PHP - Add Field To Query Incorrect
hello
in the attached code, how do i add another field to the query? ie, i have all the code to create the result and would like to add further values such as $dept? what is the correct way to code query? i must stress that i am using php 4.4.7 so json_encode is out. the code is working but just need to find a way to add the $dept to the qeury? many thanks $dept = array(); $box = array(); while ($row = mysql_fetch_array($result)) { $dept[] = $row['department']; $box[] = $row['custref']; } /*$items = rtrim($_POST['items'],","); $sql = "UPDATE `boxes` SET status = 'Out' WHERE Id IN ($items)"; $result = runSQL($sql);*/ $total = count(explode(",",$items)); $result = runSQL($sql); $total = mysql_affected_rows(); /// Line 18/19 commented for demo purposes. The MySQL query is not executed in this case. When line 18 and 19 are uncommented, the MySQL query will be executed. header("Expires: Mon, 26 Jul 1997 05:00:00 GMT" ); header("Last-Modified: " . gmdate( "D, d M Y H:i:s" ) . "GMT" ); header("Cache-Control: no-cache, must-revalidate" ); header("Pragma: no-cache" ); header("Content-type: text/x-json"); $json = ""; $json .= "{\n"; $json .= "box: [\"". implode('","', $box) ."\"]\n"; $json .= "}\n"; echo $json; $sql = "INSERT INTO `act` (`item`) VALUES ('". implode("'),('", $box) . "')"; $result = runSQL($sql); Similar TutorialsIt's been a while since I've used my queries, can anyone tell me what's wrong with this mysql query? Code: [Select] $query = mysql_query("SELECT * FROM `chicken_names` WHERE `gender` = 'Hen' AND `name` LIKE '%$s_terms%' ORDER BY `likes` DESC") or die(mysql_error()); I have the following query: $getVideos = mysql_query("SELECT catergory, COUNT(catergory) as 'catCount' FROM videos GROUP BY catergory"); Using this in PHPMyAdmin returns the correct results of: catergory | catCount 0 | 7 1 | 1 10 | 2 How would get those results into a PHP array or what not...I have done this before but along time ago and cannot for the life of me remember how to do it. Hopefully someone can point me in the right direction? Regards, PaulRyan. this is the query: $query = "INSERT INTO kamerleden VALUES ('', '$twitter_id', '$name', '$nickname', '$fraction', '$residence', '$age', '$gender', '', '', '', '')"; it used to work but i switched to a different hosting. The first '' is the id which gives this error: query failed: Incorrect integer value: '' for column 'id' at row 1 in the database id is a int, with a length of 2 and with auto increment. How can i fix it? How should it look if I want to make a query where a field value should be lower than another field? Code: [Select] ....WHERE total_points<max_points"); Hi there. I'm totally new (about a week!) with php and mysql and am encountering a problem that perhaps someone can help me with?
I've looked through to see if a similar problem has appeared or been solved, but without success, so apologies if I am repeating something.
In php I am trying to update 7 fields from a form from which a user has edited/modified any of the fields in a chosen record (except id).
