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PHP - Retrieving Database Info Based On Userlogged In
I have two tables.
Table Name:Users Fields: User_name user_email user_level pwd 2.Reference Fields: refid username origin destination user_name in the users table and the username field in reference fields are common fields. There is user order form.whenever an user places an order, refid field in reference table will be updated.So the user will be provided with an refid Steps: 1.User needs to log in with a valid user id and pwd 2.Once logged in, there will be search, where the user will input the refid which has been provided to him during the time of order placement. 3.Now User is able to view all the details for any refid 3.Up to this we have completed. Query: Now we need to retrieve the details based on the user logged in. For eg: user 'USER A' has been provided with the referenceid '1234' during the time of order placement user 'USER B' has been provided with the referenceid '2468' during the time of order placement When the userA login and enter the refid as '2468' he should not get any details.He should get details only for the reference ids which is assigned to him. Similar TutorialsI have two tables. Table Name:Users Fields: User_name user_email user_level pwd 2.Reference Fields: refid username origin destination user_name in the users table and the username field in reference fields are common fields. There is user order form.whenever an user places an order, refid field in reference table will be updated.So the user will be provided with an refid Steps: 1.User needs to log in with a valid user id and pwd 2.Once logged in, there will be search, where the user will input the refid which has been provided to him during the time of order placement. 3.Now User is able to view all the details for any refid 3.Up to this we have completed. Query: Now we need to retrieve the details based on the user logged in. For eg: user 'USER A' has been provided with the referenceid '1234' during the time of order placement user 'USER B' has been provided with the referenceid '2468' during the time of order placement When the userA login and enter the refid as '2468' he should not get any details.He should get details only for the reference ids which is assigned to him. In my script, users login with their Username & Password. However, I'd like to be able to echo the email address used on their account.
I've tried adding the email to the session I'm not having much luck... Here's a piece of the login code(untouched);
$username = $_POST['name'];
$passwd = $_POST['passwd'];
$query = "SELECT name,passwd FROM users WHERE CONCAT('0x', hex(passwd)) = '{$salt}'";
$result = mysql_query($query);
$login_ok = false;
if(mysql_num_rows($result) > 0) {
$login_ok = true;
}
if($login_ok)
{
$row = mysql_fetch_array($result, MYSQL_NUM);
$_SESSION['user'] = $row;
I've also tried messing around with this piece below in a few different ways but still nothing.
<?php echo htmlentities($_SESSION['user']['email'], ENT_QUOTES, 'UTF-8'); ?>Any help is greatly appreciated.. ok im back quicker than i thought.... i got my drop box sorted and i got it reloading the page. so it all works correctly. but how do i get the page to display information regarding the film i have selected in the drop box.? i have no code for this at the mo. also i would like the drop box to display the selected item at top of box when it refreshes code for drop box: Code: [Select] <FORM> <?php $result = mysql_query( "SELECT * FROM movie_info ORDER BY title ASC ") ; echo "<select name= Film onChange='submit()' >film name</option>"; while ($nt=mysql_fetch_array($result)){ ?> <?php echo "<option value='$nt[id]'>$nt[title] </option>"; } ?> </select> </FORM> any help would be great Hello, I have a few questions. First, this is what I am trying to accomplish - I am trying to take values that are entered into forms. And then store them into a database. My plan was to take them into an array, and create a loop that wrote the values to the database. So, the first step I thought would be to learn how to take info from a form, then display it onto the screen. Theoretically, if I could do that, I could just learn the MySQL commands to write to the database, and paste them instead of printing to the screen. So, I gave it a go. It didn't work. The way the 'Key' is chosen with arrays in PHP was different than I thought. I gave my sample code to a friend and asked for his help. He gave me a working copy back, but signed off before I could ask questions This is what I am working with: Code: [Select] <html> <body> <form name = "barInfo" method = "post"> Establishment name: <input type = "text" name = "EstablishmentName"> <br> Street Address: <input type = "text" name = "StreetAddress"> <br> City: <input type = "text" name = "City"> <br> State: <input type = "text" name = "State"> <br> Zip: <input type = "text" name = "Zip"> <br> <input type="submit" name="Submit" value="Submit"> </form> <?php if (!empty($_POST['EstablishmentName'])) { <br> print "Now, lets see if this shit works..."; <br> print "Establishment Name:" . $_POST['EstablishmentName']; <br> print "Street Address:" . $_POST['StreetAddress']; <br> print "City:" . $_POST['City']; <br> print "State:" . $_POST['State']; <br> print "Zip:" . $_POST['Zip']; <br> } ?