PHP - 2 Database Tables For Users - Php Construction
Hey All,
Been banging my head into the wall on this one. I have 2 tables one for users 'myMembers' and one for products 'products'. Each table has a auto increment id. The myMembers id is the user id and the products table id is for the product id. I have a row in the products table for agent_id. I would like the agent_id to be filled with the id from the myMembers table. I took a look at the manual; still do not understand how to take the id from the myMembers table then place that id into the agent_id; so the products(id) can be listed under the specific members id(agent_id) in the products table. So far my script for the products table inserts items correctly, but does not file under the specific agent_id. Here is the script for entering items to the products table. Thanks for the guidance! <?php // Script Error Reporting error_reporting(E_ALL); ini_set('display_errors', '1'); ?> <?php // Delete Item Question to Admin, and Delete Product if they choose if (isset($_GET['deleteid'])) { echo 'Do you really want to delete product with ID of ' . $_GET['deleteid'] . '? <a href="inventory_list.php?yesdelete=' . $_GET['deleteid'] . '">Yes</a> | <a href="inventory_list.php">No</a>'; exit(); } if (isset($_GET['yesdelete'])) { // remove item from system and delete its picture // delete from database $id_to_delete = $_GET['yesdelete']; $sql = mysql_query("DELETE FROM products WHERE id='$id_to_delete' LIMIT 1") or die (mysql_error()); // unlink the image from server // Remove The Pic ------------------------------------------- $pictodelete = ("../inventory_images/$id_to_delete.jpg"); if (file_exists($pictodelete)) { unlink($pictodelete); } header("location: inventory_list.php"); exit(); } ?> <?php // Parse the form data and add inventory item to the system if (isset($_POST['product_name'])) { $product_name = mysql_real_escape_string($_POST['product_name']); $price = mysql_real_escape_string($_POST['price']); $category = mysql_real_escape_string($_POST['category']); $subcategory = mysql_real_escape_string($_POST['subcategory']); $details = mysql_real_escape_string($_POST['details']); // See if that product name is an identical match to another product in the system $sql = mysql_query("SELECT id FROM products WHERE product_name='$product_name' LIMIT 1"); $productMatch = mysql_num_rows($sql); // count the output amount if ($productMatch > 0) { echo 'Sorry you tried to place a duplicate "Product Name" into the system, <a href="inventory_list.php">click here</a>'; exit(); } // Add this product into the database now $sql = mysql_query("INSERT INTO products (product_name, agent_id price, details, category, subcategory, date_added) VALUES('$product_name','$price','$details','$category','$subcategory',now())") or die (mysql_error()); $pid = mysql_insert_id(); // Place image in the folder $newname = "$pid.jpg"; move_uploaded_file( $_FILES['fileField']['tmp_name'], "../inventory_images/$newname"); header("location: inventory_list.php"); exit(); } ?> <?php // This block grabs the whole list for viewing $product_list = ""; $sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC"); $productCount = mysql_num_rows($sql); // count the output amount if ($productCount > 0) { while($row = mysql_fetch_array($sql)){ $id = $row["id"]; $product_name = $row["product_name"]; $price = $row["price"]; $date_added = strftime("%b %d, %Y", strtotime($row["date_added"])); $product_list .= "Product ID: $id - <strong>$product_name</strong> - $$price - <em>Added $date_added</em> <a href='inventory_edit.php?pid=$id'>edit</a> • <a href='inventory_list.php?deleteid=$id'>delete</a><br />"; } } else { $product_list = "You have no products yet"; } ?> Similar TutorialsI have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.
