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PHP - Choosing A Path - Pseudo-php Database Or Mysql Database?
At 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. Similar TutorialsHello freaks! Im new to this forum, but im not all that new to PHP and MySQL. Although there's been some years since the last time I used it, so don't go all freaky on me if I dont do this right Let's go on-topic: Im in progress of making an internal web-page for me and my colleagues to make things a bit easier for us. I am making an database of our different projects, and I need some help with the input form - as I need to upload an image to the server, and store the path in the MySQL database. In my input form, I need to store information from text fields, and I need to upload an image to the server and store the path in the database. Before I can even start to code this (although I have coded the input forum without the upload), I need to know what would be the best way to do this. I guess there are several ways.. What would the expert do (That's you right?)? Should I have the information input, and image upload in the same form, or should I make a second form (maybe on a different page) for the upload? Is it necessary with two tables, one for the info and one for the image path, and then tie them together with the imageID, or is it fine to use just one table? Any thoughts would be appreciated! <!-- TechThat --> Hi, I have a form that allows a user to enter details and also to upload a image. The form works great and inserts into the database if i dont include the file input. What i would like happen is: The user enters details and selects a file The file then gets uploaded to ../images and the path is then written to bedroom1 variable The addresses and the bedroom1 variable that now holds the uploaded file path e.g. images/test.jpg is written to the database Form Code: [Select] <form action="admin.php" method="POST"> <table> <tr><td> Address 1: </td><td> <input type="text" name="address1" id="address1" onfocus="selected(this)" onblur="notselected(this)"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> Address 2: </td><td> <input type="text" name="address2" id="address2" onfocus="selected(this)" onblur="notselected(this)"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> County: </td><td> <input type="text" name="county" id="county" onfocus="selected(this)" onblur="notselected(this)"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> Bedroom 1: </td><td> <input type="file" name="bedroom1" id="bedroom1" input name="uploadedfile"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> <input type="submit" name="submitaddapartment" value="Add" id="registerbutton"> </td></tr> </table </form> And the php code to add it to the database Code: [Select] <?php //get variables $submitaddapartment = $_POST['submitaddapartment']; $address1 = strip_tags($_POST['address1']); $address2 = strip_tags($_POST['address2']); $county = strip_tags($_POST['county']); $bedroom1 = strip_tags($_POST['bedroom1']); if ($submitaddapartment) //if submit button was pressed { if ($address1&&$address2&&$county&&$bedroom1&&) //if fields arn't blank { include 'dbase.php'; //connect to database if ((($_FILES["bedroom1"]["type"] == "image/gif") || ($_FILES["bedroom1"]["type"] == "image/jpg") || ($_FILES["bedroom1"]["type"] == "image/pjpeg")) && ($_FILES["bedroom1"]["size"] < 20000000)) { if ($_FILES["bedroom1"]["error"] > 0) { echo "Return Code: " . $_FILES["bedroom1"]["error"] . "<br />"; } else { echo "Upload: " . $_FILES["bedroom1"]["name"] . "<br />"; echo "Type: " . $_FILES["bedroom1"]["type"] . "<br />"; echo "Size: " . ($_FILES["bedroom1"]["size"] / 1024) . " Kb<br />"; echo "Temp file: " . $_FILES["bedroom1"]["tmp_name"] . "<br />"; if (file_exists("../images/" . $_FILES["bedroom1"]["name"])) { echo $_FILES["bedroom1"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["bedroom1"]["tmp_name"], "/" . $_FILES["bedroom1"]["name"]); echo "Stored in: " . "../images/" . $_FILES["bedroom1"]["name"]; $bedroom1=".'images/.' $_FILES["bedroom1"]["name"]"; } } } else { echo "Invalid file"; } mysql_query ("INSERT INTO user VALUES(NULL,'$address1','$address2','$county','$bedroom1')"); } else { } } ?