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PHP - Php Autofill Without Mysql Database
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, $body); header("Location:customize_search.php"); So, this takes the user to a more advanced form after they 'send' the first form. 1 - Is it possible to have the second form autofill with the POSTed data and NOT use MySQL or database whatever? 2 - If so, how?? Similar TutorialsAt the moment I am creating a search function for my website. The approach I have in mind is a pseudo-PHP database. To give an example: A HTML form will submit the results to a PHP file. HTML FORM - Colour: Black PHP RESULT PAGE - if ($_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. Ok, this is stumping my mind. I have a PHP/AJAX script which works almost perfectly. But its lacking one detail I absolutely need. Every AJAX demo I have seen is missing it or doesn't work right. Here is where the problem is. I have a list of countries, I want the user to spell it right when using my search, so you put in say "Virgin Islands" It should pull up the following in the list: Virgin Islands (British) St. Thomas (U.S. Virgin Islands) However all I get is: Virgin Islands (British) It has to do with the MySQL Query... Here is what came with the demo: $query = sprintf("SELECT ctry_name FROM international WHERE ctry_name '%s%%' GROUP BY ctry_name", $ctry_name); That's what pulls up only Virgin Islands I tried to mod the query to be "SELECT ctry_name FROM interntaional WHERE ctry_name LIKE '%" . $ctry_name . "%'" But it breaks the script, in other AJAX Examples, it doesn't even look it up and returns all entries in the database. Like I said the above example works fine, it just returns just the one value, I want ti to return all variables. Hi Hope someone can help me, I have a database that contain records like name , age etc. This data displays in a premade profile page. The profile page is linked to from a page that displays a short description with a image that is also populated from the database. I only want to have one profile page, and that page needs to be populated when the image link is clicked on the description page. Example, you click on short description (displaying some of row 3 info, from database for example)page link, profile page opens, and populates fields with row 3s data. I have the code to display the data on the description page and the profile page. I dont have allot of know how with coding, my code at the moment was done trough Dreamweaver Cs3. That code is working fine and displaying the data in the fields created for them. I would realy appreciate some guidance. This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=330523.0 When I click on the letter of the alphabet on my form page, it will bring up the results of my query. I got this part working. The next part is to be able to click on any of the populated results and have it autofill my form with organization, first name, last name, email address and phone number.
I know that I need to use Ajax and jQuery to accomplish this.
Here is my html
<!-- Letter Search -->
<div class="searchBox span12">
<h3>Choose the First Letter of the Person's Last Name</h3>
<ul>
<li class="alphabets" id="A"><a href="?by=A">A</a></li>
<li class="alphabets" id="B"><a href="?by=B">B</a></li>
<li class="alphabets" id="C"><a href="?by=C">C</a></li>
<li class="alphabets" id="D"><a href="?by=D">D</a></li>
<li class="alphabets" id="E"><a href="?by=E">E</a></li>
<li class="alphabets" id="F"><a href="?by=F">F</a></li>
<li class="alphabets" id="G"><a href="?by=G">G</a></li>
<li class="alphabets" id="H"><a href="?by=H">H</a></li>
<li class="alphabets" id="I"><a href="?by=I">I</a></li>
<li class="alphabets" id="J"><a href="?by=J">J</a></li>
<li class="alphabets" id="K"><a href="?by=K">K</a></li>
<li class="alphabets" id="L"><a href="?by=L">L</a></li>
<li class="alphabets" id="M"><a href="?by=M">M</a></li>
<li class="alphabets" id="N"><a href="?by=N">N</a></li>
<li class="alphabets" id="O"><a href="?by=O">O</a></li>
<li class="alphabets" id="P"><a href="?by=P">P</a></li>
<li class="alphabets" id="Q"><a href="?by=Q">Q</a></li>
<li class="alphabets" id="R"><a href="?by=R">R</a></li>
<li class="alphabets" id="S"><a href="?by=S">S</a></li>
<li class="alphabets" id="T"><a href="?by=T">T</a></li>
<li class="alphabets" id="U"><a href="?by=U">U</a></li>
<li class="alphabets" id="V"><a href="?by=V">V</a></li>
<li class="alphabets" id="W"><a href="?by=W">W</a></li>
<li class="alphabets" id="X"><a href="?by=X">X</a></li>
<li class="alphabets" id="Y"><a href="?by=Y">Y</a></li>
<li class="alphabets" id="Z"><a href="?by=Z">Z</a></li>
</ul>
<? include('search.php'); ?>
</div>
<hr style="color:#ccc; margin-bottom:20px;" />