my form page test.php
<?PHP require_once("./include/membersite_config.php"); if(!$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 am looking for help to how to build a function, that will redirect you to a construction page while they site is being updated with new updates, but allow staff only access while its being constructed on, so while i do my updates to my site, i can test the new update, before others have access When a user logs in to my members area, their online status is set to 2 in my database and they are displayed as online for everyone else to see. And when they click log out their online status is set to 1 and they are no longer shown as online. My problem is when the session expires due to inactivity, the database isnt updated and they are still shown as online. So whenever the access a page in the members area I have made it so that time also gets entered into the database. And my goal is to have the users online page auto refresh every 5 minutes to check the current time against the time stored in the database for that user and if 15 minutes has past, have the online status in the database updated to 1 again. im using pdo but im a novice at it and havnt been able to find much help on google. Is there a way i can use an if statement to check the two times and only update the online field to 1 for the inactive users while leaving the users who appear active as 2? Sorry if this was confusing, i wasnt too sure how to word it properly. Hi there everyone. I'm really confused about this as I want to email eveyone in my DB when a new Event is created by a member. I have a test code which I used to make sure it works and it sends to 5 email addresses (which is all of them) in the test table. This works fine. All members with a 0 get the mail. So I now just took the code and changed the table name (obviously!!) and put it in my live site which has 50 members so far. It now doesn't send any emails??? Not sure why this is as the code is exactly the same (except the table name). $result = mysql_query("SELECT email FROM membership WHERE send_as_email = '0'"); while ($row = mysql_fetch_array($result)) { sendMail($row[0]); } mysql_free_result($result); function sendMail($to){ $subject = 'New Event Created; $message = "Hi there,\n Just letting you know of a new Event that has been created.\n Title: $title\n Town: $town\n Date: $date_convert\n For more details or to attend this Event, please login \n Many thanks\n The Team\n \n ****************************************************************************\n THIS EMAIL IS AUTOMATED. DO NOT REPLY.\n To turn notifications off, change the notification settings in your Profile.\n ****************************************************************************\n"; $headers = 'From: members@mysite.co.uk' . "\r\n" . 'X-Mailer: PHP/' . phpversion(); mail($to, $subject, $message, $headers); } Sir/ma'am,
With the script I'm using to run my website, I've been trying to add an additional feature for the users to add/edit. I'll try to provide as much info as I can, hopefully it'll help.
Here is the code I'm using to display the user's unique info from the db.
<a class="wallet-edit"><?php echo $_SESSION['simple_auth']['INFO']?></a>That displays the user's info from the column 'INFO' perfectly. It's also a js popup to a menu to where I'm hoping to add a single textbox to edit the INFO. The script uses a similar function to edit the password with a popup. I've tried modifying the code to edit the INFO column but it doesn't work. Here is the default code it has to edit the password. I'm not sure if it can be changed to edit another column or needs a new piece of code for that. // user edit $('body').on('click', '.username-edit', function() { $('#modal').html(' '); var output = '<div class="modal-content"><h5><?php echo lang::get("Change password")?></h5><hr />'; output += '<h5><?php echo lang::get("New password:")?></h5><input type="password" name="password" id="password" value="" class="text ui-widget-content ui-corner-all" />'; output += '<h5><?php echo lang::get("Confirm password:")?></h5><input type="password" name="password2" id="password2" value="" class="text ui-widget-content ui-corner-all" />'; output += '</div>'; output += '<div class="modal-buttons right">'; output += '<button id="confirm-button" type="button" class="nice radius button"><?php echo lang::get("Change")?></button>'; output += '</div>'; output += '<a class="close-reveal-modal"></a>'; $('#modal').append(output); $('#second_modal').hide(); $('#modal').reveal(); $('#confirm-button').click(function(){ $('#password').css('border-color', '#CCCCCC'); $('#password2').css('border-color', '#CCCCCC'); var password = $('#password').val(); var password2 = $('#password2').val(); if(typeof(password) === 'undefined' || password == ''){ $('#password').css('border-color', 'red'); return false; } if(password != password2){ $('#password2').css('border-color', 'red'); return false; } password_data = encodeURIComponent(password); $.post("<?php echo gatorconf::get('base_url')?