Here is the code:
$id=$_GET['id']; ok, this is clearly 1st grade code for some, but i'm not there yet - I'm querying posts in WordPress, and the post_content will always have an image in the beginning of the post followed by the content. i don't want to get the image, just the content that's after the image, which is wrapped in anchor tags, of course. Code: [Select] <a href="http://path/to/image.jpg"><img src="http://path/to/image.jpg" /></a> <p>Post content yadda, yadda, hoowie</p> obviously a character count won't work, so i need to get anything that follows the first "</a>", say...? is this the best way, or is there an easier way? thanks for anyone's help. GN Hi, I am new to the boards and php and mysql. I have created a database and can add entries via a form. I can query the database with another form and get the results to display in a table. All good so far as that is what I was hoping to achieve but one of the fields I want to display as a hyperlink but I am having problems with the syntax as I keep getting errors when I try to wrap the variables in <a href > tag. Code: [Select] <tr> <td><?php echo $fquery['prefix']; echo $fquery['website']; ?><?php echo $fquery['website']; ?></td> </tr> now that displays e.g. http://www.example.comwww.example.com. so I feel I am nearly there I just need advice as to how to construct a hyperlink from the fields queried. Hi, I'm trying to set up a page which first queries for mySQL record results matching a country that the user selects. This works fine, but in the event of there being no records for that country, I want it to look at another field, "Region" and pick the records matching that Region instead. For example, a user searches for "Australia" but there are no matching records. So, I want it to pick all the records for the region of Australasia, and display records for Australia, New Zealand, Papua New Guinea and so on. I had created the following: $query = "SELECT * FROM specialists WHERE Country LIKE '$country' ORDER BY SpecialistName"; // specify the table and field names for the SQL query //$query .= " limit $s,$limit"; $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); if ( $numrows == 0 ) { echo '<p>We don\'t have any results for specialists in countries near to yours at the moment. Please try <a href="specialists.php" style="text-decoration:underline;">searching a different country</a></p>'; } // get results $result = mysql_query($query) or die("Couldn't execute query"); // display the results returned while ($row= mysql_fetch_array($result)) { $region = $row["Region"]; $count++ ; } // next determine if s has been passed to script, if not use 0 if (empty($s)) { $s=0; } echo '<table width="600" class="cardisplay"><tr>'; $count = 1 + $s ; echo $region; // Build SQL Query $query2 = "SELECT * FROM specialists WHERE Region LIKE '$region' ORDER BY SpecialistName"; // specify the table and field names for the SQL query //$query .= " limit $s,$limit"; $numresults=mysql_query($query2); $numrows=mysql_num_rows($numresults); // get results $result = mysql_query($query2) or die("Couldn't execute query"); // display the results returned while ($row= mysql_fetch_array($result)) { $title1 = $row["ID"]; $specialistname = $row["SpecialistName"]; $address1 = $row["Address1"]; $address2 = $row["Address2"]; $address3 = $row["Address3"]; $address4 = $row["Address4"]; $address5 = $row["Address5"]; $postcode = $row["Postcode"]; $country = $row["Country"]; $region = $row["Region"]; $website = $row["Website"]; $email = $row["Email"]; $telephone = $row["Telephone"]; $businesstype = $row["BusinessType"]; //followed by echoing out the various data etc. etc. But the problem is that $region is always blank / empty in the cases where $query is empty, so I can't pull the value out and therefore $query2 is also empty... Any ideas? I have a form that allows my client to update some products. Now the products are simple just basic info and 1 picture. I have set this up so they can edit the products and change the information, having done this many times in the past, but now hit a puzzling block that I am baffled. The client when editing is presented with the form with the