<!-- Main Form -->
<div id="mainForm">
<form method="post" id="icsForm" class="searchBox span12">
<div id="col1" class"span6">
<h3>Contact Information</h3>
<label>Church / Organization:</label><input type="text" name="organization" id="organization" class="span6 upright" /><br />
<label>First Name:</label><input type="text" name="firstName" id="firstName" class="span6 upright" />
<label>Last Name:</label><input type="text" name="lastName" id="lastName" class="span6 left upright" />
<label>Email Address:</label><input type="text" name="email" id="email" class="span6 left upright" />
<label>Phone Number:</label><input type="text" name="phone" id="phone" class="span6 left upright" />
</div>
</div>
Here is my php
if(preg_match("/^[A-Z | a-z]+/", $_POST['name'])){
$name=$_POST['name'];
}
if(isset($_GET['by'])){
$letter=$_GET['by'];
//query to sort by last name
$sql="SELECT contact_id, first_name, last_name, church_org, email_address, phone_number FROM ics_data WHERE last_name LIKE '$letter%' ORDER BY last_name ASC";
//run the query against the mysql query function
$result=mysql_query($sql);
//count results
$numrows=mysql_num_rows($result);
echo "<p>" .$numrows . " results found for " . $letter . "</p>";
//Create while loop and loop through result set
while($row=mysql_fetch_array($result)){
$first_name=$row['first_name'];
$last_name=$row['last_name'];
$church_org=$row['church_org'];
$email_address=$row['email_address'];
$phone_number=$row['phone_number'];
$contact_id=$row['contact_id'];
//display the result of the array
echo "<div id=\"search-results\">";
echo "<ul class=\"letter-results\">\n";
echo "<li class=\"result-row\">" . "<a href=\"#\" class=\"testclass\">" .$first_name . " " .$last_name . "". ", " ."" .$church_org ."</a></li>\n";
echo "</ul>";
echo "</div>";
}
}
Here is my Javascript file (Ajax)
$(document).ready( function() {
function formfill() {
var organization = $('#organization').val();
var firstname = $('#firstname').val();
var lastname = $('#lastname').val();
var email = $('#email').val();
var phone = $('#phone').val();
$.ajax ({
method: "GET",
url: "search.php",
dataType: 'json',
data: {
organization:organization, firstname:firstname, lastname:lastname, email:email, phone:phone
},
type: "POST",
success: function(data) {
$organization
$firstname
$lastname
$email
$phone
},
failu function() {
alert('fail!');
}
});
}
I know that I do not have a reference yet to JSON in my php file and that it is needed. I'm not solid on the Ajax part. That is the part that is tripping me up. I know that I need to make the form autofill when clicking on a specific result returned from my query... but i'm not sure how to do that.
Thank you in advance for any help or advice you can give!! I am relatively new to programming. Hopefully I posted this in the right forum as a lot of these technologies overlap.
Edited by kjetterman, 26 September 2014 - 10:23 AM. 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. 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); Hi there, I have been creating a website which shows products of the companys (www.theadventurestartshere.org) and I have been trying to make some filters for the products using URL Parameters and recordsheet filters... Can someone advise me on the easiest way to do this? As I have created a method to do it with, (check website) but it has to be entered manually and I was hoping there is an easier way? Please Note I like to use drop down boxes for the filters but if there is a way to do the checkbox style you see on say Amazon then that would be brilliant... I use Dreamweaver CS5, and the newest versions of PHP and MySql Many Thanks, Paul 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! 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 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. 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 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 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 I know this is a very simple and probably stupid question - but what is a patch? i've tried searching online for an explanation but I just find articles on the 'best practice' and it doesn't break it down into what it actually is!
I've just been looking into new hosting and they offer a managed service, which includes database patching.
Could someone please enlighten me?
Edited by paddyfields, 10 June 2014 - 04:49 AM. I looked into this everywhere,but the code does not make sense, is not properly documented and is outdated. Ive been trying to achieve this for the past hour. Anybody know how I can do this? 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, 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"); ?> |
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