>", { changepassword: password_data} ).done(function(data) { // flush window.location.href = '<?php echo gatorconf::get('base_url')?>'; }); }); });If the code above can be edited to work with what I'm trying to do, it of course only needs one textbox and doesn't have to be confirmed by a second input. Please help! Thanks! less than 6 characters. I think it's the way my code is ordered. I've tried switching the commands around, no luck. Help please. Code: [Select] <?php //begin register script $submit = $_POST['submit']; //form data $username= strip_tags ($_POST['username']); $email= strip_tags($_POST['email']); $pwd= strip_tags($_POST['pwd']); $confirmpwd= strip_tags($_POST['confirmpwd']); $date = date("Y-m-d"); if ($submit) { //check for required form data if($username&&$pwd&&$confirmpwd&&$email) { //check length of username if (strlen($username)>25||strlen($username)<6) { echo "<p class='warning'>username must be bewteen 6 and 25 characters</p>"; } else { //check password length if (strlen($pwd)>25||strlen($pwd)<6) { echo "<p class='warning'>password must be between 6 and 25 characters</p>"; } else { //register the user echo "<p class='success'>Thanks for signing up!</p>"; } } //check if passwords match if ($pwd==$confirmpwd) { } else { echo "<p class='warning'>your passwords do not match</p>"; } //encrypt password $pwd = md5($pwd); $confirmpwd = md5($confirmpwd); //open database $connect = mysql_connect("xxxxxxxx", "xxxxxxxx", "xxxxxxxx"); mysql_select_db("digital"); //select database //register the user $queryreg = mysql_query(" INSERT INTO users VALUES ('','$username', '$email', '$pwd') "); die("<p class='success'>Thank you for signing up you have been registered"); } else { echo "<p class='warning'>please fill in all fields</p>"; } } ?> Hello folks I am new to php and I have been trying to put together a database that a user can search and choose from the results. I have managed to make this script by copying code from google searches and trial and error. The script so far has been tested and works. The hard part is the code for choosing from the results, I have tried some things but I have been far from the mark, the thing is I can't get my head around the problem, if the first field is a number which is unique to each row, how can I pick that up in a php argument. I have tried making the first field an href link to send that number to a different table which would collect the results of the users choices, but I'm just not sure what to put in the code. Could someone throw me a lifeline here I've searched for hours on google to find any code that looks like it would work with no luck. // Get the search variable from URL $var = @$_GET['a'] ; $trimmed1 = trim($var); //trim whitespace from the stored variable $var = @$_GET['b'] ; $trimmed2 = trim($var); $var = @$_GET['c'] ; $trimmed3 = trim($var); $var = @$_GET['d'] ; $trimmed4 = trim($var); $var = @$_GET['e'] ; $trimmed5 = trim($var); $var = @$_GET['f'] ; $trimmed6 = trim($var); //connect to your database mysql_connect("localhost","root",""); //(host, username, password) //specify database mysql_select_db("a2149809_MV") or die("Unable to select database"); //select which database we're using // Build SQL Query $query = "SELECT * FROM `table` WHERE `field1` LIKE \"%$trimmed1%\" AND `field2` LIKE \"%$trimmed2%\" AND `field3` LIKE \"%$trimmed3%\" AND `field4` LIKE \"%$trimmed4%\" AND `field5` LIKE \"%$trimmed5%\" AND `field6` LIKE \"%$trimmed6%\" order by `field1`"; $result=mysql_query($query); $num=mysql_num_rows($result); mysql_close(); <table width="100%" border=2 cellspacing=2 cellpadding=2> <tr><form name="form" action="" method="get"> <td colspan="6"><input type="submit" name="Submit" value="Search" /> </td> </tr> <tr> <td><input type="text" name="a" value="" size="4" /></td> <td><input type="text" name="b" value="" size="40" /></td> <td><input type="text" name="c" value="" size="3" /></td> <td><input type="text" name="d" value="" size="10" /></td> <td><input type="text" name="e" value="" size="10" /></td> <td><input type="text" name="f" value="" size="10" /></td> </form></tr> <?php $i=0; while ($i < $num) { $f1=mysql_result($result,$i,"Field1"); $f2=mysql_result($result,$i,"Field2"); $f3=mysql_result($result,$i,"Field3"); $f4=mysql_result($result,$i,"Field4"); $f5=mysql_result($result,$i,"Field5"); $f6=mysql_result($result,$i,"Field6"); ?> <tr> <td><?php echo $f1; ?></td> <td><?php echo $f2; ?></td> <td><?php echo $f3; ?></td> <td><?php echo $f4; ?></td> <td><?php echo $f5; ?></td> <td><?php echo $f6; ?></td> </tr> <?php $i++; } ?> </table> Hello I'm trying to set up a user area for my site where it displays the current logged in users ranking and other information in the future. <? ini_set('display_errors', 1); require_once "header.php"; $sql = "SELECT * FROM users WHERE username = ?"; if($stmt = mysqli_prepare($link, $sql)){ mysqli_stmt_bind_param($stmt, 's', $_SESSION['username']); if(mysqli_stmt_execute($stmt)){ $info = mysqli_fetch_array($stmt); echo "Current rank:" . $info['rank']; } else { echo "Can't find user"; } } mysqli_stmt_close($stmt); ?> That's the code I currently have but it gives me the error "but get an error message of mysqli_fetch_array() expects parameter 1 to be mysqli_result" Hello,