information pulled from the database and the form fields loaded with that data ready to edit. The image can either be left alone or they can choose to upload a new image. They are shown the image they currently have stored in the database. The problem I have is EVEN if they decide not to upload an image and change other information, when the submit the form it must be sending a blank value for the image somewhere as it is updating the database and removing the image reference as if it has been removed. I have an if/else statement based on the form to perform 2 different queries for the update in mysql. Here is the code for the form update, as you can see the image should not update?? Please help?? if ($_SERVER['REQUEST_METHOD'] =='POST') { //This stops SQL Injection in POST vars foreach ($_POST as $key => $value) { $_POST[$key] = mysql_real_escape_string($value); } // **************************** THIS IS FOR NO NEW IMAGE ******************************** if ($_SERVER['REQUEST_METHOD'] =='POST' && empty($_FILES['product_image']['name'])) { # setup SQL statement for no new image $SQL = " UPDATE products SET product_title = '{$_POST['product_title']}', product_description = '{$_POST['product_description']}', standard_price = '{$_POST['standard_price']}', deluxe_price = '{$_POST['deluxe_price']}' WHERE product_id = '{$_REQUEST['product_id']}' "; } // **************************** THIS IS FOR A NEW IMAGE ******************************** else { // Check the image type is a jpeg or gif for the image. if (($_FILES['product_image']['type'] != "image/gif") && ($_FILES['product_image']['type'] != "image/jpeg") && ($_FILES['product_image']['type'] != "image/pjpeg")) { echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\">You have chosen not to upload a <b>Product Image</b>.<BR></SPAN>" ; } elseif ($_FILES['product_image']['size'] > 100000) { echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\">The file size is bigger than 300kb.<BR></SPAN>" ; } else { move_uploaded_file($_FILES['product_image']['tmp_name'], "/httpdocs/product_images/".$_FILES['product_image']['name']) ; echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\"><B>Your front image has successfully uploaded.</B><BR></SPAN>" ; } } # setup SQL statement for update $SQL = " UPDATE products SET product_title = '{$_POST['product_title']}', product_description = '{$_POST['product_description']}', standard_price = '{$_POST['standard_price']}', deluxe_price = '{$_POST['deluxe_price']}', product_image = '{$_FILES['product_image']['name']}' WHERE product_id = '{$_REQUEST['product_id']}' "; } #execute SQL statement $result = mysql_db_query( *****,"$SQL",$connection ); # check for error if (!$result) { echo("ERROR: " . mysql_error() . "\n$SQL\n"); } I created a drop-down menu using a MySQL statement in php for a form. My drop down menu works fine, but I want to assign a default value to it (normal text) the value will never change, thus I do not need to extract the default value from the table, I already know the value. Here is the basic snippet from the script: <?php include("../includes/xxx.php"); $cxn = mysqli_connect($host,$user,$password,$dbname); $query = "SELECT DISTINCT `plant_id` FROM `plant` ORDER BY `plant_id`"; $result = mysqli_query($cxn,$query); while($row = mysqli_fetch_assoc($result)) { extract($row); echo "<option value='$plant_id'>$plant_id</option>\n"; } ?> I have a MySQL table with a list of albums and there is a field called "views" with the number of views each album has received. I'm looking to generate an array of all the albums in the table and sort the array by the number of views (descending). I have a list of functions defined in a ContentController.php file. I created a new function called build_albumlist, which I've pasted below. The function "get_ip_log" already exists and works, and I used it as a template to create the "build_albumlist" function: public function build_albumlist(){ return $this->select_raw("SELECT * FROM albums WHERE deleted = '0' ORDER BY views DESC",array(),'all'); } public function get_ip_log(){ return $this->select_raw("SELECT * FROM sessions ORDER BY ID DESC",array(),'all'); } When I use the function, I