice. There is probably some code I can write to ommit the listing of duplicate entries, but it is probably going to be messy, so I am wondering if its better to use a centralized MySQL database, rather than a pseudo-PHP database? I've never used MySQL and don't know much about it and this is my first real attempt at using PHP. Hi, I want to make a Shoutbox in Php without using a MySQL-database. This is the code I've got: Code: [Select] <?php $nickname = strip_tags($_POST['nickname']); $message = strip_tags($_POST['message']); ?> <html> <body> <<<HERE<b><h3><center>Shout Box</center></h3></b> <iframe src="chat.txt" width="700" height="250"></iframe> <form method="post" name="form"> Nickname: <input type="text" name="nickname" value="<?php echo $nickname;?>"><br> Message: <input type="text" name="message" value="<?php echo $message;?>"> <input type="submit" value="Submit"> </form> <a href="#">Refresh</a> <?php if ($_POST['nickname'] && $_POST['message']){ $fp = fopen("chat.txt", "a"); $stuff = $nickname . ": " . $message . "n"; fputs($fp, $stuff); } else { echo "Please enter nickname and message."; } ?> </body> </html> Problem is: it isn't working. :-P It does read things from chat.txt, but I cant get the input-box working (let people add stuff to chat.txt). What do I do wrong? For some reason Im not able to connect to mysql database, when i fill in the form and select search, it just basically refreshes the page but does not come up with no error messages or any results from my database. any help is appreciated. HTML Code Code: [Select] <table id="tb1"> <tr> <td><p class="LOC">Location:</p></td> <td><div id="LC"> <form action="insert.php" method="post"> <select multiple="multiple" size="5" style="width: 150px;" > <option>Armley</option> <option>Chapel Allerton</option> <option>Harehills</option> <option>Headingley</option> <option>Hyde Park</option> <option>Moortown</option> <option>Roundhay</option> </select> </form> </div> </td> <td><p class="PT">Property type:</p></td> <td><div id="PS"> <form action="insert.php" method="post"> <select name="property type" style="width: 170px;"> <option value="none" selected="selected">Any</option> <option value="Houses">Houses</option> <option value="Flats / Apartments">Flats / Apartments</option> </select> </form> </div> </td><td> <div id="ptype"> <form action="insert.php" method="post"> <input type="radio" class="styled" name="ptype" value="forsale"/> For Sale <p class="increase"> <input type="radio" class="styled" name="ptype" value="forrent"/> To Rent </p> <p class="increase"> <input type="radio" class="styled" name="ptype" value="any"/> Any </p> </form> </div> </td> </tr> </table> <div id="table2"> <table id="NBtable"> <tr> <td><p class="NBS">Number of bedrooms:</p></td> <td><div id="NB"> <form action="insert.php" method="post"> <select name="number of bedrooms"> <option value="none" selected="selected">No Min</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> </select> to <select name="number of bedrooms"> <option value="none" selected="selected">No Max</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> </select> </form> </div> </td> <td><p class="PR">Price range:</p></td> <td><div id="PR"> <form action="insert.php" method="post"> <select name="price range"> <option value="none" selected="selected">No Min</option> <option value="50,000">50,000</option> <option value="60,000">60,000</option> <option value="70,000">70,000</option> <option value="80,000">80,000</option> <option value="90,000">90,000</option> <option value="100,000">100,000</option> <option value="110,000">110,000</option> <option value="120,000">120,000</option> <option value="130,000">130,000</option> <option value="140,000">140,000</option> <option value="150,000">150,000</option> <option value="160,000">160,000</option> <option value="170,000">170,000</option> <option value="180,000">180,000</option> <option value="190,000">190,000</option> <option value="200,000">200,000</option> <option value="210,000">210,000</option> <option value="220,000">220,000</option> <option value="230,000">230,000</option> <option value="240,000">240,000</option> <option value="250,000">250,000</option> <option value="260,000">260,000</option> <option value="270,000">270,000</option> <option value="280,000">280,000</option> <option value="290,000">290,000</option> <option value="300,000">300,000</option> <option value="310,000">310,000</option> <option value="320,000">320,000</option> <option value="330,000">330,000</option> <option value="340,000">340,000</option> <option value="350,000">350,000</option> </select> to <select name="price range"> <option value="none" selected="selected">No