> </body> </html> Ok, first off I get a parse error or something like that when I try it. My book has '<br>' thrown all over the place inside '<?php>' things. How can I do an endl; type thing in PHP? Obv <br> does not work., Code: [Select] print "\n";Does nothing, either....? Second, the reason my code didn't display anything, but his did, was because I had a loop that displayed $details[$x], and 1 was added to $x every time it looped. Am I right in saying that during the html part of the code, whatever value is assigned to 'name = ', is the key pointing to that value in my array? If that is true, how would I make that become an integer and use it with a loop. Do I have to put: Code: [Select] name = "1", name = "2", etc...? Thanks in advance for the help! Sorry if I am not referring to terms correctly, I just started looking at HTML/PHP last week... This is for band website for its event listings. The main event page list everything fine, but once selecting the "Details" link which activates the switch event. Its no longer showing anything. When I had one event in the data base it worked fine, but now I have 3 events in the database and now its not working. Page can be viewed he http://184.66.66.169/ffy/event.php Code: [Select] <?php //Event Code Here $eventid = (isset($_GET["id"])) ? intval($_GET["id"]) : 0; switch($_GET["list"]=='true') { case "0": if($_GET["id"]== $eventid) { $result = mysql_query("SELECT * FROM event ORDER by eventdate DESC"); if (!$result) { die("query failed: " . msql_error()); } while ($row = mysql_fetch_array($result)) { list($id, $eventdate, $header, $description, $image, $location) = $row; $description = nl2br($description); $eventdate = date("M j, Y",strtotime("$eventdate")); print(' <table width="680" border="0" cellpadding="14" cellspacing="0"> <tr> <td> <div class="myFont"><font size="+1" color="#4e8baf">'.$eventdate.' - </font><font size="+1"><b>'.$header.'</b></font></div> <font size="-2" color="#CCCCCC"><a href="event.php?list=true&eventid='.$id.'">Details</a></font><br /> <hr color="#FFFFFF" width="100" align="left" size="1"> </td> </tr> </table> '); }} break; case "true": if($_GET["id"]==$eventid) { $resultd = mysql_query("SELECT * FROM event WHERE id=$eventid LIMIT 1"); if (!$resultd) { die("query failed: " . msql_error()); } while ($row = mysql_fetch_array($resultd)) { list($id, $eventdate, $header, $description, $image, $location) = $row; $description = nl2br($description); $eventdate = date("M j, Y",strtotime("$eventdate")); print(' <table width="680" border="0" cellpadding="14" cellspacing="0"> <tr> <td> <div class="myFont"><font size="+1" color="#4e8baf">'.$eventdate.' - </font><font size="+1"><b>'.$header.'</b></font></div> '.$description.'<br /> <hr color="#FFFFFF" width="100" align="left" size="1"> <center> '.$location.' </center> </td> </tr> </table> '); } break; }} ?> 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");
?>
Code: [Select] $query = mysql_query("SELECT a.*, b.* FROM friendlist a INNER JOIN friendlist b ON (a.friendemail=b.friendemail) INNER JOIN users c ON (b.friendemail = c.EmailAddress) WHERE a.email = 'asdf@gmail.com' AND c.Username LIKE '%carol%' GROUP BY a.id ORDER BY count(*) DESC"); Code: [Select] while ($showfriends = mysql_fetch_array($query)) { echo $showfriends['Username']; } and I would get nothing. It produces the correct number of <div> so i know it's getting through, but it's having trouble displaying the entries? Hi all, I have a number of tables within a mysql database and I'm stuck on how to pull information from one based on the results being displayed by another. The site shows items for sale where by all the product information is held in tablea, when an individual item is being displayed I need to be able to pull a manufacturer profile from tableb based on the manufacturer name stored in tablea column profile. The manufacturer name is already being called into the page by existing code as the site has always shown the manufacturer name, I just can't work out out to pull the new profile information in as well from the new table. Any help appreciated Iain I have my database set to insert the current time stamp when an entry is made into the table, I am then trying to retrieve via the following code: $select_view_idea="SELECT * FROM $tbl_name5 WHERE message_number='$message_number'"; $result_view_idea=mysql_query($select_view_idea); while($row_view_idea=mysql_fetch_assoc($result_view_idea)){ extract($row_view_idea); } date_default_timezone_set('US/Eastern'); $date=date('l, F jS Y h:i:s A T', $date); echo $date; The above is outputting: Wednesday, December 31st 1969 07:33:31 PM EST the database contains: 2011-11-18 00:47:56 Hey all, I've written a php search feature for a mysql database. The search returns a file name like sample.gif, how would I go about displaying the actual image instead? Since the images all have the same location could I have the search return the string into a variable then make the whole thing a link? Thanks. Hi, I have managed to get the code working to store a .jpg file in the database under the longblob type. Now all i have left to do is to retrieve that image and display it. So far i have this: list.php Code: [Select] while($r = mysql_fetch_array($sql)) { //for each record ... echo " <img src= getoutside.php?id='".