> Any help would be greatly appreciated. Thanks Fred 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. after cloasing connection of database i still got the values form database. Code: [Select] <?php session_start(); /* * To change this template, choose Tools | Templates * and open the template in the editor. */ require_once '../database/db_connecting.php'; $dbname="sahansevena";//set database name $con= setConnections();//make connections use implemented methode in db_connectiong.php mysql_select_db($dbname, $con); //update the time and date of the admin table $update_time="update admin set last_logged_date =CURDATE(), last_log_time=CURTIME() where username='$uname'limit 3,4"; //my admin table contain 5 colums they are id, username,password, last_logged_date, last_log_time $link= mysql_query($update_time); // mysql_select_db($dbname, $link); //$con=mysql_connect('localhost', 'root','ijts'); $result="select * from admin where username='a'"; $result=mysql_query($result); mysql_close($con); //here i just check after closing data baseconnection whether i do get reselts but i do, why? echo "after the cnnection was closed"; if(!$result){ echo "cont fetch data"; }else{ $row= mysql_fetch_array($result); echo "id".$row[0]."usrname".$row[1]."passwped".$row[2]."date".$row[3]."time".$row[4]; } // echo "<html>"; //echo "<table border='1' cellspacing='1' cellpadding='2' align='center'>"; // echo "<thead>"; // echo"<tr>"; // echo "<th>"; // echo ID; // echo"</th>"; // echo" <th>";echo Username; echo"</th>"; // echo"<th>";echo Password; echo"</th>"; // echo"<th>";echo Last_logged_date; echo "</th>"; // echo "<th>";echo Last_logged_time; echo "</th>"; // echo" </tr>"; // echo" </thead>"; // echo" <tbody>"; //while($row= mysql_fetch_array($result,MYSQL_BOTH)){ // echo "<tr>"; // echo "<td>"; // echo $row[0]; // echo "</td>"; // echo "<td>"; // echo $row[1]; // echo "</td>"; // echo "<td>"; // echo $row[2]; // echo "</td>"; // echo "<td>"; // echo $row[3]; // echo "</td>"; // echo "<td>"; // echo $row[4]; // echo "</td>"; // echo "</tr>"; // } // echo" </tbody>"; // echo "</table>"; // echo "</html>"; session_destroy(); session_commit(); echo "session and database are closed but i still get values from doatabase session is destroyed".$_SESSION['admin']; ?> session is destroyed but database connection is not closed. thanks hello everyone, I am about to start coding my pages to display results from a database but before i do i want to know information about the following : Is it best to upload images to a database?and display them accordingly? or is it best to use images from a directory? What is most commonly used and or more reliable? Another topic i have trouble finding information on is actually positioning the output from you mysql database, is this practice done with tables?fields and rows? What is this method called? And is there more then one way to go about controlling result layout on your page? Sorry about the 1001 questions , but i am unable to find a clear answer on the topic ..especially question two . Thanks in advance. I have a very tricky php problem here. At least for me. On a page that is editing a job entry. I have a database generated group of checkboxes some of which have been checked when the job was entered. So the script has to do two things generate the complete list of checkboxes and check the ones that have already been selected in the job request. The below script is what I've done trying to figure it out. Description of what I'm trying to do: I'm generating the the boxes with a call to my database for the list of checkboxes. Then I'm building an array with all the possible items that could be checked to compare against. Ok here is where I think my first problem is. I need to make a second query to another table "worklog" in the same database. This is where the list of checked items are stored. What is the best way to do this? A second query seems wrong "or at least not efficient" Code: [Select] <div class="text">Job Type (<a class="editlist" href="javascript:editlist('listedit.php?edit=typelist',700,400);">edit list</a>):</div> <div class="field"> <?PHP $connection=mysql_connect ("localhost", "user", "password") or die ("I cannot connect to the database."); $db=mysql_select_db ("database", $connection) or die (mysql_error()); $query = "SELECT type FROM typelist ORDER BY type ASC"; $sql_result = mysql_query($query, $connection) or die (mysql_error()); $i=1; echo '<table valign="top"><tr><td>'; while ($row = mysql_fetch_array($sql_result)) { $type = $row["type"]; if ($i > 1 && $i % 26 == 0) echo '</td><td>'; else if ($i) echo ''; ++$i; $aTypes = array ("Spec Ad Campaign", "100 x 100 Logo", "100 x 35 (Featured Developer/Broker)", "100 x 40 (Featured Lender)", "120 x 180 (Home Page Auto)", "120 x 45 (Half Tile - Jobs)", "120x60 (Section Sponsor)", "125 x 40 (Profile Page logo)", "135 x 31 (Autos)", "135 x 60 (Autos)", "135 x 60 (Logotile - Jobs)", "150 x 40 (Auto Logo)", "160 x 240 (monster tile)", "160 x 400 (Skyscraper)", "160 x 600 (Tower)", "170x30 (Section Sponsor)", "240 x 180 JPG/GIF Auto Video", "240 x 180 size FLV video ", "300 x 250 (Story Ad)", "300 x 600 (Halfpage)", "468 x 60 (Banner Jpg/Gif)", "728 x 90 (Leaderboard)", "95 x 75 (Sweeps Logo) ", "Agent Profile Page", "Contest", "Creative Change", "Employer Profile", "Holiday Shop", "Home of the Week", "HP Half Banner (234 x 60 Static)", "Mobile Ads (320x53 *300x50*216x36*168x28)", "Newsletter", "Peelback", "Pencil Ad", "Print", "Real Deals", "Resize ad campaign", "Roll-Over Skyscraper", "Site Sponsor logo (170x30)", "Splash Page Production", "Travel Deals Update", "Video Ad", "Web Development", "Web Maintenance", "Web Quote (or Design)"); $dbTypes = explode(',',$row['type']); foreach ($aTypes as $type){ if(in_array($aType,$dbTypes)){ echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type' CHECKED><span style='color:#000;'>$type</span></input><br/>"; }else { echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type'><span style='color:#000;'>$type</span></input><br/>"; } } } echo '</td></tr></table>'; ?> </div> I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.
my form page test.php
<?PHP
require_once("./include/membersite_config.php");
if(!$fgmembersite->CheckLogin())
{
$fgmembersite->RedirectToURL("login.php");
exit;
}
?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?> "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?> "><br><br>
<button>Submit</button>
my editform.php
<?php
$con = mysqli_connect("localhost","root","user","pass");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");
header("Location: test.php");
?>
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 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 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 ($points >= '2'): echo 'Corporal'; else: echo 'Recruit'; endif; endif; echo '<br /><br />'; } ?> I am getting the following error when i do the query in mysql: Code: [Select] #1109 - Unknown table 'points' in field list And when i open the webpage i get the following error: Code: [Select] Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/slay2day/public_html/points/members.php on line 18 Please Help me? Hello all, first post here so i hope i'm doing this right and am putting this into the right place. Im in the process of integrating a blog into my website, mostly it's set up however i'm currently working on the code to update the posts (theres only 2 parts a title and a comments[which is in actual fact the post content itself]). The code succsefully completes without any errors but for some reason it does not actually update the mysql database.. the code i'm using is as follows: Code: [Select] <?php require_once('header.php'); include "../blog/blogconfig.php"; if(isset($_POST['submit'])){ $update="UPDATE eq_blogarticle SET title='".$_POST['title']."',comments='".$_POST['comment']."' WHERE artid='$aid'"; if(!mysql_query($update)){ echo mysql_error(); }else{ header("location:blog.php?action=listmsgs"); exit; } } ?> <?php // get value of aid that sent from address bar by blog.php?action=listmsgs $aid=$_GET['aid']; // Retrieve data from database $sql="SELECT * FROM eq_blogarticle WHERE artid='$aid'"; $result=mysql_query($sql); $row=mysql_fetch_array($result); ?> <form name="form2" method="post" action="update.php"> <script type="text/javascript">var SITE_URL="<?php echo SITE_URL;?