Max</option> <option value="50,000">50,000</option> <option value="60,000">60,000</option> <option value="70,000">70,000</option> <option value="80,000">80,000</option> <option value="90,000">90,000</option> <option value="100,000">100,000</option> <option value="110,000">110,000</option> <option value="120,000">120,000</option> <option value="130,000">130,000</option> <option value="140,000">140,000</option> <option value="150,000">150,000</option> <option value="160,000">160,000</option> <option value="170,000">170,000</option> <option value="180,000">180,000</option> <option value="190,000">190,000</option> <option value="200,000">200,000</option> <option value="210,000">210,000</option> <option value="220,000">220,000</option> <option value="230,000">230,000</option> <option value="240,000">240,000</option> <option value="250,000">250,000</option> <option value="260,000">260,000</option> <option value="270,000">270,000</option> <option value="280,000">280,000</option> <option value="290,000">290,000</option> <option value="300,000">300,000</option> <option value="310,000">310,000</option> <option value="320,000">320,000</option> <option value="330,000">330,000</option> <option value="340,000">340,000</option> <option value="350,000">350,000</option> </select> </form> </div> </td> </tr> </table> PHP Code Code: [Select] <?php $server = ""; // Enter your MYSQL server name/address between quotes $username = ""; // Your MYSQL username between quotes $password = ""; // Your MYSQL password between quotes $database = ""; // Your MYSQL database between quotes $con = mysql_connect($server, $username, $password); // Connect to the database if(!$con) { die('Could not connect: ' . mysql_error()); } // If connection failed, stop and display error mysql_select_db($database, $con); // Select database to use $result = mysql_query("SELECT * FROM tablename"); // Query database while($row = mysql_fetch_array($result)) { // Loop through results echo "<b>" . $row['id'] . "</b><br>\n"; // Where 'id' is the column/field title in the database echo $row['location'] . "<br>\n"; // Where 'location' is the column/field title in the database echo $row['property_type'] . "<br>\n"; // as above echo $row['number_of_bedrooms'] . "<br>\n"; // .. echo $row['purchase_type'] . "<br>\n"; // .. echo $row['price_range'] . "<br>\n"; // .. } mysql_close($con); // Close the connection to the database after results, not before. ?> To connect to my database I an entering mysql database details just at the first top lines server,username,password and database. is thats correct? Thank You for your help in adavance. I've tried reading through some of the threads but couldnt understand some of them. I've made a newsfeed script which works how i want it to. Now i want to add the function to delete a row from the database from an "admin panel" on the website. So far i have this: <?php include("includes.php"); doConnect(); $get_news = "SELECT id, title, text, DATE_FORMAT(datetime, '%e %b %Y at %T') AS datetime FROM newsfeed ORDER BY datetime DESC"; $result= mysqli_query($mysqli, $get_news) or die(mysqli_error($mysqli)); while ($row = mysqli_fetch_array($result)) { echo '<strong><font size="3">'. $row['title'] .' </font></strong><br/><font size="3">'. $row['text'] .'</font><br/><font size="2">'. $row['datetime'] .'</font><br/><br/><a href="delnews.php?del_id=' .$row['id']. '"> <strong>DELETE</strong></a>';} ?> then my delnews.php is: <?php include("includes.php"); doConnect(); $query = "DELETE FROM newsfeed WHERE id = "$_POST['id']""; $result = mysql_query($query); echo "The data has been deleted."; ?> I believe the problem is $_POST['id']. i've tried different things in there but none work. It displays the echo line but doesnt actually delete anything. I am new to php so this may be a stupid mistake, but try and play nice! Thanks There is a "PHP ajax cascading dropdown using MySql" at codestips.com/php-ajax-cascading-dropdown-using-mysql/ I want to use this technique but with a XML or array file instead of mysql database, but my knowledge about mysql is very low. How I can modify this code to catch the categories and products from an array, instead of mysql database? Code: [Select] $connect=mysql_connect($server, $db_user, $db_pass) or die ("Mysql connecting error"); echo '<table align="center"><tr><td><center><form method="post" action="">Category:<select name="category" onChange="CategoryGrab('."'".'ajaxcalling.php?idCat='."'".'+this.value);">'; $result = mysql_db_query($database, "SELECT * FROM Categories"); $nr=0; while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { $nr++; echo "<option value=".'"'.