$r[apartmentId]. " '> "; getoutside.php Code: [Select] <?php header("Content-type: image/jpg"); // act as a jpg file to browser $nId = $_GET['id']; include 'dbase.php'; //connect to database $sqlo = "SELECT outside FROM apartment WHERE apartmentId = $nId"; $oResult = mysql_query($sqlo); $oRow = mysql_fetch_array($oResult); $sJpg = $oRow["outside"]; echo $sJpg; ?> The result from this is a box with a red cross in it. Can anyone find a problem with this code please? Thanks I am working on a kind of CMS for my own website which no one else will be using but me as a way of improving my php skills, and am having problems with retrieving data from the database that holds both text and php code. I have searched the web and found that i should be using eval() for the code to be executed before it is send to the browser but cannot get it to work and can't find my mistake(s). the php code will always be the same, and is supposed to retrieve the id number of a page to use in a link (and works fine when tested by loading the code directly without retrieving it from the database) this is an example of data stored in the database Code: [Select] The <a href="page_builder.php?id=<?php echo $page->id('mines_DwarvenMines') ?>">Dwarven Mines</a> have a great selection of Ores,... Of course when I leave it like this, hovering over the link in my page will show exactly that and lead to nowhere Code: [Select] localhost/page_builder.php?id=<?php echo $page->id('mines_DwarvenMines') ?> Most of these links appear in tips given at the end of the page and are processed as followed Code: [Select] $questtips = $quest->getQuestTips(); $tips = ""; if ($questtips == "none") { $tips = "/"; } else { foreach($questtips as $tip) { $tips .= "<li>"; $tips .= $tip->getTip(); $tips .= "</li>"; } } and finally put on screen by the presentation layer as followed Code: [Select] <h2>Tips & Extra Info</h2> <div class="tipsList"> <ul> <?php echo $tips ?> </ul> </div> I have tried all sorts to get the code to be executed when retrieved from the database before being send to the browser so that this particular link would say "localhost/page_builder.php?id=57" but I cannot get it to work, though I suspect it is fairly easy. I suspect I would have to store the data in a different format in the database? And how exactly do I use the eval() function in my case? Could someone please adjust my code so that it does work? Thanks Hello people, currently i ran into some problem. Currently, i have a database called responses which have the fields of ID, Student_id, question_id, Answer. I had stored my results into the database which appeared to be 1 1 1 Agree 2 1 2 Disagree 3 3 4 Unsatisfied. So any recommendation on how should i do about in generating the result into a graph whereby the graph will show how many students choose "AGree" on that particular question THANKS This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=352302.0 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=318312.0 Basically i have a folder with 100+ images they are NOT all the same extension, what im wanting to do is use PHP to find all the images and put them all in a database. how would i go about doing this? thanks This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=309239.0 Hi,
I am currently working on a inventory page and a page which gets into the details of that particular product. Basically on one page i have a list of products and when i click on one of the products it takes me to another page when has more detail about the product (retrieved from the database). The problem which i am currently stuck on is when i click on say the first product (id=1) it will take me to the second page and show me product 2 (id=2) instead of the first product. here is what i have so far if anyone can assist it would be great.
this is in my first inventory page.--
<div id="viewdetails"><a href="<?php echo BASE_URL .'view/singleproda.php?id='. $info["id"];?>"> View Details</a></div>
this is how my second page looks--.:
include(ROOT_PATH ."controller/mainfunction.php"); The subject could be a bit vague, but my problem is simple. (I think so) So I made a php test site that is quite similar to a forum. Where you see a title or a subject and when you click on it you will see more details about that subject. I made my database and script for inserting data into my mySQL database. I also did my output aswell, so every topic is posted on a webpage "archive" where you can see all the subjects. But now I want to see the full details of that type of subject by clicking on it. I have no idea how to make that happen googled it but didn't find any results...just wondering if it's even possible to do that. I'm a new guy here! |
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