get this warning: Warning: mysql_real_escape_string() expects parameter 1 to be string, array given inC:\xampp\htdocs\Controllers\DBController.php on line 10 [font=cabin, 'trebuchet ms', helvetica, arial, sans-serif]The "select_raw" function that I used in "build_albumlist" is defined in the DBController.php file, and is defined as below:[/font] private function clean_array($params){ $final=array(); foreach($params as $param){ $final[]=mysql_real_escape_string($param); } return $final; } public function select_raw($query,$params,$type=''){ $query=str_replace("?","'%s'",$query); $final_query= call_user_func_array('sprintf', array_merge((array)$query, $this->clean_array($params))); if($type==''){ $result=mysql_query($final_query) or die(mysql_error()); return mysql_fetch_assoc($result); } elseif($type=='all'){ $result=mysql_query($final_query) or die(mysql_error()); $final=array(); while($row=mysql_fetch_assoc($result)){ $final[]=$row; } return $final; } Does anyone know why the "build_albumlist" function is generating this warning, while the "get_ip_log" is not? Any help would be great, as I am obviously pretty new to this. Hi all, I have the following MySQL insert query: Code: [Select] $insert= mysql_query ("INSERT INTO tablename (column1,`".$EXPfields."`) VALUES ('$something','".$EXPvalues."')"); where $EXPfields is an array of table-field-names and $EXPvalues is an array of table-field-values. Now I want to write an equivalent query, but using UPDATE instead of INSERT INTO, but I don't want to write out all the field names/values separately, but again want to use $EXPfields and $EXPvalues. So something like this: Code: [Select] $update = mysql_query ("UPDATE tablename SET (column1,`".$EXPfields."`) = ('$something','".$EXPvalues."') WHERE .... "); Is this possible? If so, what is the proper syntax? Thanks! Ever since I switched over to pagination, the ranks for the sites have been off. I use to have an $i++ increment that kept the ranks, but now with pagination, it just ranks those on the current page. Is there a way to bypass this? public function grabSites() { //amount of sites $amount = mysql_num_rows(mysql_query("SELECT * FROM sites")); if($amount > 0) { //amount per page $p_page = 13; //track page $start = $_GET['page']; //max_pages $max_pages = $amount / $p_page; //set default if(!$start) $start = 0; //new query $query = mysql_query("SELECT id,title,description,votes,date,outl,site_url FROM sites ORDER BY votes DESC LIMIT $start, $p_page"); while($fetch = mysql_fetch_assoc($query)) { ++$i; $i = $i + $p_page; echo " <div id='post-1' class='post'> <h2 class='title'><font color='white'>#". $i." ".$fetch['title'] ."</font></h2> <div class='entry'> <b>". $fetch['title'] ." currently has ". $fetch['votes'] ." votes. <a href='vote.php?id=". $fetch['id'] ."'>Vote now</a>.</b> <br/><br/> <p>". nl2br(stripslashes($fetch['description'])) ."</p> </div> <p class='meta'><a href='". $fetch['site_url'] ."'>Visit</a></p> </div> "; } //set back and forth variables $previous = $start - $p_page; $next = $start + $p_page; ?> <hr> <center> <?php if(!($start<=0)) echo "<a href='index.php?page=". $previous ."'><<</a>"; $i = 1; for($x = 0; $x < $amount; $x = $x + $p_page) { echo "<a href='index.php?page=". $x ."'> ". $i ." </a>"; $i++; } if(!($start>=$amount-$p_page)) echo "<a href='index.php?page=". $next ."'>>></a></center>"; ?> </center> <?php } else { echo " <div id='post-1' class='post'> <h2 class='title'><a href='#'>Oh NOEZ!</a></h2> <h3 class='date'>". date("M-d-Y") ."</h3> <div class='entry'> <p>There are currently no sites to display!</p> </div> <p class='meta'>View</p> </div> "; } } Hi all My site is build using php includes. the index.php file contains code like <?php if(@$_GET['page'] == "web-design"){ include("includes/web-design.php"); } else if(@$_GET['page'] == "hosting"){ include("includes/hosting.php"); } else { include("includes/home.php"); } ?