$row['ID'].'" >'.$row['Name']."</option>"; } echo '</select>'."\n"; echo '<div id="details">Details:<select name="details" width="100" >'; $result = mysql_db_query($database, "SELECT * FROM CategoriesDetails WHERE CategoryID=1"); while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo "<option value=".$row['ID'].">".$row['Name']."</option>"; } echo '</select></div>'; echo '</form></td></tr></table>'; mysql_close($connect); ajaxcalling.php is Code: [Select] include("config.php"); $ID=$_REQUEST['idCat']; $connect=mysql_connect($server, $db_user, $db_pass); echo 'Details:<select name="details" width="100">'; $result = mysql_db_query($database, "SELECT * FROM CategoriesDetails WHERE CategoryID=".$ID); while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo "<option value=".$row['ID'].">".$row['Name']."</option>"; } echo '</select>'; mysql_close($connect); Okay, I have been following this tutorial: http://www.freewebmasterhelp.com/tutorials/phpmysql/1 to achieve exactly what I wanted to get done. I also followed the way to have it formatted in tables and added extra columns that I needed, included an "Options" column which houses three links, including "edit" and "delete" Now, I have everything working fine but I am stumped on how to get the "edit" and "delete" links to work for each individual entry that is listed. I have to have these features so the entries can be edited and deleted without having to physically go into the MySQL database to do it. The tutorial explains how to do it in Step 6, but I am confused. I'm not quite sure where to place the code for the links, which are generated automatically every time a new entry is inputted into the database. Anybody available to help me out? Thanks! I'm having problems updating my database, I have 4 fields i want to change. I checked all the { on the page, that's not the problem, I tried to echo information from the database and it displayed my information so that's not the problem, i tried yelling at my computer, that didn't work, i tried to input data into the database with the insert function it worked but is not practical in my situation. I'm probably going to face palm when i find out whats wrong, help please btw, the $_SESSION['usr'] was set in another page and works. Code: [Select] <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Edit Info</title> <link rel="stylesheet" type="text/css" href="demo.css" media="screen" /> </head> <body> <div id="main"> <div class="container"> <font size="5" face="sans-serif">Change Settings <?php echo "{$_SESSION['usr']}"; ?></font> <form action="" method="POST"> <table cellpadding="3" cellspacinf="4" border="0"> <tr> <td>Name</td> <td><input type="text" name="name" /></td> </tr> <tr> <td>Age</td> <td><input type="text" name="age" /></td> </tr> <tr> <td>Gender</td> <td><input type="text" name="mf" /></td> </tr> <tr> <td>Location</td> <td><input type="text" name="loc" /></td> </tr> <tr> <td><input type="submit" name="submit" value="submit" /></td> </tr> </table> </form> <?php if ($_POST['submit']){ define('INCLUDE_CHECK',true); require 'connect.php'; $usr = $_SESSION['usr']; $sql = mysql_query("UPDATE members SET name='{$_POST['name']}', age='{$_POST['age']}, mf='{$_POST['mf']}', loc='{$_POST['loc']}' WHERE usr='{$_SESSION['usr']}'"); if($sql){ echo 'Changes Saved!'; }else{ echo 'Error'; } } ?> </div> </div> </body> </html> Hi, I cant connect to my Mysql database. I get this problem: Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'esolarch_databas'@'localhost' (using password: YES) in /home7/esolarch/public_html/new/storescripts/connect_to_mysql.php on line 21 could not connect to mysql Code: [Select] <?php /* 1: "die()" will exit the script and show an error statement if something goes wrong with the "connect" or "select" functions. 2: A "mysql_connect()" error usually means your username/password are wrong 3: A "mysql_select_db()" error usually means the database does not exist. */ // Place db host name. Sometimes "localhost" but // sometimes looks like this: >> ???mysql??.someserver.net $db_host = "localhost"; // Place the username for the MySQL database here $db_username = "esolarch_database"; // Place the password for the MySQL database here $db_pass = "Password"; // Place the name for the MySQL database here $db_name = "esolarch_admin2"; // Run the actual connection here mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql"); mysql_select_db("$db_name") or die ("no database"); ?