PHP - Retrieving Database Entries
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 $showfriends['Username']; } and I would get nothing. It produces the correct number of <div> so i know it's getting through, but it's having trouble displaying the entries? Similar TutorialsI am quite new so I am sure this is an easy fix for some of the experts around here. I am using the canned script below to add urls to the database as text. The problem is if you update one of the form text boxes it loads all the urls into the database again resulting in a lot of duplicates. My question is, How do I get the form to only post the new changes and not re-post the existing urls? <?php session_start(); if(isset($_SESSION['userSession']) && !empty($_SESSION['userSession'])) { include_once("dbc.php"); if($_POST) { $c = 0; $errMssg = ""; for($i=0;$i<count($_POST['url']);$i++) { if($_POST['url'][$i]=="") { $c++; } } if($c==5) { $errMssg = "Submission error . Please fill at least 1 url."; } else { for($j=0;$j<count($_POST['url']);$j++) { if(!empty($_POST['url'][$j])) { $sql = mysql_query("INSERT INTO images (id ,url ,user_id)VALUES (NULL , '".$_POST['url'][$j]."',".$_SESSION['userId'].")"); } } } } $sqlresult = mysql_query("SELECT * FROM images WHERE user_id =".$_SESSION['userId']); $count = 0; while($data = mysql_fetch_array($sqlresult)) { $image[$count] = $data['url']; $count++; } ?> Is there any other way of getting PHP form data into C# any other way besides calling www.downloadHandler.text I am having issues bringing all the entries into C# and breaking them all up. I can do one row fine but multiple rows isn't working. I keep getting an out range error. This is my PHP echo echo $row['userName']. '|' .$row['level']. '|' .$row['points']. '|' .$row['killRate']. '/'; And this is my C# code string nothing = "Not Placed"; string Data_string = www.downloadHandler.text; string[] DataArray; DataArray = Data_string.Split('/'); int numberOfEntries = DataArray.Length; Debug.Log(numberOfEntries); if (DataArray[0] == null || numberOfEntries == 1) { DataArray[0] = nothing; Debug.Log("Data Array [0] isn't there"); High_Points_1.text = DataArray[0]; } else { High_Points_1.text = DataArray[0]; //Debug.Log(DataArray.Length); } if (DataArray[1] == null || numberOfEntries == 2) { DataArray[1] = nothing; Debug.Log("Data Array [1] isn't there"); High_Points_2.text = DataArray[1]; } else { High_Points_2.text = DataArray[1]; //Debug.Log(DataArray.Length); } if (DataArray[2] == null || numberOfEntries == 3) { DataArray[2] = nothing; Debug.Log("Data Array [2] isn't there"); High_Points_3.text = DataArray[2]; } else { High_Points_3.text = DataArray[2]; } if (DataArray[3] == null || numberOfEntries == 4) { DataArray[3] = nothing; Debug.Log("Data Array [3] isn't there"); High_Points_4.text = DataArray[3]; } else { High_Points_4.text = DataArray[3]; } if (DataArray[4] == null || numberOfEntries == 5) { DataArray[4] = nothing; Debug.Log("Data Array [4] isn't there"); High_Points_5.text = DataArray[4]; } else { High_Points_5.text = DataArray[4]; } I have two entries in the database. But when I debugged the number int he array I get 3 strings. I want to show the top five entries in the database. I have an issue with some code I have. All the code works correctly apart from when submit is clicked not only does it update a the current club but it creates a blank entry in the database! I cant see whats wrong. Here is the code......... //gets $validation = $_GET['new_club']; //Querys $qGetClub = "SELECT * FROM clubs WHERE validationID = '$validation'"; $rGetClub = mysql_query($qGetClub); $Club = mysql_fetch_array($rGetClub); //Query for category by name $qGetCat = "SELECT * FROM club_category WHERE catID = ".$Club['cat'].""; $rGetCat = mysql_query($qGetCat); $CatName = mysql_fetch_array($rGetCat); //query for related sub categorys. $qGetSub = "SELECT * FROM sub_categorys WHERE catID =".$Club['cat'].""; $rSubCat = mysql_query($qGetSub); // query for groups created $Groupq = mysql_query("SELECT * FROM groups WHERE memberID = '".