> I have created a sub domain blog.maplewebdesign.co.uk and a sub dircetory named blog. The link to this part of my site doesn't use includes, I want it to link directly to that sub directories index.php page. This works fine, but then when I click on a different link anywhere on the site my url is as follows www.maplewebdesign.co.uk/blog/index.php?page=home For some reason it is keeping the blog/ in the URL If you want to see go to http://www.maplewebdesign.co.uk/blog/ then click on a differnt link and check out the URL What am I doing wrong? Thanks Adi The password ARE correct. My code keeps saying that the password is INCORRECT. The password is MD5'ed once a user registers, and when they type in a password at the login (as shown), the password is also MD5'ed. Why is it that it's output is incorrect password? <?php session_start(); include("includes/mysql.php"); include("includes/config.php"); ?> <title><?php echo $title; ?></title> <?php if(!$_SESSION['user']) { $username = $_POST['username']; $password = $_POST['password']; $username = mysql_real_escape_string($username); $password = mysql_real_escape_string($password); if(!$password || !$username) { echo ' <h1>Login</h1> <center><p><form action="login.php" method="POST"> <table border="0"> <tr><th>Username:</th> <td><input type="text" name="username" maxlength="20"><br/></td></tr> <tr><th>Password:</th> <td><input type="password" name="password" maxlength="30"><br/></td></tr> <tr><th></th><td><input type="submit" value="Login"></td></tr> </table></form></p></center> </div> '; } else { $query = mysql_query("SELECT COUNT(username),password,username FROM users WHERE username='$username'"); $check = mysql_fetch_assoc($query); $db_username = $check['username']; $password = md5($password); if($check['COUNT(username)'] < 1) { echo ' <p>No account exists with this username. Please go back.</p> '; } elseif($check['password']==$password && $db_username==$username) { echo ' <h1>Login Successful</h1> <p>You have successfully logged in! Return home.</p> '; $_SESSION['user']=$username; } else { echo ' <p>The password you have enetered in is incorrect. Please go back.</p> '; } } } else { echo ' <p>Your already logged in!</p> '; } ?> I have a small bit of jquery code that is not working:
$(function() { $(‘’[name=\'pet_type\']”).change(function() { $(‘’[name=\'breed\']”).removeAttr(‘disabled’); $(‘’[name=\'breed\']”).children(‘data-pet-type[name!=\'’ + $(this).val() + ‘\']’).hide(); $(‘’[name=\'breed\']”).children(‘data-pet-type[name=\'’ + $(this).val() + ‘\']’).show(); }); });My javascript is not good at all, so cannot spot errors? Any help would be much appreciated! Thanks So I am trying to make an incorrect password message but the message never appears.
My code:
<? require "./includes/config.php"; $LS->init(); if(isset($_POST['act_login'])){ $user=$_POST['login']; $pass=$_POST['pass']; if($user=="" || $pass==""){ $msg=array("Error", "Username / Password Wrong !"); } else { if(!$LS->login($user, $pass)){ $msg=array("Error", "Username / Password Wrong !"); } } } ?> I have a register page that MD5 Hash's the users password and a login which also does this. However, no matter what I try it always says incorrect password. Even when I remove the MD5. Register Code: Code: [Select] <?php error_reporting (E_ALL ^ E_NOTICE); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Member System - Register</title> </head> <body> <?php if ( $_POST['registerbtn'] ){ $getuser = $_POST['user']; $getemail = $_POST['email']; $getpass = $_POST['pass']; $getretypepass = $_POST['retypepass']; if ($getuser){ if ($getemail){ if ($getpass){ if ($getretypepass){ if ( $getpass === $getretypepass ){ if ( (strlen($getemail) >= 7) && (strstr($getemail, "@")) && (strstr($getemail, ".")) ){ require("./connect.php"); $query = mysql_query("SELECT * FROM users WHERE username='$getuser'"); $numrows = mysql_num_rows($query); if ($numrows == 0){ $query = mysql_query("SELECT * FROM users WHERE email='$getemail'"); $numrows = mysql_num_rows($query); if ($numrows == 0){ $password = md5(md5("kjfiufj".$password."Fj56fj")); $date = date("F d, Y"); $code = md5(rand()); mysql_query("INSERT INTO users VALUES ( '', '$getuser', '$password', '$getemail', '0', '$code', '$date' )"); $query = mysql_query("SELECT * FROM users WHERE username='$getuser'"); $numrows = mysql_num_rows($query); if ($numrows == 1){ $site = "http://c3221281.web44.net/"; $webmaster = "Simon <admin@simon.com>"; $headers = "From: $webmaster"; $subject = "Activate Your Account"; $message = "Thanks for registering. Click the link below to activate your account.