Here is my current code:
index.php
<html> <body> <form id="hey" name="hey" method="post" onsubmit="return false"> Name:<input type="text" name="name"> <input type="submit" name="submit" value="click"> </form> <table class="table table-bordered" id="update"> <thead> <th>Name</th> </thead> <tbody> <script type="text/javascript"> $("#hey").submit(function() { $.ajax({ type: 'GET', url: 'response.php', data: { username: $('#name').val()}, success: function (data) { $("#update").prepend(data); }, error: function (xhr, ajaxOptions, thrownError) { alert(thrownError); } }); }); </script> </tbody> </table>response.php <?php $username = $_GET['username']; echo "<tr>"; echo "<td>$username</td>"; echo "</tr>"; ?>This code works fine, it prints out the rows as what the user enters. Now what I want to do is, log all these entries to a mySQL database and also, display these rows over the site. i.e., any user who is online, should be seeing this without having to refresh the page too.. Real time updates in a way. How can I achieve that? Thanks! This topic has been moved to Other Libraries and Frameworks. http://www.phpfreaks.com/forums/index.php?topic=359217.0 Hi, I have two html made text boxes one that is called "name" and another that is called "regnum". I also have a submit button. They are both used to add data to a database. The name text box should add a name to the database under the "name" heading and the regnum should add a number under the "regnum" heading Here is the code for them: HTML Code: <form action="" method="post"> <p> Name: <input type="text" name="name"/> </p> <p> Regnum: <input type="text" name="regnum"/> </p> <p> <input type="submit" value="Add To Database" name = "submit2" /> </p> </form> Here is the PHP code that i am using for adding the user to the database: PHP Code: $host="localhost"; $username="root"; $password=""; $database="lab2"; mysql_connect("$host", "$username", "$password") or die(mysql_error()); mysql_select_db("$database") or die(mysql_error()); $name = $_POST['name']; $regnum = $_POST['regnum']; if(!$_POST['submit2']){ echo "Enter A Vaue"; }else{ mysql_query("INSERT INTO lab2('name', 'regnum') VALUES(NULL, '$name', '$regnum')") or die(mysql_error()); echo "User Added To Database"; } The problem i get with this is "Undefined Index name and regnum". I watched a video on youtube and this is how the guy did it but it worked for him and for some reason it doesn't work for me. Can anyone help?? Thanks. Hi there everyone, I have the below code (needs to be sanitized for SQL injection - i know ). Submitting to a MYSQL database. I wish to check on form submit if the users email address already exists, and if so display a simple error message (even just a windows error message) stating "the email address you have entered already exists". I don't really know where to start with this, or what the code should look like, so any help and direction would be massively appreciated. All i do know, is the email column is set to unique, and when i attempt to submit using an email i know exists the code appears to run successfully without spitting out any errors (i.e. the web url changes to the below php code) but the table doesn't update (which is correct). I just don't know how to then return the user to the form (preferably with all their info still entered) when this happens along with a nice error message... Kind regards, Tom. Code: [Select] <? $localhost="00.000.000.00"; $username="###"; $password="###"; $database="mfirst"; $firstname=$_POST['firstname']; $surname=$_POST['surname']; $dob="{$_POST['dobyear']}-{$_POST['dobmonth']}-{$_POST['dobday']}"; if (isset ($_POST['permissionnewsletter']) || (!empty ($_POST['permissionnewsletter']))) { $permissionnewsletter = "Yes"; } else { $permissionnewsletter = "No"; } $email=$_POST['email']; $userpassword=$_POST['userpassword']; $telephone=$_POST['telephone']; mysql_connect($localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $insertintocustomerdetail = "INSERT INTO customerdetail VALUES (NULL,'$firstname','$surname','$dob','$permissionnewsletter','$email','$userpassword','$telephone',now())"; mysql_query($insertintocustomerdetail); $addressline1=$_POST['addressline1']; $addressline2=$_POST['addressline2']; $cityortown=$_POST['cityortown']; $county=$_POST['county']; $postcode=$_POST['postcode']=strtoupper(@$_REQUEST['postcode']); $insertintoaddresstable = "INSERT INTO addresstable VALUES (NULL,LAST_INSERT_ID(),'$addressline1','$addressline2','$cityortown','$county','$postcode',now())"; mysql_query($insertintoaddresstable); mysql_close(); ?