> Im trying to connect to a database from php. Heres the code: <?php $dbc = mysqli_connect('192.168.0.122', 'boyyo', 'KiaNNa11', 'aliendatabase') or die('Error connecting to MySQL server.'); $query = "INSERT INTO aliens_abduction (first_name, last_name, " . "when_it_happened, how_long, how_many, alien_description, " . "what_they_did, fang_spotted, other, email) " . "VALUES ('Sally', 'Jones', '3 days ago', '1 day', 'four', " . "green with six tentacles', 'We just talked and played with a dog', " . "'yes', 'I may have seen your dog. Contact me.', " . "'sally@gregs-list.net')"; $result = mysqli_query($dbc, $query) or die('Error querying database.'); ?> I think i have a theory that my MySQL server location is wrong but i dont know. I use HostGator to do this and im using Phpmyadmin. But everytime i type in a form that i created it says Error querying database. Can someone tell me whats wrong with this code. Oh by the way im using head first into PHP and MySQL Ok, I got someone to help me fix this but he had no idea what the error was... I have 2 tables, one called points and the other called members. In members i have got: id name In points i have got: id memberid promo I have the following code: Code: [Select] <?php $con = mysql_connect("localhost","slay2day_User","slay2day"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("slay2day_database",$con); $sqlquery="SELECT Sum(points.promo) AS score, members.name, members.id = points.memberid Order By members.name ASC"; $result=mysql_query($sqlquery,$con); while ($row = mysql_fetch_array($result)) { //get data $id = $row['id']; $name = $row['name']; $score = $row['score']; echo "<b>Name:</b> $name<br />"; echo "<b>Points: </b> $score<br />" ; echo "<b>Rank: </b>"; if ($name == 'Kcroto1'): echo 'The Awesome Leader'; else: if ($points >= '50'): echo 'General'; elseif ($points >= '20'): echo 'Captain!'; elseif ($points >= '10'): echo 'lieutenant'; elseif ($points >= '5'): echo 'Sergeant'; elseif ($points >= '2'): echo 'Corporal'; else: echo 'Recruit'; endif; endif; echo '<br /><br />'; } ?> I am getting the following error when i do the query in mysql: Code: [Select] #1109 - Unknown table 'points' in field list And when i open the webpage i get the following error: Code: [Select] Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/slay2day/public_html/points/members.php on line 18 Please Help me? Hi All, Whenever I try to update any piece of PHP code to update a MySQL database, nothing happens. I have tried copying in some of the working codes of a website and tried the same, but no success. I recently installed XAMPP. I am connecting using the correct user id and pass to the database. The scripts are not giving me any error, but just not connecting, that's all. While making such a usage as noted below <FORM name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" > I get the following error Firefox can't find the file at /C:/xampp/htdocs/="<?php.. so on Why does this happen? I am pretty new to this, so please do help. Thanks, Satheesh P R Hi On my xampp server's database I get the warning that my database is read-only when I try to insert something into a table. I think I may know what cause this, because I created an export/import app that exports/import important data to the database... How can I remove the read-only from the database, because I am scared that the same thing might happen to my server's database? And is there a mysql query that test if the database is read-only? Thanks in advance Can you upload videos to a mysql database? If so how would you go about doing it? Hello all,
I haven't worked with PHP and MySQL very much, but have been working with HTML and JavaScript for years. This time, I want to build a simple website that allows users to query a database of coins. I just want to make sure that I approach this in the right direction. I'd like to describe what I'm trying to do and hopefully someone can prevent me from becoming a failure.
For the database, each coin has an ID and row, the first column of which is the date, and the following columns attributes of that particular coin. Several entries can exist with the same date.
Anyway, there will basically be three types of pages:
1. A query page from which users will click a date or enter a term in a search box.
2. A query result page displaying a list of search results, like '1909' or 'Lincoln'. Each result will be linked to an individual page for that coin.
3. An individual page. This template page will have an html table whose cells are each defined by the values for that coin in the database.
So for instance, the user searches or clicks on '1909' on the query page and sees 10 records listed in the database on the result page. The user can then click one of those 10 records and view a new page that displays more information on that particular coin.
