mallen

No edit button so I have to post again. I meant for Model 1 to show twice.

I am trying to get the item to list multiple times like this. <div class="productHolder"> <div class="mainThumbnail"> <img src="http://www......." alt="MainThumbnail"> </div> <h3>Product Name Appear here</h3> <h5>Model : Model#1 here <small>(Company 1)</small></h5> <span class="viewBtn"> <a href="?page_id=169&company=95&singleProduct=1628"><img src="http://www......./products/_images/x.png" alt="Click to View" height="18" width="137"></a> </span> </div> <div class="productHolder"> <div class="mainThumbnail"> <img src="http://www......." alt="MainThumbnail"> </div> <h3>Product Name Appear here</h3> <h5>Model : Model#2 here <small>(Company 2)</small></h5> <span class="viewBtn"> <a href="?page_id=169&company=100&singleProduct=1629"><img src="http://www......./products/_images/x.png" alt="Click to View" height="18" width="137"></a> </span> </div>

Ok sorry this might clear up my issue. If two companies appear in the query with the item, The item name and company will be a one link and the second company will be second link. Its combining the item name and company in the hyperlink.

Thank you so much. I did realize the original ifelse statement was where my problem was. I am following what someone else set up and do agree its a mess. Now here is what I got so far using your code. It will list the item like this: Image name and model. Then it will list (company 1) (Company 2) And the second company will always have a hyperlink link on it. The good news is its reading all the records now. I would like to either get the hyperlink on the second company name or better yet list the item again with name, image, and another hyperlink.

Thanks for the reply. Yes I know its not ideal. The reason the 3rd company is separate is because the data is entered in a separate form. The product numbers, description and everything has to be entered separate becuase of metric measurements and descriptions. So I need for all three listing to show if the item is contained in all three companies.

I have a search I am doing n a large list of products. When products are entered they are assigned to three different "divisions of the company such as company1, cpmpany2 and company3. When I do a search it displays the product description and includes what company is associated with. But If the same item is listed in two companies it will only display 1 company. I have tested with either company name and have decided this is the issue. Here is the code I have for the search. I don't see anything keeping duplicated from showing. $searchRequest = strtolower($_REQUEST['searchQuery']); $formQuery = "SELECT cat.cat_name, cat.cat_id, prod.prod_id, prod.prod_name,prod.prod_mainImage, prod.prod_model, prod_company1, prod_company2, prod_company3 FROM products AS prod LEFT JOIN category_assoc AS assoc ON assoc.prod_id = prod.prod_id LEFT JOIN categories AS cat ON cat.cat_id = assoc.cat_id WHERE LOWER(prod.prod_name) LIKE '%". $searchRequest ."%' OR LOWER(prod.prod_model) LIKE '%". $searchRequest ."%' OR LOWER(cat.cat_name) LIKE '%". $searchRequest ."%' OR LOWER(prod.prod_description) LIKE '%". $searchRequest. "%' ORDER BY prod.prod_name ASC"; $results = $wpdb->get_results($formQuery, ARRAY_A); get_header(); ?> <div id="mainContent"> <div id='interiorLeft'> <?php if (have_posts()) : while (have_posts()) : the_post(); ?> <h2><?php the_title();?> For <?php echo $searchRequest; ?></h2> <?php the_content(); ?> <?php endwhile; endif; ?> <hr/> <?php if(count($results) > 0) : foreach($results as $res) { $prImage = $res['prod_mainImage']; if(empty($res['prod_mainImage'])) $prImage = "noImage.png"; if($res['prod_company3'] == 1) { $theCompany = "100"; $theName = "Company3"; } elseif($res['prod_company2'] == 1) { $theCompany = "95"; $theName = "Company 2"; } else { $theCompany = "15"; $theName = "Company 1"; } ?> <div class='productHolder'> <div class='mainThumbnail'> <img src='<?php echo PRODUCT_IMAGE . "/thumbnails/". $prImage;?>' alt='MainThumbnail' /> </div> <h3><?php echo stripslashes($res['prod_name']);?></h3> <h5>Model : <?php echo $res['prod_model']; ?> <small>(<?php echo $theName;?>)</small></h5> <span class='viewBtn'> <a href='?page_id=169&company=<?php echo $theCompany;?>&singleProduct=<?php echo $res['prod_id'];?>'>

Perfect, thank you. Now that I know what they are called I can read up on them.