$User['memberID']."'"); //end of querys if(isset($_POST['insert_clubbtn1'])){ //Process data for validation $subcat = trim($_POST['subcat']); $NewSubCat = trim($_POST['NewSubCat']); //Prepare data for db insertion $subcat = mysql_real_escape_string($subcat); //find the new category //insert $result = mysql_query("UPDATE clubs SET `sub_category` = '$subcat' WHERE validationID ='$validation'") or die(mysql_error()); if ($result!=="") { $otherg = trim($_POST['other_groups']); $newg = trim($_POST['new_group']); $newg = mysql_real_escape_string($newg); //if an item other than none from the list is selected then update the club with an ID relating to the group it belongs to if ($otherg !=='None') { $groupsq = mysql_query("UPDATE `clubs` SET groupID ='$otherg' WHERE validationID ='$validation'") or die (mysql_error()); } // If none is selected then $newg must have a value so create a new group in the groups table and then on the next page I will add the group in the club table else { $groupsq = mysql_query("INSERT INTO `groups` (`memberID`, `group`, `clubID`) VALUES ('".$User['memberID']."', '$newg', '".$Club['clubID']."')")or die (mysql_error()); } } if ($NewSubCat !="") { mail("mail","New Sub Category Request","Dear Ring Master, \n\nThe club in the name of $name with a validation code of $validationID would like a new sub category called $new_cat\n\n \nTeam Arena\n\n\n\n"); } $url = "/members/create/create_clubp3.php?new_club=$validation"; header("Location: $url"); } Hello, For starters I'm not sure if what I want to do is possible, but if it is I would like your input. I have a script that will show a number of fields to fill out in a second form based on the number the user puts into the first from. the problem is that only the last one saves into the database and not all of them. Code: [Select] <form auction="index.php" method="post"> System Name: <input type="text" name="systemname"> Number of E-sites: <input type="text" name="events"> Number of Sigs: <input type="text" name="sigs"><br> <input type="reset" name="reset" value="Reset"> <input type="submit" name="start" value="Start"> </form> <form auction="index.php" method="post"> <?php $events = $_POST['events']; $system = $POST['systemname']; if (isset($_POST['start'])) { $num = $_POST['sigs']; $i = 0; While ($i < $num) { echo "Sig ID: <input type=text name=sigid>"; echo "Type: <input type=text name=type>"; echo "Name: <input type=text name=name>"; echo "Notes: <input type=text name=notes>"; echo "<br>"; $i++; } } ?> <input type="submit" name="enter" value="Enter"> </form> <?php $sigid = $_POST['sigid']; $type = $_POST['type']; $name = $_POST['name']; $notes = $_POST['notes']; mysql_connect('xt', 'x', 'x'); mysql_select_db('wormhole'); if (isset($_POST['enter'])) { $query = "INSERT INTO sites VALUES ('$system','$events','$sigid','$type','$name','$notes')"; mysql_query($query); } ?> How do I get it so all the data saves, lets say that $num = 5, I want all 5 to save not just the last one. Hello... I have a .txt database with ~100 records. I only want to show the 15 records that are at the top (referring to reading order of the file). Here is my php code to display the records. http://www.flcbranson.org/mobile.php?content=mobile-freedownloads.php if you want to see the results of the code, below. Code: [Select] <?php $index_file = 'services/series/Index-Date.txt'; $fd = fopen($index_file, 'r'); if ($fd) { while (!feof ($fd)) { $seriestitle = trim(fgets($fd, 1024)); // Series Title if(feof ($fd)) break; $seriessubtitle = trim(fgets($fd, 1024)); // Series subtitle $seriesinfo = trim(fgets($fd, 1024)); // Series Info (Church name) $serieslocation = trim(fgets($fd, 1024)); // City, State $seriesindex = trim(fgets($fd, 1024)); // Series index file $seriesstatus = trim(fgets($fd, 1024)); //Online Only? $divider = trim(fgets($fd, 1024)); // Divider $seriestitle2 = $seriestitle; $seriestitle3 = str_replace("'", "%27", $seriestitle); // Kind of makes the alt= below work "God's Will" ends up being "God%27s Will" but without is "God" if(strstr($seriestitle,"<")) { $seriestitle2 = strip_tags($seriestitle); } $seriestitle2 = urlencode($seriestitle2); if(file_exists("images/ProductCovers/".substr($seriesindex,0,-4).".jpg")) { echo "<img src='images/ProductCovers/".substr($seriesindex,0,-4).".jpg' width='115' height='150' border='0' alt='".substr($seriestitle3,7)."'