> Hi All, I've searched long and hard accross the web for an answer to this and finnally given in and requesting help. Here's what i have, i have a database setup and working fine. What i would like to do is for an administrator to be able to update my users details. It may sound odd, why don't you let your users update their own details? Well the administrators are dispatchers if you like, and my users are the 'dispatchees', for want of a better word. So i would like my administrators to be able to dispatch my users with routes and my users be able to see the routes that have been dispatched to them. I've setup a login area and a page that pulls there routes off the database, depending on their login details, i.e. jack will see his routes and jill will see her's independantly. This works by me editing the appropriate columns/rows of my database using phpmyadmin. What i'd like now is for administrators (who are directed to a seperate page, with more controls) to be able to do the same as me (updating the database) but by using a php form/script. I'd like to be able to select the routes from a second table on the same database if possible, to try and keep everything tidy. So my dispatcher would select Route001 from a drop down list, this would fill in the text fields next to the route field with From To, so my dispatcher would know what route001 actually is from/ too, choose a username (now being driven from my other table) and hit dispatch. My user would login to their area, hit view dispatched routes and it would display Route 001 with the correct information. The login area was a downloaded script i modified to suit and is called Login-Redirect_v1.31_FULL Many thanks in advance, hope you can sort of understand what i want Josh PHP/MySQL ability:Novice Hello
I am trying to work out how many regular users I have to my site and how long those users tend to be users..
So, I have a table that logs every time a user visits my site and logs in, it stores the date / time as a unix timestamp and it logs their user id.
I started by getting the id's of any user who logs in more than 5 times in a specified period, but now I want to extend that...
SELECT userID as user, count(userID) as logins FROM login_history where timestamp > UNIX_TIMESTAMP('2014-06-01 00:00:00') and timestamp < UNIX_TIMESTAMP('2014-07-01 00:00:00') group by user having logins > 5; Hi can anybody help I need to export and download tables from a database into a excel sheet. I have this code and it works what I need is to export specific fields and not just the whole table can anyone help modifying the code to export certain fields within the table please? Here is the code... Code: [Select] <?php $host = 'localhost'; $user = 'user'; $pass = 'password'; $db = 'qdbname'; $table = 'tablename'; $file = 'export'; $link = mysql_connect($host, $user, $pass) or die("Can not connect." . mysql_error()); mysql_select_db($db) or die("Can not connect."); $result = mysql_query("SHOW COLUMNS FROM ".$table.""); $i = 0; if (mysql_num_rows($result) > 0) { while ($row = mysql_fetch_assoc($result)) { $csv_output .= $row['Field']."; "; $i++; } } $csv_output .= "\n"; $values = mysql_query("SELECT * FROM ".$table.""); while ($rowr = mysql_fetch_row($values)) { for ($j=0;$j<$i;$j++) { $csv_output .= $rowr[$j]."; "; } $csv_output .= "\n"; } $filename = $file."_".date("Y-m-d_H-i",time()); header("Content-type: application/vnd.ms-excel"); header("Content-disposition: csv" . date("Y-m-d") . ".csv"); header( "Content-disposition: filename=".$filename.".csv"); print $csv_output; exit; ?> Hi guys, Is it possible to insert into two tables in a database? Thanks Hi, I am re-visiting my database table relationship for criminal incidents. Six tables are involved plus a junction table to make it seven tables in total. I need someone to review this relationship and advise whether or not it is correctly set up. The relationships are as follows: 1. A person can belong to more than one incident: t_persons (one-to-many) t)_incidents_persons 2. A person can only have one nationality by birth t_persons (one-to-one) t_countries 3. An agency can have more than one incident t_agencies (one-to-many) t_incidents 4. A status (e.g. closed incident) can belong to more than one incident t_status (one-to-many_ t_incidents 5. A keyword (i.e. offence type eg theft) t_offencekeywords (one-to-many) t_incidents How we can make the connection to ODBC in PHP If any shares the coding part then it would be great. (Table to database) I have two tables. Let's call the first one items and the second one item categories. Now the items table would look something like this: id(auto_increment id) name description categoryid The item categories table would look something like this: categoryid name description An entry in the items table would look something like this: categoryid = 1,2 name = Thing description = something id = 1 Say I want to retrieve records that contain "1" in the categoryid column. How would I do that? |