\n"; $message .= "$site/activate.php?user=$getuser&code=$code\n"; $message .= "You must activate your account to login."; if ( mail($getemail, $subject, $message, $headers) ){ $errormsg = "You have been registered. You must activate your account from the activation link sent to <b>$getemail</b>."; $getuser = ""; $getemail = ""; } else $errormsg = "An error has occueed. Your activation email was not sent."; } else $errormsg = "An error has occured. Your account was not created."; } else $errormsg = "There is already a user with that email."; } else $errormsg = "There is already a user with that username."; mysql_close(); } else $errormsg = "You must enter a valid email address to register."; } else $errormsg = "Your passwords did not match."; } else $errormsg = "You must retype your password to register."; } else $errormsg = "You must enter your password to register."; } else $errrosmg = "You must enter your email to register."; } else $errormsg = "You must enter your username to register."; } $form = "<form action='./register.php' method='post'> <table> <tr> <td></td> <td><font color='red'>$errormsg</font></td> </tr> <tr> <td>Username:</td> <td><input type='text' name='user' value='$getuser' /></td> </tr> <tr> <td>Email:</td> <td><input type='text' name='email' value='$getemail' /></td> </tr> <tr> <td>Password:</td> <td><input type='password' name='pass' value='' /></td> </tr> <tr> <td>Retype:</td> <td><input type='password' name='retypepass' value='' /></td> </tr> <tr> <td></td> <td><input type='submit' name='registerbtn' value='Register' /></td> </tr> </table> </form>"; echo $form; ?> </body> </html> Login Code: Code: [Select] <?php error_reporting (E_ALL ^ E_NOTICE); session_start(); $userid = $_SESSION['userid']; $username = $_SESSION['username']; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Member System - Login</title> </head> <body> <?php if ($username && $userid){ echo "You are already logged in as <b>$username</b>. <a href='./member.php'>Click here</a> to go to the member page."; } else{ $form = "<form action='./login.php' method='post'> <table> <tr> <td>Username:</td> <td><input type='text' name='user' /></td> </tr> <tr> <td>Password:</td> <td><input type='password' name='password' /></td> </tr> <tr> <td></td> <td><input type='submit' name='loginbtn' value='Login' /></td> </tr> <tr> <td><a href='./register.php'>Register</a></td> <td><a href='./forgotpass.php'>Forgot your password?</a></td> </tr> </table> </form>"; if ($_POST['loginbtn']){ $user = $_POST['user']; $password = $_POST['password']; if ($user){ if ($password){ require("connect.php"); $password = md5(md5("kjfiufj".$password."Fj56fj")); // make sure login info correct $query = mysql_query("SELECT * FROM users WHERE username='$user'"); $numrows = mysql_num_rows($query); if ($numrows == 1){ $row = mysql_fetch_assoc($query); $dbid = $row['id']; $dbuser = $row['username']; $dbpass = $row['password']; $dbactive = $row['active']; if ($password == $dbpass){ if ($dbactive == 1){ // set session info $_SESSION['userid'] = $dbid; $_SESSION['username'] = $dbuser; echo "You have been logged in as <b>$dbuser</b>. <a href='./member.php'>Click here</a> to go to the member page."; } else echo "You must activate your account to login. $form"; } else echo "You did not enter the correct password. $form"; } else echo "The username you entered was not found. $form"; mysql_close(); } else echo "You must enter your password. $form"; } else echo "You must enter your username. $form"; } else echo $form; } ?> </body> </html> Many thanks for your time and help, Currently I have 5 entries in my table. Entring this query displays all five in the correct order. Code: [Select] SELECT * FROM products ORDER BY count DESC LIMIT 16 However, using this code, the first result is omitted and the last four are displayed. $query="SELECT * FROM products ORDER BY count DESC LIMIT 16"; $result = mysql_query($query); $row = mysql_fetch_array($result); echo '<div>'; while ($row = mysql_fetch_array($result)) { echo '<div>'.