><br />"; } } fclose ($fd); } ?> I'm assuming that "for" would do the trick, but when I tried I got 10 copies of every record. Hehehe... I'm guessing that it's a quick fix. Thanks... JJ Hi, I want to be able to generate visitor statistics for a blog I'm creating. I'm going to be collecting numerous pieces of data when a post is viewed, including a time stamp of the visit. I need to be able to select timestamps that were within the current day, the previous day, the day before that ect.. So that I can generate the statistics. Show it for the current week (current day and 6 previous days). So it would be the entries where the timestamp was made on the days: 11/1, 10/1, 9/1, 8/1, 7/1, 6/1, 5/1. For example. Not quite sure how I could do this. Thanks. For some reason my for loop only seems to be storing some data in my fields, most of the time the data is not true but simply duplicating one of the entries. So instead of having a string of 1,2,3,4 for instance it would store all of them as 4. The entires are selected using fields in a multiple select box, in theory of someone choses 10 unique entries from the "Systems" category all 10 should be made into rows in the MySQL table. Any idea why this is happening? The form, if ($type == "games") { echo "<tr><td>".$name."</td><td><select name='".$name."_".$i."'[] multiple='multiple'><option value='' selected>--------</option>"; $sqla = mysql_query("SELECT * FROM ".$pre."games ORDER BY `name` ASC") or die(mysql_error()); while($row2a = mysql_fetch_array($sqla)) { $system = mysql_fetch_array(mysql_query("SELECT * FROM ".$pre."systems WHERE id = '".$row2a[system]."'")); echo "<option value='".$row2a[id]."'>".$row2a[name]." - ".$system[name]."</option>"; } echo "</select></td></tr>"; } if ($type == "system") { echo "<tr><td>".$name."</td><td><select name='".$name."_".$i."'[] multiple='multiple'><option value='' selected>--------</option>"; $sqlb = mysql_query("SELECT * FROM ".$pre."systems ORDER BY `name` ASC") or die(mysql_error()); while($row2b = mysql_fetch_array($sqlb)) { echo "<option value='".$row2b[id]."'>".$row2b[name]."</option>"; } echo "</select></td></tr>"; } The PHP code, while($row = mysql_fetch_array($query)) { $name = "$row[name]"; for ($i = 0; $i < count($_POST["systems_$i"]); $i++) { mysql_query("INSERT INTO ".$pre."fielddata VALUES (null, 'systems', '".$_POST["systems_$i"]."', '".$fetch[0]."', 'content')"); } for ($i = 0; $i < count($_POST["games_$i"]); $i++) { mysql_query("INSERT INTO ".$pre."fielddata VALUES (null, 'games', '".$_POST["games_$i"]."', '".$fetch[0]."', 'content')"); } Hello. Many thanks for your help. I am writing a PHP/MySQL dating-site and have hit a programming impass. I have a database full of members and a search form consisting of checkboxes. So to search, a member ticks say...gender: female; age: 21,22,23,24,25,26; height: 5'4",5'5",5'6",5'7"; county: cornwall,devon,somerset How can a run a check on the database selecting all entries that fall into the selected criteria. For example a 23 year old female of 5'5" living in Cornwall and a 26 year old female of 5'4" living in Somerset? The key index of my database is 'id' and the fields a age,height,county The names of the form checkboxes a Gender: male, female; Age: 21,22,23,24 etc; Height: 5_4,5_5,5_6 etc; county: cornwall,devon etc Hello, I am using the following code to display images managed by a MySQL database. Basically another program manages a bunch of images, but this script displays certain ones (ones with INCLUDE = 1 in the database) on my main page. My question is, is there an easy way to limit the number of images it displays, say to 5? I'm not too concerned which images actually display (ascending or descending)... or better yet, random! Most importantly, I only want five to display. Each image will be linked to the full page, which displays all the images. Any ideas? Thanks! Code: [Select] <?php $username="XXXXXXX"; $password="XXXXXXX"; $database="XXXXXXX"; mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query="SELECT * FROM ft_form_12 WHERE col_24='1'"; // $query="SELECT * FROM ft_form_12"; // SELECT * FROM ft_form_12 WHERE col_24='1' $result=mysql_query($query); $num=mysql_numrows($result); mysql_close(); ?