Hello guys. Trying to connect php with mysql database and then display results on the screen. This is my code: Code: [Select] <?php $dbhost = "localhost"; $dbuser = "username1"; $dbpass = "password1"; $db = "username1_myDB"; $connection = mysql_connect($dbhost, $dbuser, $dbpass) or die ("Could not connect"); mysql_select_db($connection, $db); $show = "SELECT Name, Description FROM people"; $result = mysql_query($show); while($show = mysql_fetch_array($result)){ $field01 = $show[Name]; $field02 = $show[Description]; echo "id: $field01<br>"; echo "description: $field02<p>"; } ?> However im getting this: Warning: mysql_select_db() expects parameter 1 to be string, resource given in /home/pain33/public_html/index.php on line 20 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/pain33/public_html/index.php on line 26 Any ideas how to fix this? Thank you. Hello playERZ' So, I have a dinky contact form that validates and uses: Code: [Select] <form name="infoget" action="<?=$_SERVER['PHP_SELF']?>" method="post"> to validate the form via an external validate.php, then generate the email. Finally: // Send the mail using PHPs mail() function mail($to, $subject, $body); header("Location:customize_search.php"); So, this takes the user to a more advanced form after they 'send' the first form. 1 - Is it possible to have the second form autofill with the POSTed data and NOT use MySQL or database whatever? 2 - If so, how?? Hi, I wonder if you could help me try to find what i'm looking for, i have a problem with bogus users on my site. I created a register page where the details are sent to the database, this is fine but someone is registering with the username: 1 and password: 1 multiple times. I have about 50 of these now and i would like to know what to actually search for (how to word it) to find out how to stop this? What would be the name of the script? i've looked for fake username script, multiple username/password prevention script, i'm just not getting it, sorry. If any of you have any ideas i'd like to hear from you, many thanks in advance for that. If you need anymore to go on please ask, once again thank you. Hello, I've been having trouble connecting to a MySQL database, I can't find the problem in the code, what am I doing wrong? Getting the database file in the config file : require_once("db_connect.php"); db_connect.php : <?php $db = mysql_connect('83.172.155.14:3306', 'username', 'password') or die(mysql_error()); mysql_select_db('databasename', $db) or die(mysql_error()); ?> I need to connect to a PhpMyAdmin database. I need this fixed asap since I'm doing this for someone and he wants the site done as quickly as possible. P.S: The database used to work in php4 and now I need it to work on php5 Thanks in advance, Hey guys, I'm working a project that requires sessions be stored within the database, as the project I'm working on is on a shared host. But I'm having a problem with getting the data of a session in the database, the other fields like session_id, session_updated, session_created are working fine. I think I've got a bug in my code, but I just can't detect it (frustrating). Database connection class db extends mysqli { private $host; private $user; private $pass; private $db; function __construct( $host='localhost', $user='user', $pass='pass', $db='website' ) { $this -> host = $host; $this -> user = $user; $this -> pass = $pass; $this -> db = $db; parent::connect( $host, $user, $pass, $db ); if( mysqli_connect_error( ) ) { die( 'Connection error ('.mysqli_connect_errno( ).'): '.mysqli_connect_error( ) ); } } function __destruct( ) { $this -> close( ); } } Session handler class sessionHandler { private $database; private $dirName; private $sessTable; private $fieldArray; function sessionHandler() { // save directory name of current script $this -> database = new db; $this -> dirName = dirname(__file__); $this -> sessTable = 'sessions'; } function open( $save_path, $session_name ) { return TRUE; } function close() { //close the session. if ( !empty( $this -> fieldarray ) ) { // perform garbage collection $result = $this->gc( ini_get ( 'session.gc_maxlifetime' ) ); return $result; } return TRUE; } function read( $session_id ) { $sql = " SELECT * FROM sessions WHERE session_id=( '$session_id' ) LIMIT 1 "; $result = $this -> database -> query( $sql ); if( $result -> num_rows > 0 ) { $data = $result -> fetch_array( MYSQLI_ASSOC ); $this -> fieldArray = $data; $result -> close(); return $data; } return ""; } function write( $session_id, $session_data ) { //write session data to the database. if ( !empty( $this -> fieldArray ) ) { if ( $this -> fieldArray['session_id'] != $session_id ) { // user is starting a new session with previous data $this -> fieldArray = array(); } } $this -> fieldArray['session_id'] = $session_id; $this -> fieldArray['session_data'] = $session_data; $this -> fieldArray['session_updated'] = time(); $this -> fieldArray['session_created'] = time(); $session_id = $this -> database -> escape_string( $session_id ); $session_data = $this -> database -> escape_string( $session_data ); $session_updated = time(); $session_created = time(); $sql = " INSERT INTO sessions ( session_id, session_data, session_updated, session_created ) VALUES ( '$session_id', '$session_data', '$session_updated', '$session_created' ) "; if( $this -> database -> query( $sql ) !== TRUE ) { return FALSE; } return TRUE; } function destroy( $session_id ) { $sql = " DELETE FROM sessions WHERE session_id=('$session_id') "; if( $this -> database -> query( $sql ) !== TRUE ) { return FALSE; } return TRUE; } function gc( $max_lifetime ) { return TRUE; } function __destruct() { //ensure session data is written out before classes are destroyed //(see http://bugs.php.net/bug.php?id=33772 for details) @session_write_close(); } } The call $session_class = new sessionHandler; session_set_save_handler( array( &$session_class, 'open' ), array( &$session_class, 'close' ), array( &$session_class, 'read' ), array( &$session_class, 'write' ), array( &$session_class, 'destroy' ), array( &$session_class, 'gc' ) ); if( !session_start() ) { exit(); } Any help at all would be appreciated. Kind Regards Mike |
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