$row[product].'</div> <div>'.$row[count].'</div>'; } echo '</div>'; Can someone help me find my error? Thanks i am writing this script for use as an include in joomla and the directphp plugin so in my joomla article i have: Code: [Select] <?php include 'test/conducttest.php'; conducttest(); ?> Ok now my issue! the form in this function is passing only the values for the last item in the loop rather than for the selected radio item when submitted. so when say for example the radio selected is: ID TEST TYPE UNIQUE TEST ID AVAILABLE 5 Adolescent Clinical 50021629 1 and the last item in the list is: ID TEST TYPE UNIQUE TEST ID AVAILABLE 4 Adult Clinical 12341629 1 When the form is submitted even if the 1st item is selected, it returns the value for the second one. I am sorta new to php so any suggestions will be appreciated. Here is my full code: Code: [Select] <?php function conducttest() { //GET JOOMLA USER ID & USERNAME FOR LOGGED USER $user =& JFactory::getUser(); $userID = $user->id; //GET JOOMLA DATABASE OBJECTS $db =& JFactory::getDBO(); //NEW TESTS QUERY (what purchased test are available for use with 'item_available' == '1' {unused test} and matching current 'user_id') $new_query = " SELECT * FROM crt_transactionHistory WHERE user_id = ".$userID." AND item_available = '1' ORDER BY item_name, id "; $db->setQuery($new_query); $new = $db->loadAssocList(); //OPEN TESTS QUERY (what purchased test are available for use with 'item_available' == '2' {resume test} and matching current 'user_id') $resume_query = " SELECT * FROM crt_transactionHistory WHERE user_id = ".$userID." AND item_available = '2' ORDER BY item_name, id "; $db->setQuery($resume_query); $resume = $db->loadAssocList(); //DISPLAY use_test FORM if(!isset($_POST['test'])) { //SELECT FORM: WHICH TEST WOULD YOU LIKE TO USE? echo ' <fieldset> <table> <form method="post" action="'.$PHP_SELF.'">'; if (empty($new)) { echo ' <th colspan="3"align="left">NO NEW TESTS AVAILABLE</th>'; } else { echo ' <th colspan="3"align="left">NEW TESTS AVAILABLE</th> <tr> <td width="75px">SELECT</td> <td width="50px">ID</td> <td width="150px">TEST TYPE</td> <td width="150px">UNIQUE TEST ID</td> <tr>'; foreach ($new as $result1) { echo ' <tr> <td><input type="radio" value="' .$result1['id']. '" name="test_id"></td> <td>'.$result1['id'].'</td> <td><input type="hidden" value="'.$result1['item_name'].'" name="item_name">'.$result1['item_name'].'</td> <td><input type="hidden" value="'.$result1['item_number'].'" name="item_number">'.$result1['item_number'].'</td> <input type="hidden" value="'.$result1['item_available'].'" name="item_available"> <input type="hidden" value="'.$userID.'" name="userID"> <tr>'; } } echo ' </table> <hr /> <table>'; if (empty($resume)) { echo ' <th colspan="3"align="left">NO TESTS TO RESUME</th>'; } else { echo ' <th colspan="3"align="left">RESUME TEST</th> <tr> <td width="75px">SELECT</td> <td width="50px">ID</td> <td width="150px">TEST TYPE</td> <td width="150px">UNIQUE TEST ID</td> <tr>'; foreach ($resume as $result2) { echo ' <tr> <td><input type="radio" value="' .$result2['id']. '" name="test_id"></td> <td>'.$result2['id'].'</td> <td><input type="hidden" value="'.$result2['item_name'].'" name="item_name">'.$result2['item_name'].'</td> <td><input type="hidden" value="'.$result2['item_number'].'" name="item_number">'.$result2['item_number'].'</td> <input type="hidden" value="'.$result2['item_available'].'" name="item_available"> <input type="hidden" value="'.$userID.'" name="userID"> <tr>'; } } echo ' </table> </fieldset> <input type="submit" name="test" value="Conduct Test" /> </form> '; } //NOW A TEST HAS BEEN SELECTED FOR USE if(isset($_POST['test'])) { $test_id = $_POST['test_id']; $item_name = $_POST['item_name']; $item_number = $_POST['item_number']; $item_available = $_POST['item_available']; $userID = $_POST['userID']; echo $test_id.'<br />'; echo $item_name.'<br />'; echo $item_number.'<br />'; echo $item_available.'<br />'; echo $userID.'<br />'; //IF THIS IS A NEW TEST... if ($item_available == "1") { echo 'new test'; } //IF WE ARE RESUMING A TEST... elseif ($item_available == "2") { echo 'resume test'; } } } ?> |