> <?php $i=0; while ($i < $num) { $f20=mysql_result($result,$i,"col_23"); //Photo file name $f21=mysql_result($result,$i,"col_24"); //INCLUDE ?> <a href="http://www.domain.com/display_whole_page.shtml"><img src="http://www.domain.com/the_file/pictures/<?php echo $f20; ?>" height="50" border="0"></a> <?php $i++; } ?> I coded the following but some parts are not working (the "curveball" mentioned at the end). How would YOU do this? I have a page (A) displaying a form with a textarea field using a WYSIWYG interface. The user will enter a list of unordered items and upon submit the string will look something like this: This is my list of food items: <ul> <li> Lettuce </li> <li> Tomatoes </li> <li> Eggs </li> <ul> Upon submit, I need to give each list item an unique id and store this id along with the text next to it in a separate table: List_item_id List_item_text List_item_state food_1 Lettuce food_2 Tomatoes food_3 Eggs The user will then land in another page with a form displaying the list. The text is now in the page itself and not in a textarea. Instead of the bullets a dropdown list appears, and the user can select "buy" or "sell" for each list item. This page (B) looks something like this: <form> This is my list of food items: <br><select name="food_1"> <option value="buy">Buy</option> <option value="sell">Sell</option> </select> Lettuce <br><select name="food_2"> <option value="buy">Buy</option> <option value="sell">Sell</option> </select> Tomatoes <br><select name="food_2"> <option value="buy">Buy</option> <option value="sell">Sell</option> </select> Eggs <input type=submit value="submit"> </form> When the user hits "submit" the table with the list items will be updated with the values selected in the dropdown list: List_item_id List_item_text List_item_state food_1 Lettuce Buy food_2 Tomatoes Buy food_3 Eggs Sell The next time the user goes to Page B the list will remember the states. Here's the curveball: At any point the user may click on "EDIT LIST" on Page B so they may add more items (at the begining, middle or end of the string). On edit mode the list items should appear as bullets again inside the WYSIWYG interface. Keep in mind that some of the text in the string may not be a list item (ie, "this is my list of food...") Im trying to create a website where users login, and then when they add a new entry to the database there name is put as the author. This is how my tables are set up. One table is named job and has the columns id, jobtext, jobdate, and authorid. Another table is called author. This table contains the columns id, username, password, and name. Authorid from the job table matches with id from the author table. When a user logins in this code is used to register the name...session_start(); $_SESSION['myusername'] = $_POST['myusername']; $_SESSION['mypassword'] = $_POST['mypassword']; header("location: index.php"); } else { echo "Wrong Username or Password"; } This is the form users use to add a new entry... if (isset($_GET['add'])) { $pagetitle = 'New Job'; $action = 'addform'; $text = ''; $authorid = ''; $id = ''; $button = 'Add job'; include $_SERVER['DOCUMENT_ROOT'] . '/jobs/includes/db.inc.php'; // Build the list of authors $sql = "SELECT id, name FROM author"; $result = mysqli_query($link, $sql); if (!$result) { $error = 'Error fetching list of authors.'; include 'error.html.php'; exit(); } while ($row = mysqli_fetch_array($result)) { $authors[] = array('id' => $row['id'], 'name' => $row['name']); } // Build the list of categories $sql = "SELECT id, name FROM category"; $result = mysqli_query($link, $sql); if (!