I came across these pieces of code and have been trying to understand this operator -> Or if it is a operator. I have searched and couldn't find explanations. In trying to learn I need to know what is is used for. Is it used only with classes? $m->myMethod(); return $this->x;

Well yeah I could but I know for example there is a plug in for WordPress that allows you to do make it look like this http://www.free-php.net/830/wordpress-custom-field-image-to-featured-image/ but I am not using WordPress. I know this board uses Code tags. This is what they are using http://alexgorbatchev.com/SyntaxHighlighter/manual/brushes/php.html

How can I post PHP code on my website and style like this? http://www.9lessons.info/2010/02/php-login-script-with-encryption.html http://www.free-php.net/830/wordpress-custom-field-image-to-featured-image/

Thanks for your replies. This is what I came up with before I saw the replies. Is there any benefit to using an array over the way I did it? (more text fields above this) $county = $_POST['county']; $to = $county; switch ($to) { case ($to =='County1' || $to == 'County2' || $to == 'County3'): $to = "[email protected]"; break; case ($to == 'County4' || $to == 'County5' || $to == 'County6'): $to = "[email protected]"; break; default: $to = "[email protected]"; }

I have a form that emails the content to a recipient. Now I want to direct the email to 6 different recipients based on the county they select. I got it to work with just two counties (cases). But my state has over 60 counties and I will be dividing by about 10 counties each. Is there a way to list 10 counties for each case? This is what I have. switch ($to) { case 'County1': $to = "[email protected]"; break; case 'County2': $to = "Bob@mysite"; break; default: $to = "[email protected]"; }

Ok I got it working, sort of... getting this error .....mysqli_real_escape_string() expects exactly 2 parameters, So I just removed the mysqli_real_escape_string() and now it is the following and working: $username = trim($_POST['user']); $password = trim($_POST['password']); Why does it throw that error?

Yes I know but I use the same connection for other pages i work on. I moved the $con = dbConnect(); line. Still not working. <?php $con = dbConnect(); $username = mysql_real_escape_string(trim($_POST['user'])); $password = mysql_real_escape_string(trim($_POST['password'])); $sql = "SELECT * FROM users WHERE username = '$username' and password = '$password'"; $result = $con->query($sql) or die(mysql_error()); echo $row['username']; $row = $result->fetch_assoc(); if(!empty($_POST['submit'])){ if($username == "") { print "Please enter your username"; }elseif($password == "") { print "Please enter your password"; }elseif($username == $row['username'] && $password == $row['password'] ){ // Check if username and password were submitted if (!isset($_SESSION['user'])){ session_regenerate_id(); //assign user's name to session } $_SESSION['user'] = $_POST['user']; } } ?> <?php if(($_SESSION['user'] == "" )|| (!isset($_SESSION['user']))){ ?> <form action="login9.php" method="post"> <table width="200" border="0"> <tr> <td width="71">User:</td> <td width="113"><label for="user"></label> <input type="text" name="user" id="user" /></td> </tr> <tr> <td>Password:</td> <td><label for="password"></label> <input type="password" name="password" id="password" /></td> </tr> <tr> <td> </td> <td><input type="submit" name="submit" value="Log in" /></td> </tr> </table> </form> <?php }else{ print "Hello, ".$_SESSION['user']; }?>

<?php function dbConnect() { $con = mysqli_connect('localhost', 'root','xxxx', 'development') or die ('Cannot connect to MySQL server'); return $con; } ?>