>";</script> <script type="text/javascript" src="<?php echo SITE_URL;?>/includes/js/nicEdit.js"></script> <script type="text/javascript"> //<![CDATA[ bkLib.onDomLoaded(function() { nicEditors.allTextAreas() }); //]]> </script> <body> <table width="100%" border="0" cellspacing="1"> <tr> <td colspan="2" class="temptitle">Equidisc Blog</td> </tr> <tr> <td width="74%" valign="top"> <table> Edit Blog Post <br> <br> Title: <br> <input name="title" type="text" class="input" id="title" value="<? echo $row['title']; ?>"> <br> <br> <span class="style1 style2 style3">Blog Post:</span> <br> <br> <textarea name="comments" cols="55" rows="12" class="input" id="comments"><? echo stripslashes($row['comments']); ?></textarea> <br> <br> <input name="aid" type="hidden" id="aid" value="<? echo $row['artid']; ?>"> <input type="submit" name="submit" value="Submit"> </form> </table> </td> <td width="26%" valign="top"><table width="100%" border="0" cellspacing="1"> <tr> <td colspan="2"><img src="../blog/images/fb.gif" width="16" height="16" /> <strong>Blog Menu</strong></td> </tr> <tr> <td><a href="blog.php">Home</a></td> </tr> <tr> <td><a href="blog.php?action=newblogpost">New Post </a></td> </tr> <tr> <td><a href="blog.php?action=listmsgs">Manage Posts </a></td> </tr> </table></td> </tr> </table> </body> <?php require_once('footer.php'); ?> Explanation: header/footer.php obvious ../blog/blogconfig.php holds my mysql connection settings and connects to the sql, this is the same config as is used for creating the new posts which i have no issues with so i dont think the issue lays there. If i run the query on phpmyadmin with dummy data it works fine and updates the entry.. Any help would be very much appreciated as i'm at the end of my tether with this!. Thanks in advance. Jo 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. 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. 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! 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?? I'm trying to create a page that allows people to backup their database on a web page but I'm having trouble with the ending of it. <?php require("../include/config.php"); $tables = array(); $qTables = mysql_query("SHOW TABLES"); while($row = mysql_fetch_row($qTables)) { $tables[] = $row[0]; } foreach($tables as $tab1) { $return.= "DROP TABLE IF EXISTS `".$tab1."`;"; $row2 = mysql_fetch_row(mysql_query("SHOW CREATE TABLE `" . $tab1 . "`")); $return.= "\n\n".$row2[1].";\n\n"; $result = mysql_query("SELECT * FROM ".$tab1) or die(mysql_error()); $num_fields = mysql_num_fields($result); $return.= "INSERT INTO `".$tab1."`"; $col = mysql_query('SELECT * FROM '.$tab1); $a = 0; $return.= " ("; while ($a < mysql_num_fields($col)) { $meta = mysql_fetch_field($col, $a); $return.= "`" . "$meta->name"; $a++; if ($a < mysql_num_fields($col)) { $return.= "`,"; } else { $return.= "`"; } } $return.= ")"; $return.=" VALUES\n("; for ($i = 0; $i < $num_fields; $i++){ while($row = mysql_fetch_row($result)){ for($j=0; $j<$num_fields; $j++){ if (isset($row[$j])) { $return.= "'".$row[$j]."'" ; } else { $return.= "''"; } if ($j < ($num_fields-1)) { $return.= ","; } if (j < ($num_fields-1)){ $return.= "),\n("; } else { $return.= ");\n"; } } } } } $handle = fopen("db-backup-".time()."-".(md5(implode(",",$tables))).".sql","w+"); fwrite($handle,$return); ?> At this part: if (j < ($num_fields-1)){ $return.= "),\n("; } else { $return.= ");\n"; } If it has finished going through all of the values it will put ); at the end and if it hasn't it will put ), and then continue with the next one. The problem I'm having is it's only doing the ), Can someone help me please? I'm trying to use php code that is stored in the sql database, but It doesn't seem to be executing the code. when I see the page source, its there but the server is not executing the command how do I accomplish this. Here is a simple code snippet to show what I am trying to do. $result = mysql_query("select * from data"); $row = mysql_fetch_array($result); echo $row['code']; In the code field in data table this is whats there. <?php echo "testing."; ?> |
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