$result) { $error = 'Error fetching list of categories.'; include 'error.html.php'; exit(); } while ($row = mysqli_fetch_array($result)) { $categories[] = array( 'id' => $row['id'], 'name' => $row['name'], 'selected' => FALSE); } include 'form.html.php'; exit(); } if (isset($_GET['addform'])) { include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php'; $text = mysqli_real_escape_string($link, $_POST['text']); $author = mysqli_real_escape_string($link, $_POST['author']); if ($author == '') { $error = 'You must choose an author for this job. Click ‘back’ and try again.'; include 'error.html.php'; exit(); } $sql = "INSERT INTO job SET jobtext='$text', jobdate=CURDATE(), authorid='$author'"; if (!mysqli_query($link, $sql)) { $error = 'Error adding submitted job.'; include 'error.html.php'; exit(); } $jobid = mysqli_insert_id($link); if (isset($_POST['categories'])) { foreach ($_POST['categories'] as $category) { $categoryid = mysqli_real_escape_string($link, $category); $sql = "INSERT INTO jobcategory SET jobid='$jobid', categoryid='$categoryid'"; if (!mysqli_query($link, $sql)) { $error = 'Error inserting job into selected category.'; include 'error.html.php'; exit(); } } } header('Location: .'); exit(); } Form.html.php = <?php include_once $_SERVER['DOCUMENT_ROOT'] . '/includes/helpers.inc.php'; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <title><?php htmlout($pagetitle); ?></title> <meta http-equiv="content-type" content="text/html; charset=utf-8"/> <style type="text/css"> textarea { display: block; width: 100%; } </style> </head> <body> <?php session_start(); ?> <h1><?php htmlout($pagetitle); ?></h1> <form action="?<?php htmlout($action); ?>" method="post"> <div> <label for="text">Type your job he </label> <textarea id="text" name="text" rows="3" cols="40"><?php htmlout($text); ?></textarea> </div> <div> <label for="author">Author:</label> <select name="author" id="author"> <option value="">Select one</option> <?php foreach ($authors as $author):?> <option value="<?php htmlout($author['id']); ?>"<?php if ($author['id'] == $authorid) echo ' selected="selected"'; ?>><?php htmlout($author['name']); ?></option> <?php endforeach; ?> </select> </div> <fieldset> <legend>Categories:</legend> <?php foreach ($categories as $category): ?> <div><label for="category<?php htmlout($category['id']); ?>"><input type="checkbox" name="categories[]" id="category<?php htmlout($category['id']); ?>" value="<?php htmlout($category['id']); ?>"<?php if ($category['selected']) { echo ' checked="checked"'; } ?>/><?php htmlout($category['name']); ?></label></div> <?php endforeach; ?> </fieldset> <div> <input type="hidden" name="id" value="<?php htmlout($id); ?>"/> <input type="submit" value="<?php htmlout($button); ?>"/> </div> </form> </body> </html> Right now, under authors, it displays all the authors in the database. I want it to just show/submit the authorid of the logged in user. 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. 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 Maybe some of the great coders here can help this noob out. So here is what I have so far: Code: [Select] //set age criteria for deletion $age = 60; //get current date $datenow = date("Y-m-d"); //set the range we want to delete $delete_range = $datenow - $age; //get old user_id from users table $oldusers_users = mysql_query ("SELECT user_id FROM users WHERE lastvisit < $delete_range "); //get user_id from images table that correspond to users table $oldusers_images = mysql_query ("SELECT user_id FROM images WHERE $oldusers_users=user_id.images "); //find folders that correspond to the usernames and delete $oldusers_files = mysql_query ("SELECT username FROM users WHERE lastvisit < $delete_range "); //print out username //$result = mysql_($oldusers_files); $foldername = mysql_result($oldusers_files, 0); $sigspath = "sigs/"; unlink($sigspath . $foldername . ".gif/index.php"); rmdir($sigspath . $foldername . ".gif"); //now delete user_id's from database mysql_query("DELETE * FROM users WHERE user_id=$oldusers_users"); mysql_query("DELETE * FROM images WHERE user_id=$oldusers_images"); unset($oldusers_users, $oldusers_images, $oldusers_files, $foldername, $sigspath ); mysql_close($link); Right now it is only returning the first entry, not the entire list that meets the criteria. It does delete that one file though but does not remover the rows from the database. I am sure the database stuff is jacked up I am really new to that part. What am I doing wrong here or is there a better way to do this perhaps 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 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. 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 |