I have this simple form checking a user and password from database. At first I was just checking a hard coded value. Now I want to check it against the database. Is there anything in my code that is keeping it form working? You can point to me and I can try to figure it out? Warning: mysql_real_escape_string(): Access denied for user ''@'localhost' (using password: NO) in ...... Warning: mysql_real_escape_string(): A link to the server could not be established in.... Warning: mysql_real_escape_string(): Access denied for user ''@'localhost' (using password: NO) in .... Warning: mysql_real_escape_string(): A link to the server could not be established in .... <?php session_start(); error_reporting(E_ALL ^ E_NOTICE); include('../includes/dev.connection.inc.php'); ?> <!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>Login Test Page</title> </head> <body> <?php $username = mysql_real_escape_string(trim($_POST['user'])); $password = mysql_real_escape_string(trim($_POST['password'])); $con = dbConnect(); $sql = "SELECT * FROM users WHERE username = '$username' and password = '$password'"; $result = $con->query($sql) or die(mysql_error()); echo $row['username']; $row = $result->fetch_assoc(); if(!empty($_POST['submit'])){ if($username == "") { print "Please enter your username"; }elseif($password == "") { print "Please enter your password"; }elseif($username == $row['username'] && $password == $row['password'] ){ // I Think this line is giving me the error // Check if username and password were submitted if (!isset($_SESSION['user'])){ session_regenerate_id(); //assign user's name to session } $_SESSION['user'] = $_POST['user']; } } ?> <?php if(($_SESSION['user'] == "" )|| (!isset($_SESSION['user']))){ ?> <form action="login9.php" method="post"> <table width="200" border="0"> <tr> <td width="71">User:</td> <td width="113"><label for="user"></label> <input type="text" name="user" id="user" /></td> </tr> <tr> <td>Password:</td> <td><label for="password"></label> <input type="password" name="password" id="password" /></td> </tr> <tr> <td> </td> <td><input type="submit" name="submit" value="Log in" /></td> </tr> </table> </form> <?php }else{ print "Hello, ".$_SESSION['user']; }?> </body> </html>

Ahh perfect. Thanks so much phpSensei. Now I will work on a mysql connection instead of a hardcoded user and password. Also will work on SQL injections. I know it would have been easier to just send the user to a "welcome" page but it was a learning experience using errors, conditions, and sessions.

When I enter the form, it should display "You are logged in me". But as it is now the screen goes blank, then if you refresh the screen the message shows.

I don't get an error, but the form goes away, then I refresh the screen and it says I'm logged in.

Notice: Undefined index: user in .....on line 41 if(($_SESSION['user'] == "" )|| (!isset($_SESSION['user']))){ ?>

Parse error: syntax error, unexpected $end on last line

Phpsensei, Thanks again for all your help. I think I'm getting closer. It prints the user's name but still shows the form if they are signed in. <?php session_start(); ?> <!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>Login Test Page</title> </head> <body> <?php if(!empty($_POST['submit'])){ $user = trim($_POST['user']); $password = trim($_POST['password']); if($user == "") { print "Please enter your username"; }elseif($password == "") { print "Please enter your password"; }elseif($user == "me" && $password == "test"){ // Check if username and password were submitted if (!isset($_SESSION['user'])){ session_regenerate_id(); //assign user's name to session $_SESSION['user'] = $_POST['user']; }elseif($_SESSION['user'] = "me"){ print '<p>You are logged in ' . $_SESSION['user'] . ' !</p>'; } } } ?> <form action="login5.php" method="post"> <table width="200" border="0"> <tr> <td width="71">User:</td> <td width="113"><label for="user"></label> <input type="text" name="user" id="user" /></td> </tr> <tr> <td>Password:</td> <td><label for="password"></label> <input type="password" name="password" id="password" /></td> </tr> <tr> <td> </td> <td><input type="submit" name="submit" value="Log in" /></td> </tr> </table> </form> </body> </html>

Thanks I didn't even occur to me I should have caught that. Yes I am just testing to work with operators, errors etc. Now that its working I have one more fix. Once I am logged I don't want to show the form.

I still get errors. I changed username to user Warning: mysql_real_escape_string(): Access denied for user ''@'localhost' (using password: NO) in C:\inetpub..... on line 13 Warning: mysql_real_escape_string(): A link to the server could not be established in C:\inetp..... on line 13 Warning: mysql_real_escape_string(): Access denied for user ''@'localhost' (using password: NO) in C:\inetp...... on line 14 Warning: mysql_real_escape_string(): A link to the server could not be established in C:\inetp..... on line 14 Please enter your username

I tried something like this but still getting errors. <?php if(!empty($_POST['submit'])){ $username = mysql_real_escape_string(trim($_POST['user'])); $password = mysql_real_escape_string(trim($_POST['password'])); if(($username == "") || ($username !== "me")) { print "Please enter your username"; }elseif(($password == "") || ($password !== "test")){ print "Please enter your password"; }else{ // Username and password havebeen submitted print '<p>You are logged in</p>'; } } ?>